Chapter The
2 Basic
Theory
of
Interest
1. (A nice inheritance) Use the "72 rule". Years = 1994-1776 = 218 years. (a) i = 3.3%. Years required for inheritance to double = Zf = 8 :'=!21.8. Times doubled= Hi = 10 times. $1 invested in 1776 is worth 210 :'=!$1,000 today. (b) i = 6.6%. Years required to double = ~ :'=!10.9. Times doubled = ~ times. $1 invested in 1776 is worth 220 :'=!$1, 000, 000 today. 2. (The 72 rule) Using (1 + r)n = 2 gives nIn (1 +r) In2 = 0.69. We have nr :'=!0.69 and thus n :'=!~
= 20
= In2. Using In (1 + r) :'=!r and :'=!PI.
Using instead In(1 + r) :'=!r- !r2 = r(1 -!r) we have nIn(1 + r) = In2 or equivalently nr :'=!~. For r :'=!0.08, we have (1 -r /2)-1 :'=!1.042. Therefore, n:'=! !(0.069)(1.042) r
= ~
r
= ~ t
3. (Effective rates) (a) 3.04% (b) 19.56% (c) 19.25%. 4. (Newton's method) We have I(") i "k 0 1 1 2/3 2 13/21 3 0.618033 4
0.618034
I("k) = -1 + " + " 2 , I , (,,) = 1 + 2" , "k+1 = "k -f'
I("k)
I' ("k) 1 3 1/9 7/3 0.00227 2.23810 -2.2 x 10-6 2.23607 0
2.23608
"k+1 2/3 13/21 0.618033 0.618034 0.618034
5. (A prize) PV = $4, 682, 460.
1
2
CHAPTER 2. mE BASICmEORY OFINTEREST 6. (Sunk cost) The payment stream for apartment A is 1,000, 1,000, 1,000, 1,000 1,000, 1,000 while for B it is 1,900, 900, 900, 900, 900, 900. At any interest rate PVA
We know that PV = -1 +
x/1.13 < -1 + 3/1.12. Hence x/1.1 < 3 which means x < 3.3. 8. (Copy machines) Assume that the maintenance payments occur at the beginning of each year. InClemental IRR from A to B: ( Cash flows in $1000) f(c)
= 0 = -24 + 6c + 6C2 + 6c3 + 6c4 + 10cs,
where c --.
1 l+r
Using Newton's method, we get c = 0.894112,
r = 0.118 Thus,
IRRA-B = 11.8% > 10%. Incremental
IRR from B to C: ( Cash flows in $1000)
f(c)
= 0 = -5 + 0.4c + 0.4c2 + 0.4c3 + 0.4c4 + 2cs , c = ~ l+r
Using Newton's method
(co = 1.1), we get c = 1.0862106,
r = -0.079
Thus, IRRB-c = -7.9% A move from A to B is justified 9. (An appraisal)
on the basis of IRR.
Consider the PV of the two following
payment
streams:
(a) Change roof now, then every 20 years: 00 1 PVl = $20, 000 x i~ (1.05)20i = $32, 097
(b) Change roof in 5 years and then every 20 years: PVl PV2 = "i"i:"OST' = $25, 149
CHAPTER2. mE BASICmEORY OF INTEREST Taking the difference
of these two we find the value of roof to be PV1 -PV2
3 =
$6, 948. 10. (Oil depletion
allowance)
Barrels Gross Net Option Option Depletion Taxable After Tax Yr Produced Revenue Income 1 2 Allowance Income Income I 80 $1,600 $1,200 $352 $400 $400 $800 $840 2 70 1,400 1,000 $308 $350 $350 $650 $708 3 50 1,000 500 $220 $250 $250 $250 $388 4 30 600 200 $100 $150 $150 $50 $178 S 10 200 50 $25 $50 $50 $0 $50 Total= $1,200 PV= $521.26 All numbersexceptyearsin thousands. (a) Depletion
= $1, 200, 000 > $1, 000, 000.
(b) PV = $521, 260. IRR = 52.8%. 11. (Conflicting
recommendations)
See the solution 12. (Domination)
NPV1
=
29.88
NPV2
=
31.84 > NPV1
IRR1
=
15.2%
IRR2
=
12.4% < IRR1
recommend
2.
recommend
1.
for (Crossing) exercise for explanation.
Equations for IRR are n Ai = Bi 2: cf i = 1, 2 j=1
where Ci = ~
which gives 1=!!.!. ""1:" c !
"'1=1 t Hence, B1/A1 > B2/A2
implies
Ai. n n 2: c{ < 2: c~ j=1 j=1
which in turn implies C1< C2 or equivalently
r1 > r2.
4
CHAPTER2. mE BASICmEORY OF INTEREST 13. (Crossing) (a) Let P(c) = Px(c) -Py(c), P(l)
a continuous
= Px(l)
function
of c. Then
n n = L Xi -L Yi > 0. i=O i=O
-Py(l)
likewise P(O) = Px(O) -Py(O) By the intermediate
value theorem
= Xo -Yo
< 0.
there is a c such that Px(c) = Py(c).
(b) We solve -100
+ 30(c + C2 + ...+
C5) = -150 + 42(c + C2 + ...+
C5)
This gives c = 0.946 and r = 5.7%. 14. (Depreciation
choice) Individual
maximizes
the PV of depreciation
(in percentage
terms) (a) 25%, 38% and 37%. PV1 = 25 + ~ (b) 331/3%.
100 PV2 = 3
38
37 + (1 +r)2.
(1 + 1""+r 1
1
+ (1 + r)2
)
Then PV1 = PV2 yields TiT = 1 which gives r = 0. when r > 0, PV2 > PV1 .Hence 15. (An erroneous
always use straight line method.
analysis) AU. DOLLAR AMOUNTS Before Tax
Yr
Cost
Revenue
Incame
IN THOUSANDS Taxable
Deprec.
Incame
After Tax
Tax
Incame
0 $1Q,(XX)
PV ($10,(XX»)
1
$310
$3;300
~
$2,(XX)
.\m)
$337
$2,653
$2.'369
2
$310
$3;300
$2.9:X>
$2,(XX)
.\m)
$337
$2,653
$2,115
3
$310
$3.'300
$2.9:X)
$2,(XX)
.\m)
$337
$2,653
$1,889
4
$310
$3;300
~
$2,(XX)
.\m)
$337
$2,653
$1,686
5
$310
$2.ID}
$2,(XX)
.\m)
$337
$2,653 Total PV=
$3.'300
$2.9:X)
$2,(XX)
.\m)
$337
$2,653
($10,
$322
$3,432
$3,110
$2,(XX)
$1,110
$377
$2,732
$2,178
$335
$3,sm
$3,234
$2,(XX)
$1,234
~
$2.814
$2,003
$349
$3,712
$3;3ffi
$2,(XX)
$1;363
$464
$~
$1,843
$E
$3fij1
$3,~
$2,(XX)
$1,~
$500
~
$1.{;00
$1Q,(XX) $310
$3;300 NO INFIATlON
496 INHATlON
Total PV=
$l,SOO ($435)
$11)
Chapter
3
fixed-1
ncome
Secu
rities
1. (Amortization) Use A = ~ = .07 x 25iOOO = $4,638.83. I -(I":j:"';jn I -I:O77 2. (Cycles and annual worth) Let d = 1/(1 + r). Then P(» =P
{ I+dn+1+d2(n+2)+...
Also
} =
P I -dn+1 .
rP A = "i-=dn .
Hence ) P(». A = r(1 I -dn+1 -dn 3. (Uncertain annuity) (a) To find the life expectancy, we multiply each age of death by its probability. Thus the life expectancy is I = 90 x .07 + 91 x .08 + ...101
x .04 = 95.13 years.
(b) To find the present value of an annuity that ends at age 95.13 we calculate the values for ages 95 and 96. From the standard formula A { I P-1-r (1 + r)n with n = 5 and n = 6 we find P95 = $39, 927 and P96 = $46, 228. Then we find P = .87 x P95+ .13 x P96 = $40, 746.
}
(c) To find the expected present value of the annuity we calculate the probabilities qi of survival to various ages i. For example, qgo = 1.0, q91 = qgo- .07 = .93, q92 = q91 -.08 = .85, and so forth. For each year greater than 90 we evaluate $10, 000 x qi/l.08i-9O. Hence the expected present value is .93 + "LO82 .85 + ...+ ~ .04 PV = $ 10, 000 LOB = $3 8, 38 7.
(
)
Note that the expected present value of the annuity is less than the present value evaluated at the expected lifetime. This will always be the case. 5
6
CHAPTER 3. FIXED-INCOME SECURI11ES 4. (APR) First find the monthly payment M at the APR of 8.083% using the annuity formula, as $203, 150 = $1, 502.41. M = ~(1+~)360 12 0808331620 (1 + -U) -1 Next find the initial balance for this monthly payment at the interest rate of 7.875%
{
12 B = 07875 .(1
1 -.:QZm +
1
360} $1, 502 = $207, 209.13.
12 )
The total fees are the difference between the initial balance and the amount of the loan Fees = $207, 209.13- $203, 150 = $4, 059.13. 5. (Callable Bond) After five years, the payment the company needs to make if exercising the call provision is pf = (1 + 0.05) x Face Value = 105. Exercising the call provision is advantageous, so
(
100 10 105 < P5 = (1 + .:\)15 + T 1 -(1
1
)
+ .:\)15 .
Therefore, the YTM then is lower than 9.366%. 6. (The bi-weekly mortgage) (a) Monthly payment: 0.1(101 + -0.1)360 m = 12 3 162 0
(1 + 12 )
x $100,000 = $877.57
-1
Total interest = 360m -$100, 000 = $215, 925.20. (b) Bi-weekly payment = T = $438.79. Let n = number of periods. Then $100, 000 = $438.79 X (1 0.1+ .Q:!)n 26 0.1-1n
26(1+26) Hence n = 545 or 20.95 years. Total interest = 545 x $438.79- $100, 000 = $139, 140.55. Savings in total interest over monthly program = $76,784.65 or 35.6%
CHAPTER 3. FIXED-INCOME SECURmES
7
7. (Annual worth) A Amortize the present value of $22,847 over four years which gives AA = $6,449 per year. B. Amortize the present value of $37,582 over six years which gives AB = $7,845. AA < AB Car A should be selected. 8. (Variable rate mortgage) (a) A = $100,000 x ~
= $8,882.74
(1.08)25-1= $94, 821.26 (b) Ps = $8882.74 x 0.08(1.080)25 (c) A' = $94, 821.26 x ~
= $9, 653.40
(d) 94, 821.26 = 8882.74 x ~. This gives n :=::38 years, which means that the total life of the mortgage is 43 years. 9. (Bond price) Straightforward use of the formula for a bond price (assuming coupons every six months) gives 91.17. 10. (Duration) Use the formula to obtain 6.84 years. 11. (Annuity duration) Using PV = ~ we have D
= =
Hence
-!!.:!:..!:1 ~ PV dr -r(l +r) .(-~2) A r
= ~
D DM=-=-. l+r
1 r
pA
=
885.84
PB =
771.68
Pc
=
657.52
PD =
869.57
12. (Bond selection) (a)
r
8
CHAPTER3. FIXED-INCOME SECURrnES (b) DA
=
2.72
DB
=
2.84
Dc
=
3.00
DD
=
1.00
(c) C is most sensitive to a change in yield. (d) VA+VB+VC+VD
=PV
DAVA +DBVB +DcVc
+DDVD
= 2PV,
where PV is the present value of the obligation. (e) Use bond D. Vc + VD = PV DcVc + DDVD = 2PV where 2,000 PV = ~
= $1, 512.29.
Solving Vc = $756.15 and VD = $756.15. (1) None 13. (Continuous
compounding) dP -= di\
n -L
e-"tktkCk k=Q
= -DP
14. (Duration limit) This follows directly from the Macaulay duration formula by setting i\ = my and noting that the second term in the formula goes to zero as n goes to infinity. 15. (Convexity
value) c-
We take T = n/m.
-P[I
I + (i\/m)
n(n+I)P m2
Hence I C = [I + (i\/m)]2
As m -00
]2
we find C = T2.
T(T + (I/m).
.
Chapter The
4 Term
Structure
of
Interest
Rates
1. (Oneforwardrate) f
2. (Spot update)
1.2
=
(1 + 52)2 (1+51
Use /1.k =
{
-1
(1
= ~ 1.063
+
5k)k
-1
} 11(k-l)
1 + 51 Hence,
= 75% ..
-1.
.
for example,
=
f
1.6
{ (1.061) 6 } 115 -1
= 06
1.05
..
32
All values are /1.2 5.60
/1.3 5.90
/1.4 6.07
/1.5 6.25
/1,6 6.32
3. (Construction of a zero) Use a combination of the two bonds: let x be the number of 9%bonds, and y the number of 7%bonds. Select x and y to satisfy 9x + 7y
=
0
x +y
=
1.
The first equation makes the net coupon zero. The second makes the face value equal to 100. These equations give x = -3.5, and y = 4.5. The price is p = -3.5 x 101.00 + 4.5 x 93.20 = 65.9.
11
12
CHAPTER4. mE TERMSTRUCTUREOF INTERESTRATES 4. (Spot rate project) All can be done on a spreadsheet
with an optimizer,
as shown
below: Maturity Coupon 15-Feb-12 6.625 15-Feb-12 9.125 15-Aug-12 7.875 15-Aug-12 8.25 15-Feb-13 8.25 15-Feb-13 8.375 15-Aug-13 8 15-Aug-13 8.75 15-Feb-14 6.875 15-Feb-14 8.875 15-Aug-14 6.875 15-Aug-14 8.625 15-Feb-15 7.75 15-Feb-15 11.25 15-Aug-15 8.5 15-Aug-15 10.5 15-Feb-16 7.875 15-Feb-16 8.875
Buying Price wl Model Acc. Int. Price $101.48 $101.49 $102.72 $102.71 $102.50 $102.50 $102.87 $102.87 $103.06 $103.06 $103.24 $103.24 $102.60 $102.59 $103.98 $103.99 $99.69 $99.69 $104.26 $104.26 $98.94 $98.94 $103.64 $103.63 $100.88 $100.90 $111.63 $111.64 $103.30 $103.30 $110.18 $110.18 $101.16 $101.17 $104.98 $104.97
Estimation Coefficients a-O 0.062009143 a-1 0.00627032 a-2 0.001099467 a-3 -0.000593607 a-4 4.95E-05 Maturity of Zero Bonds
Time from Estimated Valuation Spot Interes (Years) Rate
15-Feb-12 15-Aug-12 15-Feb-13 15-Aug-13 15-Feb-14 15-Aug-14 15-Feb-15 15-Aug-15 15-Feb-16
0.27950.06383460 0.7781 0.06729211 1.28220.07073894 1.77810.07379215 2.28220.07633260 2.77810.07813531 3.28220.07918981 3.77810.07946645 4.28220.07905432
ValuationDate 5-Nov-11 5. (Instantaneous
rates)
(a) eS(t2)t2 = eS(h)heftl't2(t2-h)
~
(b) r(t)
fh,t2 = S(t2)t t 2
t (q)q 2- 1 = s(t) + s'(t)t -S
= lim s(t)t -S(tl)q = d(s(t)t) t-h t -q dt (c) We have d lnx(t) = r(t)dt = s(t)dt + s'(t)tdt
= d(s(t)t).
Hence, lnx(t)
= lnx(O) + s(t)t.
Finally, x(t) = x(O)es(t)t. This is in agreement with the invariance property of expectation dynamics. Investing continuously gives the same result as investing in a bond that matures at time t. 6. (Discount conversion) The discount factors are found by successive multiplication. For example dO,2= dold1,2 = .950 x .940 = .893. The complete set is .950, .893, .832, .770, .707, .646.
CHAPTER 4. THETERMSTRUCTURE OFINTEREST RATES
13
7. (Bond taxes) Let t Xi Ci Pi
be the be the be the be the
tax rate number of bond i bought coupon of bond i price of bond i
To create a zero coupon bond, we require, first, that the after tax coupons match. Hence x1(1-
t)C1 +x2(1-
t)C2 = 0.
which reduces to X1C1+ X2C2= 0. Next, we require that the after tax final cash flow matches. Hence x1[100-
(100- P1)t] + x2[100 -(100-
P2)t] = [100-
(100-
Po)t].
The price of the zero will be Po = X1P1 + X2P2. Using this last relation in the equation for final cash flow, we find X1 + X2 = 1. Combining X1 + X2 = 1, C1X1+ C2X2= 0, and Po = X1P1 + X2P2, we find Po = C2P1-C1P2 . C2-C1 Plugging in the given values we find Po = 37.64. B. (Real zeros) We assume that with coupon bonds there is a capital gain tax at maturity. We replicate the zero-coupon bond after tax flows using bonds 1 and 2. Let Xi = amount of bond i required (for i = 1, 2). We require (a) 100c1(1- t)X1 + 100c2(1- t)X2 = -(~)t (b) (100- (100- P1)t + 100c1(1- t»X1 + (100- (100- P2)t+ 100c2(1- t»X2 = 100- ( !QQ=EQ. )t n (c) P1X1 + P2X2 = Po
CHAPTER 4. nIE TERM STRUCTURE OF INTERFST RATES =
28.1425.81
P
=
(28.14)(1.065)
-10(2i:6~~(.02)
r
=
( 24.06-
) = -26.09%
r Year 5
28.14
Year 6
P
r
25.81 -12.5
28.14
15
= 1751% . -12.5(.05)
= 24.06
-12.5(.07)
= 20.80
-12.5
=
(24.06)(1.085)
-10(2i:6~~(.02)
=
( 20.80 -24.06 24.06- 12.5 ) =
Money left after 6 years = 20.79-
-28 .20%
12.5-
7.5 = .80
If invested in bank = 7.5(1.06)(1.055)(1.045)(1.4)(1.06)
-7.5
= 2.64
11. (Running PV example)
(a)
dO.l .9524
dO.2 .9018
do.3 .8492
dO.4 .7981
dO.5 .7472
1
2
dO.6 .7010
=>
O -40 .9524
10 .9469
10 .9416
10 .9399
10 .9362
10 .9381
10
(b)
Year Cash Flow Discount PV(n)
9.497
51.970
44.324
36.453
28.144
19.381
10.000
3
NPV = 9.497
4
5
6
12. (pure duration)
P(A)
=
dP(A) ~
=
= 1 dP(A) --= P dA
n n -k I Xk (1 + sk/m)-k = I Xk ( (1 + sZ/m)e;\lm) . k=O k=O In Xk m -k ) ( (1 + sZ/m)e;\lm ) -k-l (1 + sZ/m)e;\lm
(
k=O n I Xk ( ---(1 -k ) k=O m Lk=oXk ( ~)
+ sk/m)-k .
(1 + sk/m)-k k Lk=oXk (1 + Sk/m)-
= D.
This D exactly corresponds to the original definition of duration as a cash flow weighted average of the times of cash payments. No modification factor is needed even though we are working in discrete time.
QIAPTER 4. THE TERMSTRUCTUREOF INfEREST RATES (d) It should be clear that v-+ 0. (Use L'Hopital's (e) WeknowP(k) = (l+r)k-I(B-rM). part (a).) So P(k) is increasing
17
rule if it is not obvious.)
ClearlyB-rM > 0. (This follows from in k and I(k) = B -P(k) must be decreasing
in k. Remember that duration is a weighted sum of the times k, with the weights being proportional to to the cash flows at those times. Hence the duration of the stream determined by P(k), which increases in k, should be the larger, because more relative weight is given to higher k's. 15. (Short-rate
sensitivity)
In general Pk-1 (,,) = Ck-1 + 1 Pk <,,) " . + rk-1 + I\
Differentiation
at"
= ° leads to Sk-I
= -+
Pk (1 + rk-I)2
Hence, ak = ~
Sk
.
1 + rk-1
' bk = r+k:;: .This process together with Pk-1 = Ck-1 +
Pk 1 + rk-1
is initiated with Pn = Cn and Sn = 0; and the two processes are worked backward to k = 0. So is the final result.
Chapter
5
Applied
Interest
Rate
Analysis
1. (Capitalbudgeting) Project Benefit-Cost Ratio 1 2 2 5/3 3 3/2 4 4/3 5 5/3 So, the approximate method based on cost-benefit ratios implies projects 1, 2, and 5 would be recommended. The optimal set of projects is the same. Note: projects 1,2, and 3 provide the same total net present value and use the entire budget. 2. (The road) The zero-one problem is the same as in Example 5.2 with the following additional constraint: (X2 + x4)(I-
(X6 + X7» = 0
Excel's Solver yields an optimal solution with a total benefit of $7,8000,000 for a cost of $4,700,000 by funding projects 4,6, and 10. 3. (Two-period budget) The problem is to
maximize
150xl + 200x2 + 100x3 + 100x4 + 120xs + 150x6 + 240x7
subject to
90xl + 80x2 + 50x3 + 20x4 + 40xs + 80x6 + 80x7 + y ~ 250 58xl + 80x2 + 100x3 + 64x4 + 50xs + 20x6 + 100x7 ~ 250 + (1.1)y Xi = 0 or 1 for each i y~o.
Excel's Solver yields a maximal NPV of 610, achieved by funding projects 4,5,6, and 7, at a cost of 220 in the first year and 234 in the second year. Another plan 19
20
CHAPTER s. APPLIED INTEREST RATEANALYSIS with NPV of 610 is to fund projects 1,4,5, and 7, at a cost of 230 in the first year and 272 in the second year. Both plans are under the budget, but the first costs less. 4. (Bond matrix)
C=
10 10 10 10 10 110
7 7 7 7 7 107
8 8 8 8 8 108
6 6 6 6 106
7 7 7 7 107
y =
5 5 5 105
10 10 110
8 8 108
7 107
100
100 200 800 100 800 1200
(a) pT and XT are identified in Table 5.3 n = 2:Cij(I+Sn)isochoosingv= i=l Sn)l, (1 + Sn)2,..., (1 + Sn)n]T solves the equation CTV = p
(b) Weknowthepriceofbondjispj
[(1+
(c) To meet the obligation of period i exactly, we require m 2:: CijX j = bi j=l or in matrix form Cx = b (d) The price of the portfolio is pTx = VTCx = vTb which shows that the present value of the portfolio must equal the present value of the liabilities. 5. (Trinomial lattice) The trinomial lattice spanning three periods (with four time points) contains 42 = 16 nodes. In general, a trinomial lattice with n time points contains n2 nodes. In a full trinomial tree spanning three periods there are 40 nodes. In general, a n-l full trinomial tree with n time points contains 2:: 3i = !(3n -1) nodes. i=O
CHAPTER5. APPUEDINTERESTRATE ANALYSIS
21
6. (A bond project) C..h
Match! C..h M.tch : R.lnv..t
G\I.rd for D .(0)
G\I.rd for D .(1)
G\I.rd for: G\I.rd for ~ G\I.rd for D .(2) i D .(3) D .(4)
Co.t: $70,723.31 :, $70,558.12$70,557.61 $70,558.01$70,559.31!$70,559.47$70,580.21 M.t\lrlty Co\lpon , Bo\lght :' Bo\lght , Bo\lght , Bo\lght , Bo\lght :' Bo\lght , Bo\lght 15-Aug-12 15-Aug-12 15-Feb-13 15-Feb-13
7,875 6,25 8.25 8.375
1e5.25! 0i 0: °i
0: 0! 0! o~
° ° ° 0
1~-Feb-1~ 15-Feb-14 15-Aug-14 15-Aug-14
8.875 8.875 6.875 8.625
0! 0] 0! 0!
0! 0! °i 0!
° ° ° °
° ° ° °
15-Feb-15 15-Aug-15 15-Aug-15 15-Feb-16
11,25 8.5 10.5 7.875
0: °i 0! 0!
° ° 0 °
0 ° 0 °
oi 4,14] 0: 144.32:
° 0 83.62 0
0! 0[ 309.81! 0i
0: 0! 249.41! 0!
73.46 ° 198.52 0
0! °l 0: 0!
0: 0! 0! 0:
92.79 ° ° °
0] 0) 0! 134.78!
7. (The fishing problem) The decisions at times after the initial on d. At time 1 the upper and lower node values are
respectively.
X2
14 + 14d
XI
7+7d
0: ° 0; 141.98!
° ° 0 147.44
time do not depend
Then the initial value is Xo = max[14d(1
+ d), 7(1 + d + d2)]
The choice depends on d. The critical value of d is d* = ~
~ .618. 2
For d < d* we choose X2. For r = 3396 we have d = -75 and for r = 2596 we have d = .8, so solution
is the
same for both. 8. (Complexico
mine)
(a) Since we mine forever, we have KK = KK+I = constant K. So K = ~
+ dK implies K = 220 every period.
Thus, the initial (b) The amount
value of the mine, Yo = 220xo = $11 million.
of gold remaining
in the mine in period n,Xn,
Zn-I where Zn equals the amount mined in period equations from Example 5.5, we find XIO = 2393.
equals Xn-I
-
n. Using Excel and the
22
CHAPTER 5. APPLIED INTEREST RATEANALYSIS Thus, by part (a), the value of the mine in period 10 is found to be 220x1O = $526,460 (at that time). (c) The optimal extraction rate in each period = ~ = 20% so, after 10 years, 5369 ounces of gold remains with a value of $1,181,116 (at that time). 9. (little Bear Oil) (a) Set up a trinomial lattice with arcs: "up" = no pumping "middle" = normal pumping "down" = enhanced pumping The reserve values can be entered on each node. (At the final time the maximum reserve is 100,000 and the minimum is 26,214 barrels.) (b) Work backward to find PV = $366,740. The optimal strategy is: enhanced pumping for the first two years, followed by normal pumping in the last year.
10. (Multiperiod harmony theorem) We can write Vo
[
Xo + -1
=
max
=
max [ XO
+
1
-
Xo achieved
by x6.
1
+ 51 1
{
VI
XI
+
X2
X3
' + (1 + 52 ' ) 2 + ...+ + 51
-
1
]
+ 51 Clearly, ~T
(Vl(x6) *)
yoXo
* = 1 + 51. -xo
Suppose Xo gives VI (xo) ~T(- ) ->
1 + 51.
yo Xo -xo
Then Vo -Xo > 0 and thus ~T yo < -Vl(XO) Xo + 1
+51
which contradicts the definition of Vo. 11. (Growing annuity) Using the hint we have S= ~ l+r implying S [ l-
+ S(l + g) (l+r)
(~
)]
l+r
~ l+r
or 1 S
=
--.
r-g
Xn (1
' + 5n-l ) n -1
}]
CHAPTER5. APPUEDINI'ERESTRATE ANALYSIS 12. (Two-stage growth) (a) This part follows
easily from (b)
(b) Let R = 1 + r. then NPV
=
Dl [ (i)
+Dl
=
Dl
o + (i)l
R
[( g
[~
l-(G)k
) k+l
+
+ ...+
( Rg ) k+2
+ ~1=<"IT
].
(i)k]
+...
]
23
Chapter
6
Mean-Variance
Portfolio
Theory
1. (Shorting with margin) The money invested is Xo. The money received at the end of a year is Xo -XI + Xo. Hence, R = 2Xo -XI
Xo
.
2. (Dice product) Let a and b be the outcomes of two die rolls. Then Z = ab. By independence, we know E[ab]
=
E[a]E[b]
and var[Z]
=
E[a2]E[b2] -(E[a]E[b])2
~
79.97
3. (Two correlated assets) For solution method, see solution to problem called Two stocks (below). (a) a equals 19/23. (b) The mjnimum standard deviation is approximately 13.7%. (c) The expected return of this portfolio is approximately 11.4%. 4. (Two stocks) Let a, {3 equal the percent of investment in stock 1 and stock 2, respectively. The problem is mina2ul IX.IJ
+ {32ui + 2a{3uI2
subject to a + {3 = 1. Setting up the Lagrangian, L, we have: L = a2ul + {32ui + 2a{3uI2 -i\(a The first order necessary conditions are: or = 2aul + 2{3UI2-i\ o = aa 25
+ {3-1)
26
CHAPTER 6. MEAN-VARIANCE PORTFOUO lHEORY oL 0 = a:ti = 2/,'O"i + 2lXO"12 -"
l=lX+/,' which
imply
[
2
0"2 -0"12 + O"i -20"12
lX-0"[
]
The mean rate of return is just lXm1 + /,'m2. 5. (Rain insurance) (a) The expected rate of return equals (.5) .;~612~5: .2 .u (b) By inspection, it can be seen that buying 3 million units of insurance eliminates all uncertainty regarding the return. So, 3 million units of insurance results in a variance of 0 and a corresponding 3
to --1 2.5
expected rate of return
equal
= 20% .
6. (Wild cats) (a) The three assets are on a single horizontal line. The efficient set is a single point on the same line, but to the left of the left-most of the three original points. (b) Let Wi be the percentage of the total investment invested in asset i. Then, since the assets are uncorrelated, we have n var (total investment) = L wlO"i2 i=l where Lf:1 Wi = 1. Setting up the Lagrangian, L = i~wlO"i2 -"
(~Wi
the first-order necessary conditions imply 2
WiO"i =2
"
.
t=
1,...,n
;\ or Wj = 2""(;."!. j Since
Lf:1 Wi = 1, we have ~" ( Lj=l
u7 ) = 1
-I )
CHAPTER6. MEAN-VARIANCEPORTFOUOrnEORY which implies
-2 Wj = ~ 0-.J
27
j = 1,2,...,n
where ""(7'2 =
(I ~ ) i=I 0-1
The minimum variance is n 2 2 n Varmin = 2: Wi o-i = 2: . I . I 1= 1=
-I.
(0- )2 -2
2
2 -2 o-i = 0- .
U.1
7. (Markowitz fun) (a) First solve for the Vi'S from 2VI VI
+ +
V2 2V2 + V2 +
= V3 = V3 =
1 1 1
This yields VI = .5, V2 = 0, and V3 = .5. This solution happens to be normalized, so also WI = .5, w2 = 0, and W3 = .5. (b) In this case we solve 2VI VI
+ +
V2 2V2 + V2 +
V3 V3
= = =
.4 .8 .8
This leads to VI = .1, v2 = .2, andv3 = .3. This solution must be normalized to get the final result WI = 1/3, W2 = 1/6, and W3 = 1/2. (c) We find the Vi'S by the formula Vi = vr -rfvf, where vf is the solution from part (a) and vr is the solution from part (b). Thus VI = V2 = V3 =
.1 -.2 x .5 .2 -° .3 -.2 x .5
= = =
° .2 .2
When normalized the solution is WI = 0, W2 = .5, and W3 = .5. 8. (Tracking) (a) var(r -rM)
=
var(r) -2 cov(r, rM) + var(rM ) n n 2: /Xi/Xjo-ij -2 2: /Xio-iM+ o-k i,j=I i=I
28
CHAPTER 6. MEAN-VARIANCE PORTFOliOrnEORY SO,to minimize var( r -rM)
subject to
2:~1 (Xi = I set up the Lagrangian
n n n L = ~ (Xi(Xj(]"ij-2 ~ (Xi(]"iM+ (]"It + ,,( ~ (Xi -I) i,j=l i=l i=l The first order necessary conditions imply n 2 ~ (Xj(]"ij -2(]"im + "
=
0 for all i
j=l n ~ (Xi = i=l
I
n (b) Similar to (a) with the added constraint ~ (Xiri = m. i=l So the first order necessary conditions imply n 2 ~ (Xj(]"ij -2(]"im + " + /.lri = 0 for all i j=l n n ~ (Xi = I ~ (Xiri = m i=l i=l 9. (Betting wheel) For every segment the payoff for a bet Bi = I/Ai
will equal $1.
Thus, the payoff is $1 independent of the wheel. Since, with this betting strategy, the reward is completely determined, the riskfree rate of return just equals r =
n I Li=11/Ai
-1.
Thus, the risk free rate of return in example 6.7 is 0. 10. (Efficient portfolio)
(rk. -rf)
(
o(tan(J) 0--
-OWk.
~n
i,j=l -
(]"ijWiWj
)
1/2 -~
(
(
n
Wiri
i=l
n
.~ (]"ijWiWj t,}=l Using
the hint,
this implies
-rf
) (n ~
~
i,j=l
)
(]"ijWiWj
)
1/2
Chapter The
7 Capital
Asset
Pricing
Model
1. (Capital market line) (a) r = .07 + ~O" (b)
= .07 + .50"
i. 0" = .64 ii. Solvewx.07+(I-w)x.23 = .39givingw = -1. Hence, borrow $1000 at the risk-free rate; invest $2000 in the market
(c) $1182 2. (A small world) O"'ft = (a)
2 O"AM = 2 0"AB
=
rA
=
rB
=
(b)
I 4(0"1 + 20"A,B+ 0"1) -2I
2
0"1 + 0"A,B
(O"A+O"A,B)
hence
!3A=
I -2 ( 0"1 + 0"A,B )
hence
!3B =
2 20"M 0"1 + 0"A,B
2 20"M
.10+~(.18-.10)=20% 3 .10 + 4(.18- .10) = 16%
3. (Bounds on returns) (a) Using the two-fund theorem and noting that the market portfolio contain assets in negative amounts, we have !w + !v
2w -v
=
=
[ "i]
[ :i]
with a rate of return of .1
with a rate of return of .16
31
cannot
32
CHAPTER7. mE CAprr AL ASSEf PRICINGMODEL so the expected rate of return .1 ~ rM ~ .16.
of the market
(b) Since rM ~ rminvarportfolio , we have 4. (Quick CAPM derivation)
portfolio
rM is bounded
by:
.12 < rM ~ .16.
From (6.9) we have n L UMii\Wi = i\ul:t = rM -rf. i=l
rm-rf Also, Hence i\ = ~ O"m n L UMii\Wi = i\ COV(rk,rM) i=l Combining,
we have
(-
-UkM rk -rf 5. (Uncorrelated
= rM -rf.
= --y rM -rf UM
)
assets) {Ji = ~UiM = nXiUi2 2 UM )=:j=l XjUj
since the assets are uncorrelated. 6. (Simpleland) The market consists of $150 in shares of A and $300 in shares of B. Hence, the market return is 150 rM = (450)rA
300 + (450)rB
1 2 = 3rA + 3rB.
(a) rM = ~ x .15 + ~ x .12 = .13 (b) UM =
[ 9(.15)2 1
4
+ 9""X3(.15)(.09)
4
+ 9(.09)2
122 + 3PABUAUB = 3(.15) 122 (c) UAM = 3UA
] 21 =
+ 9(.15)(.09)
.09 = .0105.
UAM = 1.2963 {JA = ~ UM (d) Since Simpleland satisfies the CAPM exactly, stocks A and B plot on the security market line. Specifically, r A -rf
= {JA(rM -rf).
Hence, rf=~=.0625.
CHAPTER 7. mE CAprrAL ASSET PRICING MODEL
7. (Zero-beta
assets)
(a) Let p be a portfolio 2(1-
33
a)auO1
such that p = (1-
+ a2uf
a)wo
I
.So, since 0 = ~
a)uO1 + au[
(c) The zero-beta portfolio mum variance point. (d) p = ~
~
implies
(using (a», a =
is on the minimum
ri = rz + ~(rM
-rz)
uJ = (1-
, we have 0 = -2u6 (X=O
implies A = 1. (b) 0 = U1 z = (1.Uo
+ (XW1. Then,
(~
-u1
a)2u6
+
+ 2UO1which
) < 0.
variance set but below the mini-
= .09 + .5~(.15
-.09)
That is, ri = 10%. 8. (Wizards) (a) E(r)
=
E(~-I)=E(~)E(P)-1
=
(20 .5 .5 ) + "16 24-
=
7 -= 20
9 I = 16024 -1
35%
(b) uM=E[(rp-rp)(rM-rM)]
=
E[(y)(rM-rM)]
=
E(I/c)E(p
=
Hence
(
9
160
)
20uM
-p)(rM 2
=
9 82OuM
-rM)] 2
9 2 a-~-~-~ 2 -2 j.JUM UM 8
(c) rp = rf + fJ(rM -rf)
= .09 + j(.24)
= .36 Thus the rate of return predicted
by the CAPM method exceeds the project rate of return by 1% , so the project is !1Q1acceptable-but it is close. 9. (Gavin's problem) Note: U(XM= cov(arf Q = p(arf
+ (1-
cov(Q,rM)
= cov(P(arf
+ (1-
a)rM,rM)
= (1-
a)ul:r
a)rM + 1) + (1-
a)rM + 1),rM)
= P(1 -a)ul:r.
Chapter Models
8 and
Data
1. (A simple portfolio) (a) The beta of the portfolio is a weighted combination of the individual betas: 13= 0.2 x 1.1 + 0.5 x 0.8 + 0.3 x 1 = .92. Hence, applying the CAPM to the portfolio we find rp = .05 + .92(.12-
.05) = 11.44%.
(b) Using the single-factor model, we have O"i = = 0"2 = O" =
c 2: wfO"ii = 0.22 x 0.0072 + 0.52 x 0.0232 + 0.32 x .012 i=A 0.00033725 b20"lt + O"i = 0.422 x 0.182 + 0.00033725 = 0.2776 16.7%.
2. (APT Factors) By the APT we have {\0 = rf = 10% and .15
=
.10 + 2"1 + {\2
.20
=
.10 + 3{\1 + 4{\2
This yields {\1 = .02 and {\2 = .01. 3. (Principal components) The estimated covariance matrix of the four stocks is
V=
[
90.28
50.88
79.00
40.18
50.88 79.00 40.18
107.2 105.4 30.98
105.4 162.2 56.54
30.98 56.54 68.27
]
The largest eigenvalue is 311.16 with corresponding eigenvector v = [0.217, 0.263, 0.360, 0.153] 35
36
CHAPTER8. MODELSAND DATA which has been normalized so that the components sum to one. Therefore, the first principal component is O.217T1 + O.263T2 + O.360T3 + O.153Tr. This can be considered as the weighted average of the returns bles the return of the market portfolio.
of the four stocks, and resem-
The top part of Table 8.2 is reproduced here, showing the time values of the principle component. Note that its behavior is similar to that of the market. Principal Year stock 1 stock2 stock3 stock4 market riskless Component 1 11.91 29.59 23.27 27.24 23.00 6.20 23.07 218.3715.2519.4717.0517.54 6.70 17.74 3 3.64 3.53 -6.58 10.20 2.70 6.40 .92 424.3717.6715.0820.2619.34 5.70 18.60 S 30.42 12.74 16.24 19.84 19.81 5.90 18.97 6 -1.45 -2.56 -15.05 1.51 -4.39 5.20 -6.21 720.1125.4617.8012.2418.90 4.90 19.48 8 9.28 6.92 18.82 16.12 12.78 5.50 13.16 917.63 9.73 3.0522.9313.34 6.10 11.08 10 15.71 25.09 16.94 3.49 15.31 5.80 16.76 aver 15.0014.34 10.9015.0913.83 5.84 13.36 var 90.28107.24162.1968.2772.12 84.93 4. (Variance estimate)
[
1 E(S2) = E -=-
. t=
~
i
] [
1
( Ti -r)
-!
1 i=l
[i (
1
(
.2:: 1
t=
i(Tj
-r»
)
1-
.!.)(Ti
-r)
-2::
n.]=
2:: (Ti
2
-r»
)
.
1
n
~ n- 1 0"2.
]..t
{ (1-!)2+~ n
n2
}
5. (Are more data helpful?) (a)
A A O"n O" O"(r) = O"(nr n ) = n- In = n JnJn Hence O"(:f) is independent
of n.
2
]
n.1.
= O".
2
Tj
1
] -!
)]
1 n
Ti-
n j=l
t=
=
n
~E
n-1 =
1
= E -=n
E
n
=
A
2::(Ti-r)2
n
[ =
n 1
CHAPTER 8. MODELS ANDDATA
37
(b) Assuming normality, a(u2)
= a(na2)
n
= n
./2u2n
In"-=1
= n
./2 In"-=1
a2 -= n
./2a2 In"-=1
.
Part (a) shows that by using smaller periods to get more samples does not improve the estimate ofr. Part (b) shows that using smaller periods to get more samples does improve the estimate of a2. 6. (A record) Assuming a normal population, (a) A rm
=
1 n -L Ti = 1% n. t= 1
Yyr
=
12Ym = 12%.
(b) u~
=
Uyr
-1n-1.
n L(Ti
-Ym)2
= 0.00072
t= 1
=
mUm
= 9.29%.
~ a(Tm)
=
Um ""In = .55%
a(Yyr)
=
a(12Ym)
=
~ In"-=1
=
a(12u~)
(c)
(
A 2
a am a(u2yr)
)
= 12um
= 6.6%
= ./2 X 0.00072 = 0 00021 m . = 12a(u~)
= 0.0025.
(d) From the previous exercise we know that the estimate of r will not be improved by having weekly, rather than monthly samples. All that matters is the total length of the period that is observed. However, the estimate in a2 can be improved. In fact, letting aweek(uir) denote the standard deviation in uir based on weekly data, we expect that aweek(uir) = ~a(uir) .47uir = .0012.
=
7. (Clever, but no cigar) First divide the year into half-month intervals and index these time points by i. Let Ti be the return over the i-th full month (but some will start midway through the month). We let r and a2 denote the monthly expected return and variance of that return.
38
aIAPTER8. MODELS ANDDATA Now let Pi be the return over the i-th half-month period. Assume that these returns are uncorrelated. Then Pi = rm/2 and 0"2(Pi) = 0"2/2. The return over any monthly period is a sum of two half-month returns; that is, the monthly return Ti is Ti = Pi + Pi+l. It is easy to see that COV(Ti,Ti+l) = ~0"2 and COV(Ti,Tj) = O for li-jl
> 1.
Now for Gavin's scheme we form the estimate -"- I 24 T = 24 ~ Ti. t=1 We need to evaluate 0"2(:;'0") = -liz
=
I w
=
I w
[
~ 24 (Ti -r)
]
2
t=1 24 .~ COV(Ti,Tj) t,J=1 24 ~ [COV(Ti-l,Ti) + COV(Ti,Ti) + COV(Ti,Ti+l)] . t=1
Except at the two end periods, each i will give three terms as shown. We will ignore the slight discrepancy at the ends and assume that every i gives the three terms as shown in the summation. The terms are ~0"2, 0"2, and ~0"2, respectively. Hence we have 2~ I 1 12 O" (r) = -x 242 24(-2 + I + -)0" 2 I -0"2 12 which is identical to the result for twelve nonoverlapping months of data. =
8. (General tilting) (a) Let p = pT = 1, Q = 0"2, and Q-l = 1/0"2. Then r= (b) Let P= ThenpTQ-Ip
[ ] I I
.Q=
= 0"2 1 + U2, 2 andPTQ-lp ~ T=
(
PI
P2
2+2 0"2
p.
[
O"f
O
.
= ~ 0"1 + ~. 0"2 Hence
)(
I
I
2+2 0"2
]
O
O"}
0"1
)
-1 .
0"2
Chapter General
9 Theory
1. (Certainty equivalent) The possible incomes and their utility levels (found by taking the 1/4-th power) are Income Utility
80 2.99
90 3.08
100 3.16
110 3.23
120 3.31
130 3.38
140 3.44
The total utility is the average of these, which is 3.23. We must find C such that Cl/4 = 3.23. Using an iterative process we find C = $108, 61. 2. (Wealth independence) The investment will be made if: E[U(W -w
+ x)] > E[U(W)] ,
for our case we have: E [ -e-aW e-a(x-w) ] > E [ -e-aw] or equivalently, -e-a(W-w)E [e-ax] > -e-aW. Dividing the expression by -e-aW, the investment will be made if: eawE [e-ax] < 1 which is independent of W. 3. (Risk aversion invariance) The risk aversion coefficient for a utility function, U (x ) is: U1'(X)
a(x)
= -~
v' (x)
=
bU' (x)
V1'(X)
=
bU1'(X)
For V(x) = c + bU(x)
so the risk aversion coefficient for V (x) is given by: -~V'1(x) = -~U'1(x) = a(x).
39
40
CHAPTER 9. GENERAL rnEORY 4. (Relative risk aversion) Given U(x), the relative risk aversion coefficient 11is defined as: (x) l1(x) = -U'xU" (x) (a) U(x) = log(x)
U' (x) U" (x) =>l1(x)
= = =
~ --IT x 1
(b) U(x) = )'X)'-l
U' (x) U" (x) =>l1(x)
= = =
)'2X)'-1 )'2 ()' -1) X)'-2 1 -)'
Relative risk aversion coefficients, 11,are constant for both utility fW1ctions. 5. (Equivalency) If results are consistent, we have that V(x) = aU(x) + b, and since V(A') = A' and V(B') = B' we must have A'
=
aU(A') + b
B'
=
aU(B') + b
So solving both equations simultaneously we find parameters a and b: a
=
A' -B' U(A') -U(B')
b
=
B'U(A') U(A') -A'U(B') -U(B')
6. (HARA) The hyperbolic absolute risk aversion function is given by: U(x)
=
y 1-)'
(~+b ax
)' , b>O.
(a) Linear: We can write the HARA as: U(x)
=
(
y1 ax(1 -)')
l=l.
)' + b(l-
1
)')y
)'
Choosing)' = 1 and a = 1 and using L'Hopital's rule we can write: U(x)
= =
-
IJ!!-
CHAPTER9. GENERALnIEORY (b) Quadratic:
By choosing
y = 2, HARA takes the form U(x)
Choosing
41
=
1
--a2x2 2
+ abx
--b2
1 2
a > 0 and b = 1/ a > o we have an equivalent
form
of the re-
quired quadratic form. Furthermore, by adding b2/2, which is a legitimate transformation, we get the precise desired form: U(x)
1 2CX2,
= x-
where c = a2 (c) Exponential:
By choosing
b = 1 and y = -00 and using L'Hopital's
can write U(x)
=
lim .!..=..r y--oo y
=
-1
(1 -y ~
=
lim eYIn(R+l) y--oo ~+1 )I -lim e In( 'i/~
=
y--oo -e-ax
rule we
)y +
1
(d) Power: If we let b = 0, HARA takes the form: U(x)
= (1-
which for y < 1 is of the required
y)l-YaYxY y
form
U(x) (e) Logarithmic:
= cxY
Let b = 0 and a = 1, then HARA is given by: U(x)
We can subtract utility
,
= (1 -y)l-Y
the constant c = (1-
xY -. y
y)(l-y)/y,
and obtain an equivalent
function U(x)
Now letting
= (1 -y)l-Y
y = 0 and using L'Hopital's U(x)
=
( -y XY-l
).
rule we get
lim ~ y-O y
42
CHAPTER9. GENERALrnEORY
= =
a(x)
)' lnx
lim )'-0 lnx.
= The Arrow-Pratt
e)'lnx -1
lim )'-0
1
risk aversion coefficient
a(x)
U"(x) -if("XT
=
=
a
e)'lnx
for HARA is
(~
-a2
+ b
= -a
(~
I-)'
(-+bax I-v
)-1 =
+
b
) y-2
) '-1
ax a
-+b I-)'
#
1
-I
b' r=yx+a
which is of the form l/(cx
+ d), as required.
7. (The venture capitalist) The expected value e is given by e = p + (1- p)9 = 9- 8p so, 9-e p=-
8
On the other hand since U(x) = .jX: U(C) = -.JC =
So
C=
(3-2
pU(l)
+ (1- p)U(9)
=
p + 3(1- p)
=
3 -2p
=
3-2
( -S 9-e ))
(-S 9-
e
)
2 =16(3+e)2 1
Solving for e we get: e=4-.JC-3 which agrees with the values of the table on example 9.3 8. (Certainty approximation) By definition, the certainty equivalent c is such that E[U(x)] = U(c). Substituting in the given approximations, we have U(x)
+ ~U"(x)var(x)
~ U(x) + U'(x)(c
-x).
CHAPTER 9. GENERAL tHEORY
43
Solving for c yields (x) c ::::x-U" + u;-(-x}var(x) as required. 9. (Quadratic mean-variance) In general E[U(y) ]
=
E[ay-
~by2]
=
aE[y] -~bE[y2]
=
aE[y] -~b(var[y]
+ E[y]2)
If the random payoff of the portfolio of the investor with unit wealth is R, it would maximize E[U(R)] = aE[R] -~b(var[R] + E[R]2) Now, if the investor with wealth W purchases the same portfolio, its payoff must be WR and R should maximize: E[U(y2)]
=
aE[RW] -~b2(var[RW]
+ E[RW]2)
=
aWE[R] -~b2(W2 var[R] + W2E[R]2)
=
W [ aE[R] -~b2W(var[R]
+ E[R]2) ] .
If the second investor has b' = ~, the same R will solve this as the R using unit wealth. 10. (Portfolio optimization) The first-order conditions for portfolio optimization are: E[U' (x* )di] = "Pi
for all i.
Dividing by Pi we get E[U'(x*)(l+ri)] E[U'(x*)]
+ E[U'(x*)ri]
=
"
foralli
=
"
for all i.
or,
Taking the difference between asset i and the risk-free asset yields E[U'(x*)]
+ E[U'(x*)ri]
-E[U'(x*)]
+ E[U'(x*)rf]
E[U'(x*)ri]-E[U'(x*)rf] E[U'(x*)(ri
-rf)]
=
" -"
=
0
=
0.
44
CHAPTER 9. GENERAL nIEORY
11. (Money-back with
price
guarantee)
0 6
1~
We have the three
1 and the new money-back 1 3
1~
0
investment
guarantee 1~
1.2 1.2
0
Residual Rights
states
cated by combining that the price no arbitrage
p
and three
the existing
'securities',
guarantee
9.6
~
p3,000
Money-Back Guarantee the new alternative
ones in certain
of the Money-back
example
p
Risk Free
amounts
security
can be repli-
A, B, and C. It follows
is A + B + C, so that
there
are
opportunities: 6A + 3B + 1.2C
=
3,000
B + 1.2C
=
p
1.2C
=
p
=
p
A+B+C Solving
from
1.2
Film Venture
Since we had three
options
alternative:
the
system
gives
us the price
of the Money-back
guarantee
deal:
p
=
$1,500 Alternatively we could have used the state prices from example 9.9 where we had I/Jl = 1/6, 1/J2 = 1/2 and 1/J3 = 1/6 so that the price of the Money back guarantee is given by the single
equation: p=
with
solution
12. (General
results
state prices
in S states.
is characterized its payoff
result)
Security
by DT x E Es.
appending
vector
N elements
while
[
of the vector
the (n + 1) -th
element
matrix
are given value
Now
the negative
A=
The first
Let the NxS
prices
by x E EN. The market
is given
from
1
p = $1, 500.
positive
N securities
1 1 63,000+2P+6P,
or cost c of a portfolio
consider
matrix
of
x is q .x and
the (S + 1) x N matrix
of the price
vector
--~~--
]
p given
D be the payoff
by some q E EN and a portfolio
by Ax
is the negative
q as a row
= p correspond
A which
to DT :
to the payoff
of the cost of the portfolio
x.
CHAPTER 9. GENERAL lHEORY
45
If there is no arbitrage, then we must have no p ~ ° except p = 0, since otherwise we could have non-positive cost (c oS0) with a positive probability of yielding positive payoff or negative cost (c < 0) with zero payoff which is instant money. Using the stated matrix theory result it follows that there exists a vector y > ° such that A Ty = 0. Note that this is the same as: DIIYl D21Yl
+ +
D12Y2 D22Y2
+ +
DNIYl
+
DN2Y2 +
D13Y3 D23Y3
+ +
...+ ...+
D1SYS -'llYS+l D2SYS -'l2YS+l
= =
0 0
DN3Y3 +
...+
DNSYs -'lNYS+l
=
0
Note that by dividing each element of y by YS+l we get positive state prices tfJi = Yi/Ys+l such that: D 11tfJl D21tfJl
+ +
D12tfJ2 + D22tfJ2 +
D13tfJ3 + D23tfJ3 +
...+ ...+
DlstfJS -'11 D2stfJS -'12
= =
0 0
DNltfJl
+
DN2tfJ2 +
DN3tfJ3 +
...+
DNstfJS -'IN
=
0
Therefore, if there is no arbitrage, there are positive state prices. 13. (Quadratic pricing) From the earlier exercise we have E[Ut(x*)(ri U(x) = x -C/2X2 then, Ut (x) = 1- cx, so in this case we have E[(l-
cWRM)(ri -rf)]
= 0
E[(l-
CWRM)(Ri -R)]
= 0.
=
cW[E(RMRi) -HMR]
=
cW[COV(RM,Ri) + HM(Ri -R)].
or equivalently Written out we have Hi -R
This implies, Hi -R = )1COV(RM, Ri) for )1 = 1 -cwHM. AppIing this to RM yields HM -R = )1var(RM) which shows that RM-R ) )1 = var(RM
and hence finally Ri -R = fJi(RM -R).
-rf)]
= 0. If
46
CHAPTER 9, GENERAL THEORY
14. (At the track) Gavin Jones will choose the fraction (Xof his money m to bet on the horse so as to maximize his expected utility: max E[U] = ~~
+ ~~,
(a) The first order necessary condition is: 1 m ---=0 2 .Jm + 4(Xm which yields:
3 m 4 ..;(1- (X)m
7 = .1346 (X= 52
Gavin's maximizing choice is to bet 13.46% of his money and keep the rest in his pocket. (b) We can summarize Gavin's world by the following three alternatives:
1<: Keepmoneyin pocket
1<: Betfor NoArbitrage
1<~ BetagainstNoArbitrage
Unear pricing holds if there aren't any arbitrage opportunities. Thus, we could replicate a four dollar bet against No Arbitrage by 'shorting' the bet in favor and keeping five dollars in the pocket. Dividing by four we get the implied payoff for a one dollar bet against No Arbitrage: y = 45 = 1.25 15. (General risk neutral pricing) From log-optimal pricing we have p = E (~
)
Now, using the expectation operation E defined by ~ E(x) = E ( R* RX ) where R is the risk free rate, we can multiply the log-optimal pricing formula by ~ = 1 without affecting it. This yields R.E(;.) p -R which is risk neutral pricing.
-R
E(~)
-R
E(d)
Chapter
10
Forwards,
Futures,
and
Swaps
1. (Gold futures) Applying equation 10.2 we have 5 F
=+ =+++
M-I c(k) L k=Od(k,M)
d(O,M) 412
2/4
2/4
2/4
d(0,9m)
d(0,9m)
d(3m,9m)
d(6m,9m)
where (with m = "months") d(0,9m)
=
d(3m,9m)
=
d(0,9m)
=
1 09 = .9354 (1 + 4)3 109 2 = .9565 (1 + 4 ) 1 09 = .9780. (1 + 4 )
Combining these, we find F = 440.45 + .5345 + .5227 + .5112 = $442.02. 2. (Proportional carrying charges) Suppose at time zero you take out a loan for 5(0), purchase 1 unit of the commodity, and short (1- q)M units of the commodity at a forward price of F per unit. The total intial cash outlay of these transactions is zero. Each period you pay the carrying cost of the commodity by selling a fraction q of your commodity holdings. Hence the amount of commodity held at the end of M periods is ( 1- q )M.At the final time you deliver your commodity holdings to make good on your short, receiving F ( 1- q )M.You also repay your loan by paying 5(0) 1d(O, M). The total profit from these transactions (which clear all accounts) is F(I- q)M -5(0)/d(0,M). To avoid arbitrage, this profit must be zero. Hence F = 5(0)(1 ~ q)-M /d(O,M). 3. (Silver contract) ill general the spot and forward prices are related by: M-I 5 = Fd(O,M) -L c(k)d(O, k) k=O 47
CHAPTER 10. FORWARDS, FuroRES,ANDSWAPS
49
The transactions yield no net cashflow until the final period where we recieve S/d(O,M) + Lr=-i c(k)/d(k,M), from our investments and pay E for the asset, as required by our forward contract, which in turn we will deliver to Mr. X who lent us the asset in time 0. Thus if E < S/d(O,M) + ~r=ol c(k)/d(k,M), we have an arbitrage profit so the inequality must be false under the no arbitrage condition, which completes the proof of the forward price formula with canying costs in section 10.3. 6. (Foreign currency alternative) In Example 10.12 the company hedges by shorting forward contracts for 500,000 Deutsche marks, assuring dollar receipts in 90 days of $ 500, OOOET,where ET is the forward dollar price for Deutsche marks in 90 days. Based on the no arbitrage condition we have that: E-~S T -1
+ TG 0,
where TUSis the 90 day U.S.risk-free rate, TGis the 90 day interst rate in Germany, and So is the spot price for Deutsche marks at time 0. To verify this consider taking a long position on a 90 day Deutsche mark forward, which has zero cost at time 0 and a value of ST -ET at the maturity time T . On the other hand consider borrowing ET/ ( 1 +TUS) dollars at time zero for 90 days and at the same time purchasing 1/ ( 1 +TG) worth of risk free german bonds with a dollar cost of S0/(l+TG). The total cost of this strategy is S0/(l+TG)-ET/(l+Tus) with a total net payoff of ST -ET at time T which is equal to the value at time T of the 90 day forward. It follows that if there is no arbitrage, the cost of both alternatives must be equal, therefore we must have So/ ( 1 + TG) -ET / ( 1 + TUS) = 0, which gives us the the Deutsche mark forward price: ET = So(l + Tus)/(l + TG) So the dollar value of the receipts at time T for the firm by hedging with forward contracts is: 1 + Tus 500, 000 1 So dollars +TG If instead it borrows 500, 000/(1 + TDM) Deutsche marks (to be repaid with the receivables at time T) and sells them into dollars, the receipts at time 0 are 500, OOOSo/(l + TG) dollars, which invested for 90 days in U.S. T-bills will pay 1 + TUS 500, 000 1 So dollars. +TDM Hence, the two procedures are equivalent.
50
CHAPTER10. FORWARDS.FuroRES, AND SWAPS
7. (A bond forward) Ft
We solve first for Ft the current =
S ~ (O 2)
price of the bond:
2 d(O, k)c(k)
+L
, =
forward
d
k=1
(O , 2
920( 1.1.035 04 ) 2 -80(1.04)2
) -80(1.04)2
(1.04)2
=
$831 ..47
Now we solve for the value of the forward contract: 831.47- 940 it = (Ft -Fo)d(O, 2) = (1.04)2 = -$100.34. 8. (Simple formula) X / C units of the bond will pay X in coupons plus a final prindpal payment of 100X/C. Hence a stream (X,X,X.. .,X) is worth B(M,C)X/C d(0,M)100X/C.
The result follows immediately.
9. (Equity swap) (a) The market price at time i -1 Vi-I(Si + Vi) = Si-lo We have Vi-I(Ti)
for the cash flow Si + di is Si-l.
=
Vi-I([Si + di -Si-I]/Si-l)
=
1- Vi-l[l]
=
1 -d(i
Hence,
-1, i).
(b) We just discount Vi(Ti) back to time 0. Hence Vo(Ti) = d(O, i -1)[1 1, i)] = d(O, i -1) -d(O, i) because d(O, i -l)d(i -1, i) = d(O, i).
-d(i
-
(c) M L VO(Ti) = i=1 =
[d(O,O) -d(O, 1)] + [d(O, 1) -d(0,2]
+ ...+
[d(O,M -1) -d(O,M)]
1- d(O,M).
(d) Value = { Lf-1 d(O, i)T -[1-
d(O,M)] } N. The first term can be reduced
using the formula of the previous exerdse. 10. (Forward vanilla) We have d(0,i+1)=
1 . (1 +To)(l +TI) ..0 (1 +Ti)
Therefore d(O, i + l)Ti
= = =
Ti (1 +To)(l +TI) 0. .(1 +Ti) 1 + Ti -d(O i + 1) (1 +To)(l +TI) ...(1 +Ti) , d(O, i) -d(O, i + 1)
CHAPTER 10. FORWARDS, FUfURES, AND SWAPS
51
Then we find
M-I L [d(O, i) -d(O, i + 1)] = 1- d(O,M), i=O which agrees with the text.
11. (Specific vanilla) (a) The floating side of the swap is worth VFloat =
[1-d(0,6)]x$10million
=
[1-
~]
x $10 million
=
[1- .6029] x $10 million = $3.971 million
(b) The fixed side of the swap is worth VFixed = =
M L d(O, i)r x $10 million 1=0 4.606r x $10 million
where the 4.606 was obtained by SlJmming the discount rates implied by the term structure. Setting VFixed= VFloatwe find r = .0864 = 8.64%. 12. (Derivation) We obtain the mean-variance hedge formula by solving: max E[x + h(ET -Eo)] -r h
var[x + hET ]
which can be written as: max E[x] + h(ET -Eo) -r(var[x] h
+ 2hcov[x,ET ] + h2var[ET ])
Taking the derivative with respect to h we get the first order condition: ET -Eo -2r
cov[x, ET] -2hrvar[ET]
=0
Solving for h we find the mean-variance hedge formula: h =
ET -Eo -COV[X,ET ] 2r var[ET ] var[ET ]
13. (Grapefruit hedge) The mjDimum-variance hedge is h
=
UGSG -{3W = -p--150,000 uoSo
=
-(.7)
=
-131,250
$1.50/lb (~.2 ) ($1.20/lb
orange,
grapefruIt
Ibsorangejuice.
) 150, 000
CHAPTER 10. FORWARDS, FUfURES, ANDSWAPS
53
where as before x = 1, 000, 000 x K. Hence, Uy = .7211 stdev[x]. As expected the standard deviation with the equal and opposite hedge is greater than with the minimum variance hedge were we had Uy = .6 stdev[x ] 15. (Immunization as hedging) First we calculate the present value of the portfolio. (We shall use the continuous-time formula for discount factors in this exercise; however, the discrete-time version would give similar results.) PV
f I! partaa-
-'\:"'
c .e -r(ti)ti
Lo-t i
=
-(I)d-(1.0S)1
--.
0
-(2)e-(1.OS3)2
-(l)e-(1.OS6)3
+ 4.253e-(1.OS3)2
Hence the portfolio is balanced in the sense of haVing assets equal to liabilities. The current spot price of the bond considered for the futures contract is S = e-(.061)6. The corresponding forward price is then F = Se(.OS)l= e-(.061)6e(.OS)1. A futures contract will not change the PV of the portfolio. One way to express the present value is to assume that we actually take delivery of the contract. This means that there will be cash flows at the end of one year (to buy at the contract price) and at the end of six years (when the bond pays its principal). Hence a contract to purchase $x par value zero-coupon bonds has present value PVFutures = xe-(.061)6 -xe-(.061)6e(.OS)le-(.OS)1= 0. We now assume that the term structure is of the formf(t) a parallel shift. In this case PV
-'\:"'
tf I! par
a
0-
c.
e
Lo-
+ a, where a represents
-(!(ti)+a)ti
t
.
i We will set the derivative ~
da
Ia-O -=
equal to zero. We have
(1)(I)e-(.OS)l + (2)(2)e-(.OS3)2+ (3)(I)e-(.OS6)e =
-(2)(4.253)e-(.oS3)2 -6xe-(.061)6 -.566- 3.468x.
+ (1) (x)e-(.061)6e.OSe-.os
SolVing, we find x = -.1632. Hence the fund should short $163,200 worth of the Treasury futures.
54
CHAPTER 10. FORWARDS, furoRES, ANDSWAPS
16. (Symmetric probability) Given the future wealth W = a + hx + CX2, the investor's problem can be written as max E[U(a + hx + CX2)] where U is strictly concave. h (a) The first order necessary condition for the problem is E[U' (a + hx + CX2)X] = 0 Let f(x) be the probability density function of x, such that f(x) Then the first order condition takes the form ['"' U'(a + hx + cx2)xf(x)
= f( -x).
dx = 0
or equivalently, r,", U'(a + hx + cx2)xf(x) Io'"'-U'(a
-hx
+ cx2)xf(-x)
dx + Io'"'U'(a + hx + cx2)xf(x)
dx
=
o
dx + I:
dx
=
0
U'(a + hx + cx2)xf(x)
However, since f (x) is symmetric, f (x) = f ( -x) , so we can writte the above condition as -Io'"' u' (a -hx
+ CX2)X f(x)
dx + Io'"'u' (a + hx + CX2)X f(x)
dx = 0.
Rearranging by moving the first term to the right hand side of the equation we get Io'"'U'(a + hx + cx2)xf(x)
dx = Io'"'U'(a -hx
+ cx2)xf(x)
dx
Clearly, this condition holds if h = 0. Furthermore, since U is strictly concave h = 0 is the only solution. (b) The farmer's revenue is given by C2 + CC R = 10C -1600 1OOOh where c = 3, 000. If C has a symmetric distribution, then the distribution of x = C -C is such that f(x) = f( -x), and we can write the farmer's revenue as R
=
(x + 3000)2 x 10(x + 3, 000) -1000 -16OOh
=
21,000
+
(4-
LOOO h
) x-
LOOO X2
CHAPTER10. FORWARDS,FUTURES,AND SWAPS From part (a) we know that for an investor with strictly optimal solution is: h 4 -IOOO = 0.
concave utility
55 the
Hence, h * = 4,000.
17. (Double symmetric var(R)
probability)
=
var(Axy
=
A2var(xy)
We have + Ex -hy) + E2var(x)
+2ABcov(xy,x) Setting the derivative
-2Ahcov(xy,y)
-2Ehcov(x,Y).
of the above (with respect to h) to zero we have
2hvar(y) Hence
+ h2var(y)
-2A
cov(xy,y)
h* = E cov(x,y)
-2E
cov(x,y)
+ A cov(xy,Y)
= 0.
.
var(y) Now we show that cov(xy,Y) cov(xy,y) because E(y)
Let s = -x,
= E(xy2)
h*=~.
-E(xy)E(y)
= E(xy2),
= 0 by symmetry.
ds = -dx,
E(xy2)
= J J xy2 f(x,y)dxdy.
t = -y,
dt = -dy.
E(xy2)
Since E(xy2)
= 0.
= -E(xy2)
Then
=
f f ( -S)t2 f( -s, -t)ds
=
-f
=
-E(xy2).
it follows
dt
f st2 f(s, t)ds dt
that E(xy2)
= 0. Thus cov(xy,y)
Uy
18. (A general farm problem)
In general the farmer's R = PC + h(P -Po)
substituting
for P in terms of D we now find
revenue will be
= 0 and
56
OIAPTER 10. FORWARDS,FUTURES,AND SWAPS
R=
(10 -100,000 D ) C + wo:oooh tJ-D
after some algebraic manipulation we can write the equation as
(
R = -100, 000-tJ) + 10C -100, tJ000 (C- (:)(D
) (C-
(
C- ) -100, h000 + 100,(:000
) (D-D)-
then, substituiting values for (: and tJ we get
(
(C-(:)(D-tJ) R = -100,000
+ 7(C -C) -100,000 h
3
+ 100
) (D -D),-
which is of the formR = Axy+Bx-hy and where the distributions ofx = C -(: and y = D -tJ are symmetric about (0,0) as required in Exercise 13. Thus, we can find the minimum variance position by applying the result h = Baxy/a; which gives us h 3 = ~7aCD 100,000 + 100 Solving for h we find
(
3 + ~7acD h = 100, 000 -100
)
as required. (a) For the case where D = 100C we know that PCD = aCD/aCaD = 1 so then ~ -~ -ac --1afj -aD -100ac -100 Thus, the minimum variance hedging position is h
=
100,000 ( -~
+ ~)
= 4, 000 (b) If the crop size is not correlatedwith the total demand,Iiamely aCD= 0, the minimum variancehedgeis given by h
=
100,000 ( -~)
=
-3, 000
Varianceis minimized by taking an opposite position to the expected crop size.
Chapter Models
11 of
Asset
Dynamics
1. (Stock lattice) If we consider that ~t is small, then by the formulas u
=
d
=
p
=
eu,;M = 1.105 1
-= u
0.905
~ + ~ ( ~ ) .fM = 0.65
149.1
1
(11.1), we set
1S1.834
9
2
.1
21
4
0
°
1
°
11
16
.9
55
11
16
.1
61
0.01S
0.016
Figure 11.1 The binomial lattice. The numbers above the nodes are the stock prices. The numbers below the final nodes are the probabilities of achieving those nodes. The binomial lattice for the stock is shown in the left in Figure 11.1. If we do not consider that ~t is small, then by the formulas (11.25), we get p = 0.64367, u = 1.11005, and d = 0.90086. For comparison, this binomial lattice is shown to the right of the first one. 57
58
CHAPTER 11. MODElSOFASSET DYNAMICS The probabilities of the various final nodes are shown in the above figure under the nodes. For example, the probability of the top node is p4 = .644 = .179. 2. (Time scaling) Each movement in k corresponds to a month, and each movement in K corresponds to a year. Let kK denote the first month of year K. Then 11 W(K) = 2: W(kK-l + i) i=O So,
11 E[W(K)] = E[ 2: w(kK-l + i)] = 12v i=O 11 Var[W(K)] = E [ 2: w(kK-l + i)]2 = 120"2 i=O
3. (Arithmetic and geometric mean) (a) Proof: For n = 2, (VI -V2)2 ~ 0 ==> vf + 2VlV2 + v~ ~ 4VlV2. That is:
1 2 4(Vl + V2) ~ VIV2
So, VA ~ vG (b)
rl = 50%, r2 = -20% Arithmetic mean is
I 2(r1 + r2) = 15%.
Geometric mean is [(I +rl)(1
1 +r2)]2 -I
= 9.54%
(c) The arithmetic mean rate of return essentially assigns a return based on simple interest, while the geometric mean rate of return is a measure of compound interest. Usually, the geometric mean rate of return is the most appropriate for measurement of investment performance.
CHAPTER11. MODELSOF ASS£I' DYNAMICS 4. (Complete w -(w
the square)
-W)2 2a2
=
-~[ 2a2 1
=
--w-
-2a2w [
+ W2 -2ww
( -2 w+a
)]
2
+ w2 + 2a2w
-2a2w
+ a4 -a4]
-a2 +w+-
2a2
2
So, -1 u
f +OO w =
e ~
=
e'W+T2
=
-u2 ew+T
-00 1
- (w-'W ) 2 / 2u2
dw
e
f +OO e-[w-('W+u2)]2/2u2 dw
~
-00
5. (Log variance) Suppose that u = ew, where w is distributed
E(U2)
=
E(e2W)
=
1
as N(w,
f +OOe2we-(W-'W)2/2U2dw.
~ Following
59
-00
the method in exercise (4), we have 2w -~
= -2k[W
-(w+
2a2)]2 + 2a2 + 2w.
Then E(U2) = e2u2+2'W So, var(u) 6. (Expectations)
= E(U2) -U2
= e2'W+u2(eU2 -1)
We have 1 v = J.l -2a2
= 0.2 -2
1
x 0.16 = 0.12
So, E[lnS(l)] Stdev[lnS(l)] E[S(l)] Stdev[S(l)]
= 0.12 = 0.40
= eO.2= 1.22
= eO.2(eO.16-1)~
= 0.51
a2)
60
CHAPTER 11. MODELS OFASSETDYNAMICS 7. (Application of Ito'slemma) We have G(t) = F(s,t) = st(t), and o2F/oS2 = -ls-t.
Therefore according to Ito's lemma
dG(t)
=
(ts-taS
-ls-~b2S2)dt
=
(ta -lb2)Gdt
8. (Reverse check) We have a = 11-to"2, dS
then oF/aS = ts-t,
+ tbGdz
b = O",S = F(Q) = eQ. Thus
=
(aQa+at+2~b
=
(S(11-to"2)
=
I1Sdt + O"Sdz.
oF
+ tS-tbSdz
oF
1 02 F
2
) dt+aQbdz oF
+ tS0"2) dt + SO"dz
9. (Two simulations) Using eX = 1 + x + tx2 + ..., expressed as 1 S(tk+l) = [1 + (v + 20"2f2(tk»~t
the equation in (11.20) can be
+ O"f(tk).Jt;i]S(tk)
Obviously, it is different from the expression in (11.19). But the expected values of the two expressions are identical to the first order: E[S(tk+l)] = [1 + (v + to"2)~t]S(tk)
= [1 + l1~t]S(tk)
So, over the long run the two methods should produce similar results. 10. (A simulation experiment) (a) A simulation shows that convergence is achieved only after a few thousand years. (SeeFig. 11.2 for an example.) We know that 1nS(t) has a normal distribution with mean vt = .10- .302/2 = 0.055t and variance 0"2t = .09. Hence tlnS(t) has mean v and standard deviation '5t. As t goes to infinity, it is clear that the standard deviation goes to zero and 1 tlnS(t) -+ v. (b) We have O"= .30. We would like the standard deviation of the simulation to be .005. Hence we must have .30/ .jf = .005. In other words, t ~ 3, 600 years; or equivalently, about 43,000 months. This is consistent with the simulation experiment. (c) This does not converge. The expected value is 0"2 = .302 but the simulation moves around that value. [For those who have studied statistics: The distribution of the quantity approaches a chi-squared distribution.]
Chapter Basic
12 Options
Theory
1. (Bull spread) The initial cost of the spread is nonnegative since C(Kl) ~ C(K2) for Kl < K2 (see Exercise 4).
s
Figure 12.1 Bull spread. This is the combination of a call and a put. 2. (Put-call parity) Use the same portfolio as in the text: buy one call, sell one put, and lend an amount dK. This will reproduce the payment of the stock, except that it will be short by an amount with present value D. Hence C -p + dK = 5 -D. 3. (Parity formula)
Q
=
max[O,5 -K]
=
(5 -K)
=
0- (K -5)
-max[O,K
-O + K = 5 +K =5
Therefore, Q = 5.
63
-5]
if 5 ~ K if 5 ~ K.
+K
64
CHAPTER 12. BASICOPnONSTHEORY 4. (Call strikes) (a) Assume K2 > KI, and suppose to the contrary that C(K2) > C(KI). Buy option 1 and short option 2. Use option 1 to cover the obligations of option 2, since max[O, 5-KI] ;:: max[O, 5-K2] for all5. Keep profit of C(K2) -C(KI). (b) Assume K2 > KI and suppose to the contrary that K2 -KI < C(KI) -C(K2). Buy option 2 and short option 1 to obtain K2 -KI + E profit (where E > 0). Use option 2 and profits K2 -KI to cover option 1 since max[0,5 -K2]
+ (K2 -KI)
= max[K2 -KI,5
-KI]
;:: max[[0,5 -KI].
Hence you still keep E profit. This arbitrage opportunity assumed inequality cannot hold. (c) Assume K3 > K2 > KI and suppose to the contrary that C(K2) >
( K;""=K;: K3-K2 ) C(KI)
Buy (~ ) of option 1 and (~ 2. the profit is some E > 0. Notice that
+
shows that the
( K3 K2-KI-K1 ) C(K3).
) of option 3 and short one unit of option
( K:i=K4 K3-Kl) C (K I ) + ( ~K3-Kl ) C (K 3) ;:: (~)(5-Kl)+ =
(~
=
5 -K2.
(~)(5-K3)
) {(5-
K2) + (K2 -KI»
+ (~
) {(5-
K2) + (K2 -K3»
Also (~)C(KI)
+ (~)C(K3)
> 0.
Therefore it is possible to cover option 2 and make a profit of E > 0. This is an arbitrage opportunity, so the original inequality cannot hold. 5. (Fixed dividend) The intitial present value of the dividend is 3e-(.IO)(3.S)/I2) = 2.9138. Hence we set 5*(0) = 50- 2.9138 = 47.086. We assume (and this is an approximation) that 5* has the same volatility as 5. Hence we assume that 5* follows a binomial process with u = eO".[&f= e.2/.J[2= 1.0594 and d = .9439. The monthly return is R = 1 + 0.1/12 = 1.0083. We find the risk-neutral probability q = ~ = 0.5577. The process for 5* is shown in the upper part of Fig. 12.2. The stock price itself is expressed as 5(t) = 5*(t) + 3e-(.10)(3.S-t)/I2)for t < 3.5 and 5(t) = 5* (t) for t ;:: 3.5. Hence we find the lattice for 5 by adding the appropriate amount to each node in the 5* lattice. The result is shown in the figure. The call option value is found by the normal backward process on the stock lattice. We find $2.83 and $2.51 for American and European options, respectively. See Fig. 12.2.
CHAPTER
47.09
Random
12.
BASIC
OPTIONS
49.88
52.85
55.99
59.32
62.84
66.58
44.44
47.09
49.88
52.85
55.99
59.32
41.95
44.44
47.09
49.66
52.85
39.60
41.95
44.44
47.09
37.38
39.60
41.95
35.28
37.36
Component
of
Stock
Price
(S.)
rnEORY
65
33.30
50.00
Stock
Price
52.82
55.81
58.98
59.32
62.84
66.58
47.38
50.05
52.87
52.85
55.99
59.32
44.91
47.43
47.09
49.88
52.65
42.59
41.95
44.44
47.09
37.38
39.60
41.95
35.28
37.38
(S)
33.30
2.83
American
4.23
6.23
6.98
10.14
13.26
16.58
1.11
1.80
2.87
4.23
6.40
9.32
0.27
0.48
0.87
1.58
2.85
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
Call
Option
0.00
2.51
European
3.70
5.32
7.47
10.14
13.26
16.58
1.07
1.72
2.72
4.23
6.40
9.32
0.27
0.48
0.87
1.58
2.85
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
Call
Option
0.00
Figure
6.
(Call to
inequality) the
price
7.
of
off
(A) the
max[O,S
-KB(T)].
(Perpetual we
price
(A
of
be
From
call
a
(A)
a
stock
than call
is
=
must
stock
to
C(S,
T) ~ C
~
always
greater
and of
than
sell (B).
or
K
equal
oS
S.
or
C to
S
~
S
zero.
Hence,
-KB(T) in
]
the
the
-KB(T). Thus
C
~
B(T)
=
limit
equal
Hence
SincelimT-~
max[O,
I T)
than bonds.
Thus
max[O,S-KB(T)].
limT-~ (S
is
that
greater
C(S)
satisfy
call
of
equal
always
dividend.
one share
or
6, T)
with
purchase one
Exercise C(S,
option
on
purchase greater
of
limT-~ a
of
(B)
value
call) have
Call
payoff
of
must
However
0,
8.
The
payoff
12.2
S. C
Clearly =
S
= the
.
surprise)
(a)
PV(r)
(b)
The
will
value
changing that are
ofr.
increase
of
r predicted increased;
if
the
in
r
is
Simplico
the by whereas
decreased.
mine
with
spreadsheet. part
r
=
Hence
(a). the
The value
it
reason of
4%
is the
is
$22.2
million,
moves
in
that
when
gold
income
the
obtained
opposite r
is is
reduced, largely
by
just
direction the independent
as costs
66
CHAPTER 12. BASICOPTIONS rnEORY 9. (My coin) This is like the example in the text. Draw a lattice with three stages. The payoffs of the four final nodes are 27, 27, 0, 0. Roll back one stage. The three nodes there have implied values 27, 9, 0. Roll back one more stage. The implied values there are 15, 3 which can be written as 12+3, 0+3. Hence the implied value of initial node is 4+3=7. The result can also be found by direct riskneutral valuation (without rolling back) using q = 1/3 and R = 1. The risk-neutral probability of the two nodes with 27 are 1/27 and 6/27. Hence the total value is 7.
10. (The happy call) The payoff is max[.5S,S -K]
= .5S + max[O, .5S -K]
= .5S + .5max[0,S -2K].
Hence, by linear pricing, we add the prices of the individual pieces to obtain CH = .5P + .5C2. Thus £x= .5, 13= 0, )' = .5. 11. (You are a president) Notice that a $1 increase in the S&Pindex index corresponds to a rate of return of 4Ii74 in the index. Hence, for each dollar invested in the spedal offer, the payoff is $1 plus ~ x 4Ii74 of a call option on the S&P 100 index with strike price $414.74 and expiration in November. The price of the $1 portion of the payoff is its present value. Assuming 3.5 months until maturity, this value is PI =
1 3.5 = .991010707. 1 + .031112"
The value of one call with strike price 414.74 is found by interpolation to be C = 13 -(13-
i~)4.74
= 10.393.
The value of the fractional call that is offered is therefore 1 P2 = 41 x 4"l4J4 x 10.393 = .0662647. Hence the total value of the offer (per dollar invested) is V = PI + P2 = .991010707 + .0662647 = $.9972754. We conclude that, from the data we have, the offer is low by about 0.3%-which is not bad. 12. (Simplico invariance) Note that changing u or d amounts to changing the standard deviation of gold price and therefore would have an influence on the lease value of the Simplico gold mine. However, the initial gold price $400 is relatively very
68
CHAPTER12. BASICOPnONS nIEORY
0 1 2 3 305.78 371.05 450.07 545.93 256.62 310.56 375.08 212.01 255.59 171.11 K-Value
4 5 6 7 8 9 662.68 805.98 984.441212.451516.031940.64 451.97 543.14 650.54 775.99 921.25 1091.61 306.83 366.28 433.65 506.54 576.57 614.03 204.76 243.00 284.76 326.57 358.26 345.39 133.83 157.87 183.13 206.32 218.80 194.28 100.36115.23127.64131.29109.28 71.05 77.47 77.62 61.47 46.26 45.34 34.58 26.24 19.45 10.94
10 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
Figure 12.4 Average value Complexico. The overall value does not change much, but the computation is slightly more complicated than the original version.
14. (Average value Complexico) The Complexico gold mine of Example 12.8 is modified to reflect the average value of gold price in adjacent periods and the discounting of revenue received at the end of each period. As before V i (xi) = KiXi for all i with KlO = 0 .However, the following adjustments are made
(~!i.t.l. .. Ki=
)2
2R -R 2,000+R
~ Ki+l
. 1=1,2...,9.
Gold prices follow a process described by u = 1.2, d = 0.9, and q = 0.6667 the K-values, shown in Fig. 12.4, are calculated by working back from period 9. The value of the lease is 50, 000 x 305.78 = $15, 288, 786. the value in the original version was $16,220,000. 15. ("As you like it" option) We use the solutions for the call and the put in the examples in the text. At the end of the third year, we know that the value will be the maximum of the call or the put value. Hence, we just copy those values from the examples and then work backward with the standard risk-neutral discounted valuation process. The table below shows the details. The value is A = $6.73, which is greater than either the put or the call value. 6.73
8.34 4.83
11.18 4.92 4.81
14.71 6.96 2.45 7.86
CHAPTER12. BASICOPTIONS11IEORY
69
16. (Tree harvesting) Grcmth
5
1.6 1.6
1.5 2.4
6 4.5
7.2 5.4 4.05
Lumber Price
1.4 3.36
1.3 4.368
1.2 5.242
1.15 6.028
1.1 6.631
1.05 6.962
1.02 7.101
1.01 7.172
8.64 10.37 12.44 14.93 6.48 7.776 9.331 11.2 4.86 5.832 6.998 8.398 3.6454.3745.2496.2997.558 3.281 3.937 4.724 2.952 3.543 2.657
17.92 13.44 10.08
21.5 25.8 16.12 19.35 12.09 14.51 9.0710.88 6.802 8.163 5.102 6.122 3.826 4.592 2.87 3.444 2.152 2.583 1.937
30.96 23.22 17.41 13.06 9.796 7.347 5.51 4.132 3.099 2.325
5.669 4.252 3.189 2.391
1.743 23.1331.1340.6451.9565.41 81.45 100.6123.4150.7 20.68 28.06 36.81 47.2 59.55 74.23 91.7 18.7325.5733.6743.2654.6568.1684.19103.1124.9 17.27 23.64 31.17 40.07 50.62 16.32 22.28 29.32 37.59 Optimal Value 15.88 21.49 28.19 Bold=cut 16.02 21.14 15.86
112.5 63.15 47.36 35.52 26.64 19.98 14.98
183.2 137.4
222 166.5
77.29 93.68 57.97 70.26 43.48 52.69 32.61 39.52 24.46 29.64 18.34 22.23 13.7616.67 12.5
The first two rows show the growth rate and the cumulative growth by year. The first lattice shows the price of lumber using an up factor of u = 1.2 and a down factor of d = .9. The bottom lattice shows the optimal value as a function of position in the lattice. The risk-neutral probability of an up move is q = ~ (1.1- 0.9)/(1.20.9) = .6667. The value at a typical node is the maximum
= of
either (1) the market value of the trees or (2) the risk-neutral discounted value of continuing to the next time. For example, the value at the top node of the second to last column (in millions) is max[25.8
x 7.101, (q222 + (1-
q)166.5)IR
-2]
= 183.2.
In this case the maximum was obtained in the first portion of the maximization, meaning that the trees should be harvested. The value is entered in bold face to indicate that.
Chapter
1 3
Additional
Option
Topics
1. (Numerical evaluation of normal distribution) The equation is best implemented on a calculator, spreadsheet, or other computing device. The answer for the call stated in the exercise is C = $2.57. 2. (Perpetual put) (a) For the expression P(S) = aIS + a2S-Y we have p' (S)
=
aI -)'a2S-Y-I
p" (S)
=
)'()' + l)a2S-Y-2.
Substituting in the Black-Scholes equation !0"2)'()' + l)a2S-Y + raIS -r)'a2S-Y
-aIrS
-a2rS-Y
= 0.
Canceling terms we find !0"2)'()' + 1) -r)'
-r
= 0.
Hence, )' = 2r /0"2 satisfies the equation. Since a I and a2 are arbitrary, this represents two independent solutions to the second-order differential equation; and hence is the general solution. (b) P(oo) = ° implies aI = 0. P(G) = K- G implies a2G-Y = K -G leading to a2 = (KP(S) = (K- G) (S /G)-Y .
G)/G-Y. Hence
(c) It makes sense to maximize P(S) since the maximization is independent of S. We maximize (K- G) GY.Differentiation with respect to G gives the condition -G-Y + )'(K -G)GY-I
= 0.
Thus G = )'K/()' + 1) and P(S) = K
(1 -21+)'
71
)(
)
~ )'K
-y
.
72
CHAPTER 13. ADDmONALOPnONTOPICS 3. (Sigma estimation) Using the spreadsheet implementation of the previous exerdse, we adjust O"by trial and error to obtain the give call premium. The result is O"= .251. 4. (Black-Scholes approximation) For S = Ke-rT we find dl
= --2
d2
=
In(e-rT) + (r + 0"2/2)T O".jf O".jf .
-~.O".jf
Hence C
~ =
S [ ! + u..j'f ] -SerT 2~ SO".jf ~ .4SO"vT. I;;; ~
e-rT
[!
-u..j'f
2"272"7r ]
For delta we have 6 = N(dl)
~ ~+ ~
~ ~ + .20".jf.
For the call of the example Ke-rT = 60e-.5/l2 = 57.55. The value at that value of S is (by the above approximation) .4x.2x 57.55x.J57U = 2.972. The value of 6 at the base point is 6 ~ .5 + .2 x .2..rsTf2 = .5258. The difference in S is 62- 57.55 = 4.44. Hence the final value is C ~ 2.97 + 4.44 x .5258 = $5.30. 5. (Delta) Using the options calculator the price of the call at $63 is $6.557. Hence, 6 ~ 6.557- 5.798 = .759. Changing T to T = 5/12 + .1 we obtain a call price of $6.490. Hence e ~ (6.490 5.798) /0.1 = 6.02. 6. (A spedal identity) This is just the Black-Scholes equation.
CHAPTER 13. ADDmONALOP11ON TOPICS
73
7. (Gamma and theta)
r
=
02C 06 aS2=-as
=
~
=
oS N'(dl) -su:!f .
=N'(dl)~ oS
For theta, use Exercise6 to write e
=
rC -rS6
=
rSN(dl)
=
-~
-~u2s2r -rKe-rTN(d2)
-rSN(dl)
-~uSN'(dl)/JT
-rKe-rTN(d2).
8. (Great Western CD) It is possible to set up a lattice of possible S&Preturns in the standard way; however the payoff of the CD depends on the path through this lattice. Hence it is necessary to use a full tree representation. We use two-month periods so that the tree will not be too large. In the tree a move across corresponds to an up move; a move down to the nearest nonzero entry corresponds to a down move. The parameters are u = e.2.Ji76 = 1.085, d = l/u. shown in Fig. 13.1.
The tree of S&P returns is
The tree can be expressed in a filled in form as shown in Fig. 13.2. In this version, each row represents a distinct path trough the tree. This version simplifies the computation of the average value along a path (formed from the cells beginning in the second column). The risk-neutral valuation is shown in Fig. 13.3. The final column is the total return of the CD, defined as the average of the S&P returns of the path, or 1, whichever is larger. The risk-free rate is set arbitrarily at first. It is adjusted (with a solving program or by trial and error) until the original value is 1.0. Note that the risk-neutral probabilities will vary as the risk-free rate is adjusted. The final equivalent interest rate is 7.63%.
74
CHAPTER
13.
ADDmONAL
sig=
0.2
u= d=
TOPICS
0.55741874
~
0.44258128
0.92159478
RR.
1.01272205 0.99999999 8.491
Rete=
(Two-month
risk-free
(Risk-neutrel E-09
(RR
1.08507580
return)
result
(Dillerence
7.63%
1.00000000
q=
1.0850758
Vel= Error= Yeerly
ornoN
of
from
Vel
converted
from
to
1.17738905
yeerly
finel 1.
sheet.)
Drive
to
zero.)
rete.)
1.27755812
1.38824497
1.50418059
1.83214985 1.38824497
1.27755812
1.38824497 1.17738905
1.17738905
1.27755612
1.38824497 1.17738905
1.08507560
1.17738905 1.00000000
1.08507580
1.17738905
1.27755812
1.38824497 1.17738905
1.08507580
1.17738905 1.00000000
1.00000000
1.08507580
1.17738905 1.00000000
0.92159478
1.00000000 0.84933893
1.00000000
1.08507580
1.17738905
1.27755812
1.38824497 1.17738905
1.08507580
1.17738905 1.00000000
1.00000000
1.08507580
1.17738905 1.00000000
0.92159478
1.00000000 0.84933693
0.92159478
1.00000000
1.08507580
1.17738905 1.00000000
0.92159478
1.00000000 0.84933893
0.84933893
0.92159478
1.00000000 0.84933693
0.78274448
0.84933693 0.72137322
0.92159478
1.00000000
1.08507580
1.17738905
1.27755812
1.38624497 1.17738905
1.08507580
1.17738905 1.00000000
1.00000000
1.08507580
1.17738905 1.00000000
0.92159478
1.00000000 0.84933893
0.92159478
1.00000000
1.08507560
1.17738905 1.00000000
0.92159478
1.00000000 0.84933893
0.849338930.92159478
1.00000000 0.84933693 0.78274448
0.84933693 0.72137322
0.849338930.92159478
1.00000000
1.08507580
1.17738905 1.00000000
0.92159478
1.00000000 0.84933693
0.849338930.92159478
1.00000000 0.84933693 0.78274448
0.84933893 0.72137322
0.78274448
0.849338930.92159478
1.00000000 0.84933693 0.78274448
0.84933693 0.72137322
0.721373220.78274448
0.84933893 0.72137322 0.68481379
0.72137322 0.61288892
Figure
parameters
13.1
Great
and
Western
the
solution
CD.
This
are
is
shown
the
at
tree
the
of
top.
possible
returns
on
the
S&P
500.
The
CHAPTER
!
C"PV 01pr.viciu. .h..t but with dupllcel8dnumbers.
13.
ADDmONAL
i
OPTION
1 i 1.0860768! 1.17738906 1.27766812! 1.38824497! 1.27766812! 1.38824497 1 1.0860768! 1.17739906i 1.27766812! 1.38824497 1.27766812! 1.17738906
; ,
1!
1.0860768;
; !,
l' 1
1.0860758' 1.17738906 1.0850758! c 1.17738905i
!
1,1.0850758;
; !
1; 1.0860768; 1.17738905 1.0850768' 1!1.0850758!1.177389051.0850768:
L:l j ;
::~::~~::11,1773890~ 1 ~ 1.0850768!+ l' 1.0850768'
1 1:
::~::~~::11.1ii3890~ ~::~~:::~:1~::::~=::~ 1.0850768;" 1.1ii38905i 0.921594i8i1.1i738905 r 1.0850768; 1.17738905' 1.0850768' 1.17738905
! j,
1 1
1 1
1.0850758! 1.0850758!
1.17738906
1.27766812!
1.17738905:
1.0860758! 1.0850768; c
1.17738906
,
1.0850768; 1.0850768i
,
1.17738905 1.17738905
1.0860768'
1.27765812; , 1.27766812!
1.0860758! 1.17738905
1
,
1.38824497 , 1.17738906
1.0850758!1:17i38905
1 1.0860768; 10.92159478!
1 1
"
1.0850768i 1.0850758i
1.17738905 1
,
1 1.0860758; )J.),~~~~!~~L 1: 1.0850758;
1 1.0860768; !.r!:~~~~ 10.92159478'
j ;" ; ., !
1 1;~ 1; 1!
1 i 0.92169478! 1: 0.92159478j 1 1 c 0.92159478!" 1c 0.92159478; , 0.84933893 1 , 0.92159478; 0.84933893! 0.92159478! 1 1 0.92159478i 0,84933893! 0.92159478i 0.84933893
! j ! ! !
1i 1.0850758i 1! 0.92169478! 1!0.92159478! I:Q,92159~78! 1!0.92159478!
10.92169478iO,84933893;0.78274448!0.i2137322 1 1.0850758! 1.17738906i 1.27755812! 1.38824497 1 1.0850758!1.177389061.27755812!1.17738905 11,08507$8(r,)7738905! 1.0860758! 1.17738905 1 1.0860758;1.17738905i 1.0850i58i 1
! !
1! 0.92159478; 1 0.92159478!
l' 1.0860758; 1 i 1.0860758!
;
1
1
! ! ! .,
1! 0.92159478! 1! 0.92159478; 1! 0.92159478!
1! 0.92159478i 1 0.92159478i 1 i 0.92159478i
! !
r,Q,9?169478l 1! 0.92159478!
1 0,92r~9478jO.~~93~~93,Q,782744~8! Q,84933893 1 0.92159478! 0.84933893; 0.78274448!0.72137322
, !
1! 0.92159478; 0.84933893 0.92159478' 1: 0.92159478! 0.84933893i 0.92159478!
! ; ! : !
1 i 0.92159478! 0.84933893 0.92159478! 0.84933893i 1! 0.92169478! 0.84933893 0.78274448; 0.84933893; 1 i 0.92159478! 0.84933893! 0.78274448! 0.84933893 1; 0:92169478: 0.84933893 0.78274448: 0.84933893' 1 i 0:92159478!0:84933893!0.i82i4448jo:84933893 i
! ! :
1 0.92159478! 0.84933893 0.78274448! 0.72137322: 0.78274448io:7213i322 1; 0.92159478! 0.84933893 0.78274448! 0.72137322i 0.88481379! 0.72137322 1 i 0.92159478! 0.84933893' 0.78274448! 0.72137322 0.88481379! 0.81258892
'+
0.92169478'
Figure
13.2
shown
in duplicated
Great
Western form.
1,0.92159478;
1 1
; : ;
1.0850758i 1.0850768! , 1.0860758! 1.0850758!
75
!
! !
,
TOPICS
1
1! 1.0860758;1.17738905
1 i 0.92159478! 1 1! 0.92159478! 0.84933593
0.92169478!
l'
"
0.84933893i 0.84933893
0.92159478'
1
r
1 0.92169478; 0.92159478! 0.92169478! c
0.84933893 1 0.84933893
l' 0.92159478' 1 1 0.92159478! 0.84933893
CD.
This
is
the
tree
Each
row
of
the
sheet
0.i8274448:ij:i2137322 0.92159478; 1 0.92159478i 0.84933893 0.78274448: 0.84933893 0.782i4448: 0.72137322
of
possible
represents
returns
on
a path
through
the
S&P the
500, tree.
but
76
CHAPTER
13.
ADDmONAL
Ri.k.Neutrel
ere
0.99999999
ornoN
Valuetion.
averege
of
1.05651643
Celi.
p.th
from
in
finel
previou.
TOPICS
column
.heef.
1.12874264
or
1,
1.19883352
which
ever
1.25972483
i.
maximum.
1.30897438
1.343766
1.30278188
1.23390736
1.26501114
1.23020182
1.15860322
1.199535321.23020182
1.1953925
1.13577814
1.18331241
1.13374757
1.072910971.12532407
1.187858231.19812173
1.16331241
1.104101051.13123232
1.10166748
1.03773908
1.07490761
1.10166748
1.07210264
1.020756281.04485584
1.01974532
0.99591923
1.03741115
1.08222685
1.11273766
1.18855689
1.07442068
1.07490761
1.10186748
1.07210264
1.01078778
1.045714171.07210264
1.0425378
0.99585418
1.01529099
1
0.972289120.99230094
1.01880965
1.04495564
1.01529099
0.98743777
1
1
0.97503335
0.98743777
1
1
0.98743777
1
1
0.957564230.98487685
1.02256418
1.08993082
1.11176028
1.14131009
1.10850077
1.04800309
1.07442068
1.04465584
0.99230094
1.01880965
1.04485584
1.01529099
0.98743777
1
1
0.96572111
0.96036815
0.99713007
1.01760903
1
0.98743777
1
1
0.97503335
0.98743777
1
1
0.98743777
1
1
0.95069003
0.962784750.97503335
0.98743777
1
1
0.98743777
1
1
0.97503335
0.98743777
1
1
0.98743777
1
1
0.962784750.97503335
0.98743777
1
1
0.98743777
1
1
0.97503335
0.98743777
1
1
0.98743777
1
1
Figure
adjusted
13.3
Great
until
Western
the
value
CD.
at
the
This
originating
is
the
risk-neutral
node
valuation.
is
1.0
The
interest
rate
used
is
CHAPTER 13. ADDmONALOrnON TOPICS
77
9. (The control variate method) We have x = Xavg+ a(yavg -y). Let 0-2 denote the variance of x. Then 0-2 = var(xavg) + 2acov(xavg,Yavg) + a2var(yavg). Minimization leads immediately to COv(Xavg, Yavg) a---var(yavg) . The average values of x and y each contain factors of l/n (where n is the number of samples). These factors cancel out, giving cov(x,y) a-- -var(y)
.
10. (Control variate application) The simulation can be carried out by moving through the lattice, selecting the subsequent nodes according to the risk-neutral probabllities. As the simulation is carried out, the sample averages of the unknown x and the control variate y are calculated The covariance between x and y and the variance of y are estimated from the samples as well. For instance the estimate of var(Y) is ~ If:l (Yi -y)2, where n is the number of samples. We use these estimates to select the parameter a (which may change with n, but in practice tends to be quite stable). The results of such a simulation using the standard call option as control variate are shown in Fig. 13.4. The column headed x is the average of the call price itself, formed without using the control variate. The column headed by y is the corresponding estimate of the control variate. The column headed by x estimate is the estimate of x corrected for the control variate. The final two columns give the standard deviation (as computed from the formula using sample covariances and variances) of the two estimates. Notice that using the control variate reduces the standard deviation by about half. The same experiment was repeated using the normalized average price as a control variate. Specifically, we used the form 1 Y= {6(SO+SI+S2+S3+S4+SS)-K
} /R s.
The expected value of the one period return is qu + (1 -q)d = R. Hence the expected value of Y is E(y) = {(1 +R +R2 +R3 +Rs +R6)62 -60}/Rs = 3.171732. The results are shown in Fig. 13.5.
80
CHAPTER 13. ADDmONALornoN TOPICS
11. (Pay-later options) (a) We set up a standard price lattice using the parameters u = eu..jKf = e.2Ji7I2 = 1.06. This lattice is shown at the top of Fig. 13.6. ! :""" !)
10.7! 11.3! 12.0 12.7 13.5! 14.3 15.1 16.0! 17.0
! :
! [StockPrIce
! :
!
9.5! 10.1 ! 9.0
!
! 0.25 0.38! 0.57 0.85! 1.23 1.75! 2.40! 3.20 4.09! 5.04!
! !
! -0.05
-0.06! -0.06! -0.04! 0.04 -0.04! -0.07! -0.10! -0.14
10.7! 11.3! 9.5! 10.1
0.23! 0.59 -0.19! -0.22
,
0'
12.0 10.7
12.7! 11.3!
1.19 -0.17
2.07! 0.11!
0
0
13.5 12.0!
3.00! 0.93
0
Figure 13.6 Pay-Iateroption. The pay-Iater premium is set so that the original value is zero. Next we calculate R = 1 + 0.10/12. This gives q = 0.44 as the risk-neutral probability for an up move. The standard lattice using discounted riskneutral valuation is shown as the second lattice in the figure. The resulting price of the option is $0.53. (b) The paylater option lattice is set up exactly the same way except that the final values are S -K -C, for those cells where S -K is positive, and zero otherwise. The value C is unknown. We use a solving routine in the spreadsheet (or trial and error) to adjust the value of C so that the initial price is zero. In this case that value is $2.04. The corresponding lattice with that value is shown as the third lattice in the figure. (c) Obviously the pay-later option premium is higher than that of a standard option: The premium is not paid until later (meaning there is interest rate advantage, and more importantly, no premium is paid if the option does not end up in the money.
CHAPTER
!,
!, ! !
13.
ADDmONAL
OPnON
TOPICS
81
69.8! 83.5 100! , 120! , 143! , 172, 205 246! , 294! 353, , 422 505 605! , 724 , :; : 58.3 69.8! ; 83.5: 100! " 120: " 143 172!, 205!; 246: 294 353 422!, 505 !HousePrta : 48.7! 58.3,69.8!83.5! 100 120! 143, 172! 205 246 294! 353 ! : ! 40.7 48.7! 58.3 69.8 83.5, 100 120: 143, 172 205' 246
,
;
!
, 28.4: 34 40.7!48.7 58.3,69.8 83.5 100, 120
Figure 13.7 California housing put. The put value is remarkably small.
12. (California
housing put)
The calculation is shown on the spreadsheet in Fig. 13.7. The up and down factors for housing prices are u = eO", d = e-O", using the small6t approximation (although it is not fully justified
here). This produces the house price lattice.
The mortgage interest rate of 1096defines an amount A that is the yearly payment per dollar of loan. This is A = .13. This defines the yearly balance which is initially recorded in the first full row of the spreadsheet. The value of the put is calculated in the usual backward style using the appropriate risk-neutral probabilities (q = .73) and the risk-free rate of 1096. As an example, the last entry in the third to last column is max[20.7 0 + (1- q)2.38)/1.1] = 10.5. The put value is .72 for a loan of $90; which is quite small.
-9.63(1.05),
(q x
82
CHAPTER 13. ADDmONALOPnONTOPICS We then assume that the bank would like to charge for the put. Hence the new loan is for $90.72. At 10% the payments of this loan are $11.9 per year. These payments spread over fifteen years is equivalent to an interest rate of 10.1%. (See the example on APR in Chapter 3.) This new rate will change the balance structure of the first row. In fact, the values shown in the spreadsheet correspond to this 10.1% rate. In theory, these new balances will change the put value and the whole process must be iterated until convergerice. However, only this single step is required in this example, because the put value is so small, and the value does not change except in far out decimal places. (We have also not accounted for the fact that the full $.72 may not be recovered if the put option is exercised. The value of this "put on put" is likely to be extremely small, however, and can safely be ignored.)
13. {Forest value) This solution is identical to that of Exercise 16 in Chapter 12, except that the risk-neutral probability is q = ~ = (1.1- .9 + .05)/(1.2- .9) = .833. The value obtained from the modified spreadsheet is $42.42 millions. 14. (Mr. Smith's put) This is straightforward. It turns out that it is never optimal to exercise the put early. The appropriate lattices are shown in Fig. 13.8. ,
,
,
;! 0;! 1i 2;i' : 0.625: 0.644: 0.664: : : 0.607: 0.625: : : : 0.589: : : : :
I
15tockPriCe
: 0.017: 0.007 : : 0.023 : : : :
I
0.002: 0.01: 0.031:
i PutOption
I
,
3; 0.684: 0.644: 0.607: 0.571:
,
4; 0.7051 0.664: 0.625: 0.589:
5,
0.726: 0.684: 0.644: 0.607:
I 0.5541 g:~~~1 g:~i~1
0: 0: 0: 0.003: 0: 0: 0.015: 0.004: 0: : 0.041: 0.023: 0.007:
:
6 ;; 0.748: 0.705: 0.664: 0.625:
0: 0: 0: 0:
I 0.053\ g:g~~i g:g~i
Figure 13.8 Mr. Smith's put. The lattice calculations are standard.
Chapter
14
Interest-Rate
Derivatives
1. (A callable bond) The calculation of the bond value in part (a) is straightforward The calculation is shown in the two upper lattices of the spreadsheet in Fig. 1. The value is 91.72.
,
,
,
,
0.21
0.19,
0.17
I ,
i
I !
I 0.1Si
0.18; 0.13;
0.16 0.12
0.1S! 0.11i
0.13, 0.l0
0.00!
0.00
O.OS,
0.04,
0.04
0.03,
0.03,
71.SSl 81.91
72.68 81.99 90.22
84J9 91fJ)l 9722:
Short rate lattice!
,
:
, r ,
,Bond v..u.tionJ
! ,
! ,
! !
!
, !
0.07
O.OS
, :
91.n!
! :
84.9S!
, 1 n.77'
l !
91.76!
,
97.6S
10237!
",.
10S.77!
90.98 103.47
97.24! 10823!
102.S4! 100.69 1ll.84! 1!4.26
100.581 11!.l8! l!S.46! l!S.50!
90.71
84.90! 96.ni
91.~! 97.44 101.S4: 104.78
101.93: 100.00i
Figure
Callable
107.78
108.43:
107.79
100.00
1ll.52 1!4.46
1!0.70! 1!2.4S!
108.81: 100.59
106.00 106.00
100.00, 100.00: 100.00! 100.00,
100.00 100.00
104.88! 100.00 100.00 100.00
L.~~I..!~:!.L.!~,~i!91:~L.!91:~
14.1
83
%.28 106.00 99.7S 100.00 102.S4 100.00
bond.
84
CHAPTER 14. INTEREST-RATE DERIVAnVES To find the value with the call feature, we construct a third lattice in a similar way, except that at each step we compare the discounted risk-neutral value of the next period with the option of calling the bond, which gives 106. The value is 90.95. 2. (General adjustable formula) First find the payment require, form Chapter 3, p
ks
= (rks + p)(l + rks + p)100 (l+rks+p)n-k-l .
After one period the remaining principal will be loaned again. This amount is Lks = 100[1 + rks + pJ -Pks. Then the recursion for the value is found by discounted risk-neutral valuation as
Vks =
[ LkS ( 1 + iio
]
[.5V(k+l).S + .5V(k+l),(s+1)J) + Pks 1 + rks
-100.
3. (Bond futures option) You could set up a futures price lattice as shown in Example 14.4. Then use this as the underlying price lattice and carry out a standard option backward evaluation. 4. (Adjustable-rate CAP) The same spreadsheet as in the example can be used. It is only necessary to change the formula for the interest rate used to determine the loan payments, so that it is capped at 11%. The formulas in the evaluation lattice do not change, since the bank should still discount by the actual rate. The resulting two lattices are shown in Fig. 14.2. The answer is $4.20.
! ,
i!, i
!
! ,
58.39
Ill!
,
100;
, ": 40.92 58.39 1Iii " 100! c i 32.23i 40.35 57.13 108.6i 100i
!
!-6.073!-2.496!
,
-2.635: , , 0.033 , 1.294! : 1.76857 2.7~i 2.634, 1.876!
0
0 0
Figure 14.2 Auto loan with a CAP of 11%.
5. (Forward construction) The elementary price lattice is constructed using the equations in the text. The value of a zero-coupon bond is found by summing the elementary prices corresponding to the maturity date. The spot rate is then determined from the value of the zero. The results are shown in Fig. 14.3
alAPTER
! ! i! i!, i
! ! , 0.060i
i
DERIVA11VES
0.104 , O.078i 0.0S8i
0J1)3! , omoi 0.052i,
0.084
0.076
0.068
0.063 0.047
0.OS7 0.043
0.051 0.038~
0.061! ! O.046i O.034i ,
0.055
0.086 0.06S
0.041 0.031
! ! i i,
0.054i
0.049
0.044~
0.039i
0.03S
0.032
0.029
O.026i
0.023
i
0.020
0.OS7
0J})4
0.118i
0.127
0.122
O,I~ 0.223
0.151 0.212
0,167 0,l77!
O.lror 0.136!
0.138 0.(1)8
O,lll 0.068i
i
i!
!
i 0.101!
9.046r 0,l90 !
of
forward
I !
Figure
INTEREST.RATE
! , i 0.072i!
~
I, !
14.
14.3
Construction
prices
and
determination
of
spot
rates.
85
86
CHAPTER
!a's,
14.
.1.!i1
INTEREST.RATEDERIVAnVES
1I.lIlY!
Y:IW,
, !
10, 9!
i : ; ! i
7! 6: 5, 4! 3!
i : , ! !
!
1!
8.92!
! !,
8! 7! 6'
Equations! :;
!
4!
! i
2! ti
i'
; ShortRate
ll.~!
, ,
10.77
;
,
:
!
11.'IJ!
! !
: : , ! ! to.99 9.90!
! !
1:L./~
1~.U:I! i~.:l/'
, !
i i 14.52
i~.:1i
i~.:ll!
l~.!i/,
1~:I:l
!
14.93 14.78
15.00! 14.85!
15.11 14.96
15.16 15.01
, !
! f ; 11.76! 11.64!
! ! fl~,~: 12.54, 12.96 12.42! 12.83 12.29! 12.70
13.66 13,5Z 13.39 13.25 13.12
13.99! 13.85[ 13.72; 13.58 i 13.44!
14.23 14.09 13.95; 13.81 13.67!
14.49 14.34 14.20 14.06 13.92
14.56! .4.41: 14.27; 14.13! 13.99:
14.66 14.52 14.37 14.23 ! 14.09
14.71 14.57 14.42 14.28 14.13
: : ; , !
11.41!
12.05!
12.86
13.18!
13.40!
13.64
13.71!
13.81
13.85
!
0.002!
0.007 0.015 0.027 0063" 0.039 0069
0.024! 0.047! 0067
0.031 0.05 00 c9
0.035 0.047 0048
0.036! 0.041! 0037 "
;
12.45
! !'
!,
0.004 0.014!' ivvi' , 00270048 ; o.\1\17 ..;
;
...;
'
."
..!
0.044! 0.098! 0.131! 0.136 0.12! 0.095! 0.07 0.048! 0.032: 0.02 0.012!
14.4 Term match using
volatility)
From node
( k -1,
or ak + bk(S + 1). The variance var
i:L~~,
! 0.213' 0.291 0.263! 0.196! 0.131 0.082 0.048! 0.027 0.015 0.008! 0.004! 0.002 0.001 0.464! 0.427: 0.291 0.175: 0.098! 0.053 0.027 0.014! 0.007 0.003 0.002! 8E-04;4E-04 2E-O4!
Figure
6. (Ho-Lee
iU.bb
!(ak lb2 2
k
+ bkS)2 + !(ak -lb2
4
the Black-Derman-
s) the short
Toy model.
rate next period
is either
ak + bkS
is + bk(S + 1»2-
~[ak
+ bk(S + 1) + ak + bkS]2
k
1b2
4 k. Hence the standard deviation is bk/2.
\
7. (Term match) This can be solved by a simple modification of the spreadsheet in the Ho-Lee example. We merely change the formula for the short rate to ake'OlS. The new values of the ak'S are found by optimization as in the earlier example. The resulting spreadsheet is shown in Fig. 14.4 8. (Swaps) Fig. 14.5 shows three binomial lattices. The first is the short-rate lattice. The second is the lattice ofvalues for the floating rate payment stream (in hundred thousands of dollars). Each node value is equal to the amount of interest to be paid at the end of the year plus the risk-neutral sum of the values in the next two
CHAPTER 14. INTEREST-RATE DERIVAnVES
87
nodes-all of this discounted by the risk-free rate. For example, the value at the top of the last column is 100 x .260/1.260 = 20.629; and the value at the top of the second to last column is [100x.20+ .5 x 20.629+ .5 x 15.250]/1.20 = 31.612. The final value is $3.979 million. ,, , ~Q~!'~~~c~~~~~ !
,; , 1:
0.154
T... ,, 0.260 0.138: 0.125
1 : 0.091 0,07°1 0.063
9 0.082! 0.057!
0.074 0.051
0.066: 0,046l
I , ! Floatingratevalue! 37.308 !, 39.874! , 28.214 ! 40,474 30,240: 20.838
i
25.751!
21.002
0.060 0.041
; ZO.629 31.612! 15.250 23.72)l 11.077 17.421: i 7.939 12,586! 5,634
15.019[
7.954
32.558 28.012: 22.330 15.654: 39.790! ": 35.313 29.785: , 23.338 16.124! 8.153 8.297
Figure 14.5 Determination of swap interest rate. The third lattice is constructed with a given fixed interest rate A. The values are the present values of the fixed-rate payments. For example the value at the top of the second to last column is [100A + .5 x 6.857 + .5 x 7.322] /1.2. The value of A is adjusted until the initial value of this lattice is equal to that of the one above. This value is A = 8.64%. 9. (Swaption) To find the value of the swaption, we merely put the minimum of the two value lattices (from the previous exercise) in the third column. (The case where this minimum indicates exercise of the swaption is shown in bold in Fig. 14.6.) We then value the lattice backward in the usual way. The value of the ,!
ic 25.751
1 32,5§81
28,0)2
!
1.628
jSwaptionValue
Figure 14.6 Evaluation of a swaption
CHAPTER 14. INTEREST-RATE DERIVAnVES
89
12. (Continuous zero) Since the cash flow is all at time T we have V(O) = F.[ exp (I:
-r(s)
ds )]
First find that r(s) = ro + uz(s). Hence r(s) is a normal random variable. Next we need J~ -r(s) ds = -roT- u Jl z(s) ds = -roT- u Z(T). The variable Z(T) is basically a sum of normal random variables, so it is normal. It has zero mean, since each z(s) has zero mean. We need the variance of Z(T). Let S(t) = var Z(t). Then S(t) = F.[I: z(s) ds ]2. Hence ~
= 2F.[ z(t) I: z(s) ds ]
And
[
]
dZ rt d2S(t) = 2F.[z(t)2] + 2F. dt ~ Jo z(s) ds .
Since changes in z(t) are independent of z(s) for s < t the second term on the right of the above is zero. Hence ~ dt2
=
2var
z(t)
=
2t
.
And we find S(t) = t3/3 + at + b. Since S(O) = 0 and S'(O) = 0 it follows that a = b = 0. Hence S(T) = T2/3. Since -Z(T)
is normal, with mean zero and variance T3/3 we have F.[exp(-uZ(T»]
= exp[u2T3/6].
Hence V(O) = exp [ -roT + u2T3/6 ] which agrees with the Ho-Lee example.
Chapter
1 5
Optimal
Portfolio
Growth
1. (Simple wheel strategy) At each turn, your money will either be multiplied by 3)' + (1- )') = I + 2)' or by 1- )', each with probability one-half. Hence over the long run, the factor is (1 +2)')n/2(1-)')n/2. We wish to maximize (1 +2)')(1- )') = I + )' -2)'2. This is easily found to give)' = 1/4. 2. (How to play the state lottery) (a) Suppose Victor buys one ticket. His expected logarithm is then EIn
=
10-6 In[107 + (105 -1)]
+ (1-10-6)In[105
~
10-6In107 + (1- 10-6) [In 105- 10-5]
~
10-6 In 107 + In 105- 10-6 In 105- 10-5
=
In 105 + [7In10 -5In10
<
In 105.
-1]
-10]10-6
Hence, Victor should not buy a lottery"ticket (even at the 10 to I odds in his favor!). (b) Let IXbe the fraction of tickets purchased. The optimal solution maximizes 10-6 In[ 107IX+ 105 -IX] + (1 -10-6) In[105 -IX] This implies 10-6(107107IX
Approximately,
1)
+ 105 -IX
I -10-6
+=0
105 -IX
10 -- I 107IX+ 105 -105
which has solution IX = 11/100. Victor should invest only 11 cents of his $100,000 wealth, despite the tempting odds. 3. (Easy policy) The expected logarithm is ~In(2IX + (I-IX») 91
+ ~In(I
+ (I-IX»).
92
CHAPTERIS. OYrIMAL PORTFOUOGROwrn Differentiation
and setting to zero gives 1
! 1-= IX
1+(X
1--
0.
2
This has solution (X= !. 4. (A general betting wheel) (a) If sector j occurs on the wheel, the total return is rj(Xj+l~f:l (Xi. where 1Lf=1 (Xi is the fraction of wealth not invested. So, maximizing the expected log of returns for all sectors gives the problem n max 2: pjln(rj(Xj al,...lXn .. 1=1
n + 1- 2: (XJ. t=I
(b) These are the first-order conditions of differentiating spect to (X.
the objective with re-
(c) Setting (Xi = Pi in the equation from (b) implies n Pkrk -, n -L. rkPk + 1- Li=1 Pi j=1 rjpj
Pj ,\:,n . + 1- L.i=1 Pj
But since ~f=1 Pi = 1, this requirement reduces to n ~=2:J!L Pkrk
j=1 rjpj
which holds if, as assumed, Lj=1 ~ = 1. J (d) Since rl = 1, r2 = 2, r3 = 6, we have L;=I ~ = 1. Thus by part (c) an optimal strategy is (Xi = Pi; that is, (XI = ~' (X2= ~ 1 (X3= ! .So the optimal growth rate is em = 1.06991; (see Example 15.5). 5. (More on the wheel) From the previous problem, the first-order conditions imply n ,
Pkrk
Pj
=L. f or k =1,2,...n-l rk(Xk + (Xo j=1 rj(Xj + (Xo where (Xo = 1- ~f=1 (Xi. Dividing both sides by rk and summing over k = 1, 2, ..., n1 we have
n-l 2:
Pk
k=1 rk(Xk + (Xo
=
( )( n-l 2:
..!.
i=1 rk
2: n
Pj
)
j=1 rj(Xj + (Xo
.
CHAPTERIS. OPTIMALPORTFOUOGROWTH (a) Using the fact that Lk-::;t :k = 1-
!; and lXn = 0, we can simplify
93
to get
n L rklXkPk+ lXO. ~Pnrn = k=l Thus, we have Pkrk = ~ rk lXk + lXO lXO which implies
(
lXk =
~
-J:.-
Pnrn Finally,
since
Lk=llXk
= 1 -lXO we find
lXk = Pk --orPnrn rk
f
rk
)
lXO.
lXO = Pnrn
and
k = 1, 2, ..., n -1.
(b) For the specific wheel lXl
=
1
2
pumping)
lX2
=
°
lX3
=
---=
Since Vi = Jli -!Ui2 v
=
n L WiJli -! i=l
=
.23-
-
3.3
1
6
6. (Volatility
2
---=
5
18
2
-
3.6
1
18
for each stock, we have Jli = .23. Thus, L~j WiUijWj
-2 12 [ Ui~j(n2 -n)
+ nu ii] .
n Since Ui~j = .08 and Uii = Ui2 = .16, we have v = .19- ~. 7. (Dow Jones Average puzzle) Keeping the certificates in his drawer is equivalent to following a "buy and hold" strategy. The Dow Jones Average, on the other hand, "sells" some portion of stocks when they get high (and split) and the proceeds of the "sale" are used to buy other stocks in the Average. This is a weak form of pumping, and so we expect that the DowJones Average will out-perform the buy and hold strategy (and it does). 8. (power utility) (a) We know that returns combine according to weight in the portfolio. Hence, letting B(t) be the value of a bond at time t, we have ~X(t)
-~ -w
S(t)
+ (1- w )~B(t)
=
(r + w(Jl-
r») dt + wudz.
94
CHAPTER15. OPTIMAL PORlFOUO GROWfH It follows that X(t) is lognormal with mean Tt + w(J1 -T)t standard deviation WO"Jf. In other words, X(t) (b) We have immediately U(X(t»
-!W20"2t
and
= X(0)ert+w(IJ-r)t-~w2u2t+wnu..ff. that
= !X(y»' )1
= ~e>'[rt+w(IJ-r)t-~w2u2t+wnu..ff]. )1
Hence the expected value is E [U(X(t)]
= !E[X(y»'] )1
(C) The first-order
conditions
= ~e>'[rt+w(IJ-r)t-~w2u2t]+~>'2w2u2t. )1 are dE [U(X(t) dw
] =
0.
Or equivalently 0 = E [U(X(t)
) ] [)1t(J1 -T)
-W0"2yt
+ W0"2)12t].
Hence )1t(J1- T) = wyt0"2(1
-y)
which yields w=
J1-T 0"2(1- y)
as stated. 9. (Discrete-time
log-optimal
(a) The log-optimal
pricing
portfolio
formula)
is defined by the problem
n max E[ In (1 + 2:: Ti£Xi -(1i=l The first-order
conditions
[
Ti
are -T
E T+roThis is equivalent
f
]=0
for t. = 1,2,...,n.
to E[TiPO] -TfE[Po]
Using the relation
n 2:: £Xj)Tf ) ]. j=l
= 0.
E[TiPO] = COV(Ti,PO) + riE[Po],
as -COV(Ti,PO) T.t -Tf = which is the stated result.
E[Po ]
the above can be written
CHAPTER15. OPTIMALPORTFOUOGROWfH
95
(b) We have ( .-r Jlt
f
)At = -COV(nim, 1/(1 + JloAt + nom)) E[I/(I+JloAt+nom)]
Using, in the numerator, the approximation that the denominator approaches 1, we find (Jli -rf)At
= O"i,oAt.
Hence, II. ,..t
-r
1/(1 + x) ~ 1-
f
= (T. t, o.
. x, and noting
Chapter
16
General
1. (A state
Investment
Analysis
tree)
Security
a
b
c
1
1.2
1.0
0.8
2
1.2
1.3
1.4
To find the short-term riskless asset, we note that for any portfolio with weights (XI, (X2with (XI + (X2= 1, the payoff factor for state 1 is 1.2. Hence if there is a riskless return it must be 1.2. We therefore solve the equations (XI + 1.3(X2 =
1.2
.8(XI + 1.4(X2 =
1.2
This has solution (XI = 1/3, (X2 = 2/3. Since (XI + (X2 = 1 it is a portfolio, and hence defines a riskless asset. Yes, there is an arbitrage: buy security 2 and subtract the same amount of security 1. 2. (Node separation) 1.21
1.21 1.1
1.0
1.1 A tree must be used to represent the growth of $1 because the middle node separates.
97
98
QIAPTER16. GENERAL INVESTMENT ANALYSIS 3. (Bond valuation) (a) The appropriate lattice is
100,{
The value at the top node at time 1 is VII = (.5 + .5)/1.1 = .9091. The value at the lower node is VIO = (.5 + .5)/1 = 1.0. Finally, the value at the initial node is Voo = .5 x .5(.9091 + 1.0)/1.1 = $0.8678. (b) I $1 + 4L2l I $1 + 4U I $1 + 4U I $1 = $0.8678. voo = 4L2l 4. (Optimal option valuation) In each period we have the the following maximization problem m:x { PI ~(Xu + (1- (X)Ro+ P2~(X+ (1- (X)Ro+ P3~(Xd + (1- (X)Ro} . The solution is (X = 1.01. The corresponding risk-neutral probabilities are qI = .218, q2 = .635, and q3 = .148. The option lattice is almost identical to that of the example in the text. The value of the call is found to be $5.8070. 5. (Gold correlation) We wish to find the qij'S. The gold fluctuation is modeled as a binomial lattice with u = 1.2, d = .9. The interest rate has u = 1.1 and d = .9 with risk-neutral probabilities of .5. The initial interest rate is 4%. The risk-neutral probability for gold is qUg = l+r-d d = .46667.
uTo find the risk-neutral probabilities for each of the four successornodes, we need a total of four equations,but we only have three equations so far (the two individual risk-neutral probabilities and the fact that the sum of all four probabilities is 1.) Wemust first find the real probabilities and then use the invariance theorem to get the final equation.
CHAYI'ER16. GENERALINVESTMENTANALYSIS Let S' and r' denote the S(k + 1) f S(k) and r(k + 1) fr(k),
respectively.
99
Then we
calculate £(InS']
=
.6In(I.3)
+ .4 In(0.9)
= .067
£(Inr']
=
.7In(I.I)
+ .3 In(.09)
= .035
var(InS']
=
.6(Inl.2)2
var(Inr']
=
.7(Inl.l)2
cov(InS',Inr')
-£(InS']2
= .02
+ .3(InO.9)2 -£(r']2 = .008 1 p(var(InS')var(Inr')}2 = -.005
=
We can now solve for the Pij'S (letting correspond
+ .4(InO.9)2
i = gold, j = interest
rate).
We let "1 "
to an "up" move. Pll + P12
=
.6
Pll + P21
=
.7
P21 + P22
=
.4
Pll(Inl.2)(Inl.l) +P22(InO.9)2
+
P12(Inl.2)(InO.9)
=
cov(InS',Inr')
+ P21(InO.9)(Inl.l) +£(InS')(£(Inr')
This yields Pll = .33 P12 = .27 P21 = .37 P22 = .03 We now have the following
set of equations for the qij'S qll + q12
=
qua = .46667
qll + q21
=
qur = .5
qll + q12 + q21 + q22
=
1
=
PIIP22
qllq22 q12q21 The solution
is qll
P12P21
= .1 q12 = .36 q21 = .4 q22 = .14.
6. (Complexico mine) This problem can be solved by combining the methods for solving example 16.4 and example 12.8. A set of spreadsheets is required. The answer is $14.898 (in millions). 7. (Simultaneous solution) We use a numerical Black-Scholes calculator and find S and u so that the prices of the two options match the given data. Using a time to maturity of T = .25 and interest rate of 7% we find S = $16.81 and u = 20.6%. It would be more appropriate the standard
to use a formula
for options
on futures,
rather than
option formula.
8. (Default risk) We may treat the default risk as risk-neutral risk because it is independent of the interest-rate process and because we seek the zero-level price.
100
CHAPTER16. GENERALINVESTMENTANALYSIS (a) 10
10
10
10
110
0
0
0
0
0
~~~~~
In this case p = 100 X
( .::i!:.- ) 1.1
5
5
+ L
k=l
(~
1.1
)
k
= $65.17.
(b) Evaluation can be done recursively using Vt,s = Ct,s+
1
1 -21 (.9Vt+l,s + .9Vt+l,S+1). + rt,s
This produces vo,o = $63.25. 9. (Automobile choice) We set ao = 111000. Hence ak = ao(1.05)-5 = 11(1000. (1.05)k). We know from the CE formula that for uncertain cash flows at time j, the CE at time j -1 is equal to the discounted CE at time j . For car B: CEB= -35, 000- ~
In (.5e-~12,OOO + .5e-~8,OOO)= -$27, 761.78.
For car A: We must use two stages. The certainty equivalent of the second car, as evaluated when that car is purchased is CEA2= -20,000-
~In(.5e-~lO,OOO
+ .5e-~5,OOO)= -$14,718.54.
The total certainty equivalent at year 2 is therefore either-$14, 177.19+$10, 000 = $4,177.19 or -$14, 177.19 + $5, 000 = $9,177.19 each with probability .5. Hence the overall certainty equivalent of car A is CEA = -20, 000- -a!w In ( e-a210,OOO+CEA2 + .5e-a25,OOO+CEA2 ) = $28, 132.52. Hence car B is preferred and the difference in certainty equivalence is $370.74. 10. (Continuco mine simulation) The results of one simulation are shown below. The corresponding (average) value is $22.5 million.
