SolvedProblems[1]

5.5. SOLVED PROBLEMS

81

Example 5.5.10. Let X be uniformly distributed in [0, 2π] and Y = sin(X). Calculate the p.d.f. fY of Y . Since Y = g(X), we know that fY (y) =

X

1 |g 0 (x

n )|

fX (xn )

where the sum is over all the xn such that g(xn ) = y. For each y ∈ (−1, 1), there are two values of xn in [0, 2π] such that g(xn ) = sin(xn ) = y. For those values, we find that q |g 0 (xn )| = | cos(xn )| =

1 − sin2 (xn ) =

p 1 − y2,

and fX (xn ) =

1 . 2π

Hence,

1 1 1 = p fY (y) = 2 p . 2 1 − y 2π π 1 − y2 Example 5.5.11. Let {X, Y } be independent random variables with X exponentially distributed with mean 1 and Y uniformly distributed in [0, 1]. Calculate E(max{X, Y }). Let Z = max{X, Y }. Then P (Z ≤ z) = P (X ≤ z, Y ≤ z) = P (X ≤ z)P (Y ≤ z)   z(1 − e−z ), for z ∈ [0, 1] =  1 − e−z , for z ≥ 1. Hence,

  1 − e−z + ze−z , for z ∈ [0, 1] fZ (z) =  e−z , for z ≥ 1.

82

CHAPTER 5. RANDOM VARIABLES

Accordingly, Z

Z

∞

E(Z) = 0

zfZ (z)dz =

Z

1

z(1 − e

−z

+ ze

−z

)dz +

0

∞

ze−z dz

1

To do the calculation we note that Z

1

0

Z

1

zdz = [z 2 /2]10 = 1/2, Z

ze−z dz = −

0

1

0 −1

= −e

Z

1

Z z 2 e−z dz = −

0

= −e Z

∞

1

0 −1

Z zde−z = −[ze−z ]10 + −

[e−z ]10

ze

.

Z

+ 2(1 − 2e

dz = 1 −

1

1

−1

e−z dz

0

z 2 de−z = −[z 2 e−z ]10 +

Z −z

= 1 − 2e

−1

1

) = 2 − 5e

1

0 −1

2ze−z dz

.

ze−z dz = 2e−1 .

0

Collecting the pieces, we find that E(Z) =

1 − (1 − 2e−1 ) + (2 − 5e−1 ) + 2e−1 = 3 − 5e−1 ≈ 1.16. 2

Example 5.5.12. Let {Xn , n ≥ 1} be i.i.d. with E(Xn ) = µ and var(Xn ) = σ 2 . Use Chebyshev’s inequality to get a bound on α := P (|

X1 + · · · + Xn − µ| ≥ ²). n

Chebyshev’s inequality (4.8.1) states that α≤

1 X1 + · · · + Xn 1 nvar(X1 ) σ2 var( )= 2 = 2. 2 2 ² n ² n n²

This calculation shows that the sample mean gets closer and closer to the mean: the variance of the error decreases like 1/n.


83

Example 5.5.13. Let X =D P (λ). You pick X white balls. You color the balls independently, each red with probability p and blue with probability 1 − p. Let Y be the number of red balls and Z the number of blue balls. Show that Y and Z are independent and that Y =D P (λp) and Z =D P (λ(1 − p)). We find µ ¶ m+n m P (Y = m, Z = n) = P (X = m + n) p (1 − p)n m µ ¶ λm+n m+n m λm+n (m + n)! m = p (1 − p)n = × p (1 − p)n (m + n)! m (m + n)! m!n! (λp)m −λp (λ(1 − p))n −λ(1−p) = [ e ]×[ e ], m! n! which proves the result.


6.7

95

Solved Problems

Example 6.7.1. Let (X, Y ) be a point picked uniformly in the quarter circle {(x, y) | x ≥ 0, y ≥ 0, x2 + y 2 ≤ 1}. Find E[X | Y ]. Given Y = y, X is uniformly distributed in [0, E[X | Y ] =

p 1 − y 2 ]. Hence

1p 1 − Y 2. 2

Example 6.7.2. A customer entering a store is served by clerk i with probability pi , i = 1, 2, . . . , n. The time taken by clerk i to service a customer is an exponentially distributed random variable with parameter αi . a. Find the pdf of T , the time taken to service a customer. b. Find E[T ]. c. Find V ar[T ]. Designate by X the clerk who serves the customer. P P a. fT (t) = ni=1 pi fT |X [t|i] = ni=1 pi αi e−αi t P b. E[T ] = E(E[T | X]) = E( α1X ) = ni=1 pi α1i . c. We first find E[T 2 ] = E(E[T 2 | X]) = E( α12 ) = i P P E(T 2 ) − (E(T ))2 = ni=1 pi α22 − ( ni=1 pi α1i )2 .

Pn

2 i=1 pi α2i .

Hence, var(T ) =

i

Example 6.7.3. The random variables Xi are i.i.d. and such that E[Xi ] = µ and var(Xi ) = σ 2 . Let N be a random variable independent of all the Xi s taking on nonnegative integer values. Let S = X1 + X2 + . . . + XN . a. Find E(S). b. Find var(S). a. E(S)] = E(E[S | N ]) = E(N µ) = µE(N ).

96

CHAPTER 6. CONDITIONAL EXPECTATION

b. First we calculate E(S 2 ). We find E(S 2 ) = E(E[S 2 | N ]) = E(E[(X1 + X2 + . . . + XN )2 | N ]) X 2 = E(E[X12 + · · · + XN + Xi Xj | N ]) i6=j

=

E(N E(X12 )

+ N (N − 1)E(X1 X2 )) = E(N (µ2 + σ 2 ) + N (N − 1)µ2 )

= E(N )σ 2 + E(N 2 )µ2 . Then, var(S) = E(S 2 ) − (E(S))2 = E(N )σ 2 + E(N 2 )µ2 − µ2 (E(N ))2 = E(N )σ 2 + var(N )µ2 . Example 6.7.4. Let X, Y be independent and uniform in [0, 1]. Calculate E[X 2 | X + Y ]. Given X + Y = z, the point (X, Y ) is uniformly distributed on the line {(x, y) | x ≥ 0, y ≥ 0, x + y = z}. Draw a picture to see that if z > 1, then X is uniform on [z − 1, 1] and if z < 1, then X is uniform on [0, z]. Thus, if z > 1 one has Z E[X 2 | X + Y = z] =

1

z−1

x2

1 1 x3 1 1 − (z − 1)3 dx = [ ]z−1 = . 2−z 2−z 3 3(2 − z)

Similarly, if z < 1, then Z

z

E[X 2 | X + Y = z] = 0

1 1 x3 z2 x2 dx = [ ]z0 = . z z 3 3

Example 6.7.5. Let (X, Y ) be the coordinates of a point chosen uniformly in [0, 1]2 . Calculate E[X | XY ]. This is an example where we use the straightforward approach, based on the definition. The problem is interesting because is illustrates that approach in a tractable but nontrivial example. Let Z = XY . Z E[X | Z = z] = 0

1

xf[X|Z] [x | z]dx.


97

Now, fX,Z (x, z) . fZ (z)

f[X|Z] [x | z] = Also,

fX,Z (x, z)dxdz = P (X ∈ (x, x + dx), Z ∈ (z, z + dz)) = P (X ∈ (x, x + dx))P [Z ∈ (z, z + dz) | X = x] = dxP (xY ∈ (z, z + dz)) z z dz dz = dxP (Y ∈ ( , + )) = dx 1{z ≤ x}. x x x x Hence,

  fX,Z (x, z) =

1 x,

if x ∈ [0, 1] and z ∈ [0, x]

 0, otherwise.

Consequently, Z fZ (z) =

0

Z

1

fX,Z (x, z)dx =

1

z

1 dx = −ln(z), 0 ≤ z ≤ 1. x

Finally,

f[X|Z] [x | z] = −

1 , for x ∈ [0, 1] and z ∈ [0, x], xln(z)

and Z E[X | Z = z] =

1

x(− z

1 z−1 )dx = , xln(z) ln(z)

so that E[X | XY ] =

XY − 1 . ln(XY )

Examples of values: E[X | XY = 1] = 1, E[X | XY = 0.1] = 0.39, E[X | XY ≈ 0] ≈ 0. Example 6.7.6. Let X, Y be independent and exponentially distributed with mean 1. Find E[cos(X + Y ) | X].

98


We have Z E[cos(X + Y ) | X = x] =

∞

cos(x + y)e

−y

Z dy = Re{

0

= Re{

∞

ei(x+y)−y dy}

0

eix cos(x) − sin(x) }= . 1−i 2

Example 6.7.7. Let X1 , X2 , . . . , Xn be i.i.d. U [0, 1] and Y = max{X1 , . . . , Xn }. Calculate E[X1 | Y ]. Intuition suggests, and it is not too hard to justify, that if Y = y, then X1 = y with probability 1/n, and with probability (n − 1)/n the random variable X1 is uniformly distributed in [0, y]. Hence, E[X1 | Y ] =

n−1Y n+1 1 Y + = Y. n n 2 2n

Example 6.7.8. Let X, Y, Z be independent and uniform in [0, 1]. Calculate E[(X + 2Y + Z)2 | X]. One has, E[(X + 2Y + Z)2 | X] = E[X 2 + 4Y 2 + Z 2 + 4XY + 4Y Z + 2XZ | X]. Now,

E[X 2 + 4Y 2 + Z 2 + 4XY + 4Y Z + 2XZ | X] = X 2 + 4E(Y 2 ) + E(Z 2 ) + 4XE(Y ) + 4E(Y )E(Z) + 2XE(Z) = X 2 + 4/3 + 1/3 + 2X + 1 + X = X 2 + 3X + 8/3. Example 6.7.9. Let X, Y, Z be three random variables defined on the same probability space. Prove formally that E(|X − E[X | Y ]|2 ) ≥ E(|X − E[X | Y, Z]|2 ). Let X1 = E[X | Y ] and X2 = E[X | Y, Z]. Note that E((X − X2 )(X2 − X1 )) = E(E[(X − X2 )(X2 − X1 ) | Y, Z])


99

and E[(X − X2 )(X2 − X1 ) | Y, Z] = (X2 − X1 )E[X − X2 | Y, Z] = X2 − X2 = 0. Hence, E((X −X1 )2 ) = E((X −X2 +X2 −X1 )2 ) = E((X −X2 )2 )+E((X2 −X1 )2 ) ≥ E((X −X2 )2 ). Example 6.7.10. Pick the point (X, Y ) uniformly in the triangle {(x, y) | 0 ≤ x ≤ 1 and 0 ≤ y ≤ x}. a. Calculate E[X | Y ]. b. Calculate E[Y | X]. c. Calculate E[(X − Y )2 | X]. a. Given {Y = y}, X is U [y, 1], so that E[X | Y = y] = (1 + y)/2. Hence, E[X | Y ] =

1+Y . 2

b. Given {X = x}, Y is U [0, x], so that E[Y | X = x] = x/2. Hence, E[Y | X] =

X . 2

c. Since given {X = x}, Y is U [0, x], we find Z 2

E[(X − Y ) | X = x] = 0

x

21

1 (x − y) dy = x x

E[(X − Y )2 | X] =

Z 0

x

y 2 dy =

x2 . Hence, 3

X2 . 3

Example 6.7.11. Assume that the two random variables X and Y are such that E[X | Y ] = Y and E[Y | X] = X. Show that P (X = Y ) = 1. We show that E((X − Y )2 ) = 0. This will prove that X − Y = 0 with probability one. Note that E((X − Y )2 ) = E(X 2 ) − E(XY ) + E(Y 2 ) − E(XY ).

100


Now, E(XY ) = E(E[XY | X]) = E(XE[Y | X]) = E(X 2 ). Similarly, one finds that E(XY ) = E(Y 2 ). Putting together the pieces, we get E((X − Y )2 ) = 0. Example 6.7.12. Let X, Y be independent random variables uniformly distributed in [0, 1]. Calculate E[X|X < Y ]. Drawing a unit square, we see that given {X < Y }, the pair (X, Y ) is uniformly distributed in the triangle left of the diagonal from the upper left corner to the bottom right corner of that square. Accordingly, the p.d.f. f (x) of X is given by f (x) = 2(1 − x). Hence, Z E[X|X < Y ] = 0

1

1 x × 2(1 − x)dx = . 3

108

CHAPTER 7. GAUSSIAN RANDOM VARIABLES

7.4

Summary

µ, Σ ) both in terms of We defined the Gaussian random variables N (0, 1), N (µ, σ 2 ), and N (µ their density and their characteristic function. Jointly Gaussian random variables that are uncorrelated are independent. If X, Y are jointly Gaussian, then E[X | Y ] = E(X) + cov(X, Y )var(Y )−1 (Y − E(Y )). In the vector case, X | Y ] = E(X X ) + ΣX,Y ΣY−1 (Y Y − E(Y Y ), E[X when ΣY is invertible. We also discussed the non-invertible case.

7.5

Solved Problems

Example 7.5.1. The noise voltage X in an electric circuit can be modelled as a Gaussian random variable with mean zero and variance equal to 10−8 . a. What is the probability that it exceeds 10−4 ? What is the probability that it exceeds 2 × 10−4 ? What is the probability that its value is between −2 × 10−4 and 10−4 ? b. Given that the noise value is positive, what is the probability that it exceeds 10−4 ? c. What is the expected value of |X|? Let Z = 104 X, then Z =D N (0, 1) and we can reformulate the questions in terms of Z. a. Using (7.1) we find P (Z > 1) = 0.159 and P (Z > 2) = 0.023. Indeed, P (Z > d) = P (|Z| > d)/2, by symmetry of the density. Moreover, P (−2 < Z < 1) = P (Z < 1)−P (Z ≤ −2) = 1−P (Z > 1)−P (Z > 2) = 1−0.159−0.023 = 0.818. b. We have P [Z > 1 | Z > 0] =

P (Z > 1) = 2P (Z > 1) = 0.318. P (Z > 0)


109

c. Since Z = 104 X, one has E(|Z|) = 104 E(|X|). Now, Z E(|Z|) =

Z

∞

Z

∞

|z|fZ (z)dz = 2

zfZ (z)dz = 2 −∞ 0 r Z ∞ r 2 1 2 2 . = − d[exp{− z }] = π 0 2 π

Hence,

r −4

E(|X|) = 10

∞

0

1 1 √ z exp{− z 2 }dz 2 2π

2 . π

Example 7.5.2. Let U = {Un , n ≥ 1} be a sequence of independent standard Gaussian random variables. A low-pass filter takes the sequence U and produces the output sequence Xn = Un + Un+1 . A high-pass filter produces the output sequence Yn = Un − Un+1 . a. Find the joint pdf of Xn and Xn−1 and find the joint pdf of Xn and Xn+m for m > 1. b. Find the joint pdf of Yn and Yn−1 and find the joint pdf of Yn and Yn+m for m > 1. c. Find the joint pdf of Xn and Ym . We start with some preliminary observations. First, since the Ui are independent, they are jointly Gaussian. Second, Xn and Yn are linear combinations of the Ui and thus are also jointly Gaussian. Third, the jpdf of jointly gaussian random variables Z is 1

fZ (zz ) = p

(2π)n det(C)

1 exp[− (zz − m )C −1 (zz − m )] 2

where n is the dimension of Z , m is the vector of expectations of Z , and C is the covariance Z − m )T ]. Finally, we need some basic facts matrix E[(Z   Z − m )(Z  from algebra. If C = a b d −b  , then det(C) = ad − bc and C −1 = 1  . We are now ready to det(C) c d −c a answer the questions. U. a. Express in the form X = AU  

 Xn Xn−1



=

 0 1 2

1 2 1 2

1 2



Un−1



    Un    0 Un+1

112


Then det(C) =

1 4

−

1 14

=

3 16

and  C −1 =

fXn Yn (xn , yn ) =

2 √ π 3



 Xn



16  3 − 14



U ] = 0. Then E[[Xn Yn+1 ]T ] = AE[U  U U T ]AT =  C = AE[U

− 12

1 2



− 12



Un+1 

1 0



0 1 

C −1 =  1 π

 Un



 1 2 1 2

and

fXn Yn+1 (xn , yn+1 ) =



 1 2

1 2 1 2

=

Yn+1

1 4

− 14 1 2

exp[− 43 (x2n − xn yn + yn2 )]

ii. Consider m=n+1.

Then det(C) =

 1 2

 1 2 1 2

1 2

− 12



=

 1 2

0

0

1 2



 2 0



0 2

2 )] exp[− 14 (x2n + yn+1

iii. For all other m.

  

 Xn Ym



=

1 2

1 2

0

0 − 12

0

Un

 0  Un+1   1  Um−1 2  Um

      

U ] = 0. Then E[[Xn Ym ]T ] = AE[U   U U T ]AT =  C = AE[U

Then det(C) =

1 4

1 2

1 2

0

0 − 12

0

and

0 1 2

1 π



2 )] exp[− 14 (x2n + ym



1 2 1 2

0

   0 1 0 0  0      0 0 1 0   0 −1 2   1 0 0 0 1 0 2

C −1 =  fXn Ym (xn , ym ) =

1 0 0 0

 2 0 0 2



   1    2 =   0 

 0 1 2




113

Example 7.5.3. Let X, Y, Z, V be i.i.d. N (0, 1). Calculate E[X + 2Y |3X + Z, 4Y + 2V ]. We have

 E[X + 2Y |3X + Z, 4Y + 2V ] = a Σ−1 

 3X + Z



4Y + 2V

where a = [E((X + 2Y )(3X + Z)), E((X + 2Y )(4Y + 2V ))] = [3, 8] and  Σ=

 var(3X + Z)

E((3X + Z)(4Y + 2V ))

E((3X + Z)(4Y + 2V ))

var(4Y + 2V )



=

 10

0

0

20

.

Hence,  E[X+2Y |3X+Z, 4Y +2V ] = [3, 8] 

 10−1

0

0

20−1



 3X + Z

 = 3 (3X+Z)+ 4 (4Y +2V ). 10 10 4Y + 2V

Example 7.5.4. Assume that {X, Yn , n ≥ 1} are mutually independent random variables ˆ n = E[X | X + Y1 , . . . , X + Yn ]. Find the with X = N (0, 1) and Yn = N (0, σ 2 ). Let X smallest value of n such that ˆ n | > 0.1) ≤ 5%. P (|X − X ˆ n = an (nX + Y1 + · · · + Yn ). The value of an is such that We know that X ˆ n )(X + Yj )) = 0, i.e., E((X − an (nX + Yj ))(X + Yj )) = 0, E((X − X which implies that an =

1 . n + σ2

Then ˆ n ) = var((1 − nan )X − an (Y1 + · · · + Yn )) = (1 − nan )2 + n(an )2 σ 2 var(X − X =

σ2 . n + σ2

114


2

ˆ n = N (0, σ 2 ). Accordingly, Thus we know that X − X n+σ ˆ n | > 0.1) = P (|N (0, P (|X − X q where αn =

σ2 . n+σ 2

σ2 0.1 )| > 0.1) = P (|N (0, 1)| > ) 2 n+σ αn

For this probability to be at most 5% we need 0.1 = 2, i.e., αn = αn

r

σ2 0.1 = , 2 n+σ 2

so that n = 19σ 2 . The result is intuitively pleasing: If the observations are more noisy (σ 2 large), we need more of them to estimate X. Example 7.5.5. Assume that X, Y are i.i.d. N (0, 1). Calculate E[(X + Y )4 | X − Y ]. Note that X + Y and X − Y are independent because they are jointly Gaussian and uncorrelated. Hence, E[(X +Y )4 | X −Y ] = E((X +Y )4 ) = E(X 4 +4X 3 Y +6X 2 Y 2 +4XY 3 +Y 4 ) = 3+6+3 = 12. Example 7.5.6. Let X, Y be independent N (0, 1) random variables. Show that W := X 2 + Y 2 =D Exd(1/2). That is, the sum of the squares of two i.i.d. zero-mean Gaussian random variables is exponentially distributed! We calculate the characteristic function of W . We find

Z iuW

E(e

∞

Z

∞

2 +y 2 )

eiu(x

) = −∞ −∞ 2π Z ∞

Z

eiur

= Z0 ∞ = 0

Z =

0

∞

0 2

eiur e−r

2

1 −(x2 +y2 )/2 e dxdy 2π

1 −r2 /2 e rdrdθ 2π

2 /2

rdr

1 1 2 2 d[eiur −r /2 ] = . 2iu − 1 1 − 2iu


115

On the other hand, if W =D Exd(λ), then Z E(e

iuW

) =

∞

eiux λe−λx dx

0

=

λ 1 = . λ − iu 1 − λ−1 iu

. Comparing these expressions shows that X 2 + Y 2 =D Exd(1/2) as claimed. Example 7.5.7. Let {Xn , n ≥ 0} be Gaussian N (0, 1) random variables. Assume that Yn+1 = aYn + Xn for n ≥ 0 where Y0 is a Gaussian random variable with mean zero and variance σ 2 independent of the Xn ’s and |a| < 1. a. Calculate var(Yn ) for n ≥ 0. Show that var(Yn ) → γ 2 as n → ∞ for some value γ 2 . b. Find the values of σ 2 so that the variance of Yn does not depend on n ≥ 1. a. We see that var(Yn+1 ) = var(aYn + Xn ) = a2 var(Yn ) + var(Xn ) = a2 var(Yn ) + 1. Thus, we αn := var(Yn ), one has αn+1 = a2 αn + 1 and α0 = σ 2 . Solving these equations we find var(Yn ) = αn = a2n σ 2 +

1 − a2n , for n ≥ 0. 1 − a2

Since |a| < 1, it follows that var(Yn ) → γ 2 := b. The obvious answer is σ 2 = γ 2 .

1 as n → ∞. 1 − a2

116


Example 7.5.8. Let the Xn ’s be as in Example 7.5.7. a.Calculate E[X1 + X2 + X3 | X1 + X2 , X2 + X3 , X3 + X4 ]. b. Calculate E[X1 + X2 + X3 | X1 + X2 + X3 + X4 + X5 ]. a. We know that the solution is of the form Y = a(X1 + X2 ) + b(X2 + X3 ) + c(X3 + X4 ) where the coefficients a, b, c must be such that the estimation error is orthogonal to the conditioning variables. That is, E((X1 + X2 + X3 ) − Y )(X1 + X2 )) = E((X1 + X2 + X3 ) − Y )(X2 + X3 )) = E((X1 + X2 + X3 ) − Y )(X3 + X4 )) = 0. These equalities read 2 − a − (a + b) = 2 − (a + b) − (b + c) = 1 − (b + c) − c = 0, and solving these equalities gives a = 3/4, b = 1/2, and c = 1/4. b. Here we use symmetry. For k = 1, . . . , 5, let Yk = E[Xk | X1 + X2 + X3 + X4 + X5 ]. Note that Y1 = Y2 = · · · = Y5 , by symmetry. Moreover,

Y1 +Y2 +Y3 +Y4 +Y5 = E[X1 +X2 +X3 +X4 +X5 | X1 +X2 +X3 +X4 +X5 ] = X1 +X2 +X3 +X4 +X5 . It follows that Yk = (X1 + X2 + X3 + X4 + X5 )/5 for k = 1, . . . , 5. Hence, 3 E[X1 + X2 + X3 | X1 + X2 + X3 + X4 + X5 ] = Y1 + Y2 + Y3 = (X1 + X2 + X3 + X4 + X5 ). 5 Example 7.5.9. Let the Xn ’s be as in Example 7.5.7. Find the jpdf of (X1 + 2X2 + 3X3 , 2X1 + 3X2 + X3 , 3X1 + X2 + 2X3 ).


117

These random variables are jointly Gaussian, zero mean, and with covariance matrix Σ given by



14 11 11



   Σ=  11 14 11  . 11 11 14 Indeed, Σ is the matrix of covariances. For instance, its entry (2, 3) is given by E((2X1 + 3X2 + X3 )(3X1 + X2 + 2X3 )) = 2 × 3 + 3 × 1 + 1 × 2 = 11. We conclude that the jpdf is x) = fX (x

1 (2π)3/2 |Σ|1/2

1 exp{− x T Σ−1x }. 2

We let you calculate |Σ| and Σ−1 . Example 7.5.10. Let X1 , X2 , X3 be independent N (0, 1) random variables. Calculate Y ] where E[X1 + 3X2 |Y 







 X1  1 2 3     Y =   X2   3 2 1  X3 By now, this should be familiar. The solution is Y := a(X1 + 2X2 + 3X3 ) + b(3X1 + 2X2 + X3 ) where a and b are such that 0 = E((X1 +3X2 −Y )(X1 +2X2 +3X3 )) = 7−(a+3b)−(4a+4b)−(9a+3b) = 7−14a−10b and 0 = E((X1 +3X2 −Y )(3X1 +2X2 +X3 )) = 9−(3a+9b)−(4a+4b)−(3a+b) = 9−10a−14b. Solving these equations gives a = 1/12 and b = 7/12. Example 7.5.11. Find the jpdf of (2X1 + X2 , X1 + 3X2 ) where X1 and X2 are independent N (0, 1) random variables.

118


These random variables are jointly Gaussian, zero-mean, with covariance Σ given by   5 5 . Σ= 5 10 Hence, 1 1 exp{− x T Σ−1x } 1/2 2 2π|Σ| 1 1 T −1 exp{− x Σ x } 10π 2

x) = fX (x = where



 Σ−1 =

10

1  25 −5

−5 5

.

Example 7.5.12. The random variable X is N (µ, 1). Find an approximate value of µ so that P (−0.5 ≤ X ≤ −0.1) ≈ P (1 ≤ X ≤ 2). We write X = µ + Y where Y is N (0, 1). We must find µ so that g(µ) := P (−0.5 − µ ≤ Y ≤ −0.1 − µ) − P (1 − µ ≤ Y ≤ 2 − µ) ≈ 0. We do a little search using a table of the N (0, 1) distribution or using a calculator. I find that µ ≈ 0.065. Example 7.5.13. Let X be a N (0, 1) random variable. Calculate the mean and the variance of cos(X) and sin(X). a. Mean Values. We know that 2 /2

E(eiuX ) = e−u

and eiθ = cos(θ) + i sin(θ).

Therefore, 2 /2

E(cos(uX) + i sin(uX)) = e−u

,


119

so that 2 /2

E(cos(uX)) = e−u

and E(sin(uX)) = 0.

In particular, E(cos(X)) = e−1/2 and E(sin(X)) = 0. b. Variances. We first calculate E(cos2 (X)). We find 1 1 1 E(cos2 (X)) = E( (1 + cos(2X))) = + E(cos(2X)). 2 2 2 Using the previous derivation, we find that 2 /2

E(cos(2X)) = e−2

= e−2 ,

so that E(cos2 (X)) = (1/2) + (1/2)e−2 . We conclude that var(cos(X)) = E(cos2 (X)) − (E(cos(uX)))2 =

1 1 1 1 −2 + e − (e−1/2 )2 = + e−2 − e−1 . 2 2 2 2

Similarly, we find E(sin2 (X)) = E(1 − cos2 (X)) =

1 1 −2 − e = var(sin(X)). 2 2

Example 7.5.14. Let X be a N (0, 1) random variable. Define    X, if |X| ≤ 1 Y =   −X, if |X| > 1. Find the pdf of Y . By symmetry, X is N (0, 1). Example 7.5.15. Let {X, Y, Z} be independent N (0, 1) random variables. a. Calculate E[3X + 5Y | 2X − Y, X + Z]. b. How does the expression change if X, Y, Z are i.i.d. N (1, 1)?

120


a. Let V1 = 2X − Y, V2 = X + Z and V = [V1 , V2 ]T . Then E[3X + 5Y | V ] = a Σ−1 V V where V T ) = [1, 3] a = E((3X + 5Y )V and

 ΣV = 

 5 2 2 2

.

Hence,  E[3X + 5Y | V ] = [1, 3] 

−1 5 2



2 2

  1  2 −2  V = [1, 3] V 6 −2 5

2 13 1 V = − (2X − Y ) + (X + Z). = [−4, 13]V 6 3 6 b. Now, 1 V − E(V V )) = 8 + [−4, 13](V V − [1, 2]T ) E[3X + 5Y | V ] = E(3X + 5Y ) + a Σ−1 V (V 6 13 26 2 − (2X − Y ) + (X + Z). = 6 3 6 Example 7.5.16. Let (X, Y ) be jointly Gaussian. Show that X − E[X | Y ] is Gaussian and calculate its mean and variance. We know that E[X | Y ] = E(X) +

cov(X, Y ) (Y − E(Y )). var(Y )

Consequently, X − E[X | Y ] = X − E(X) −

cov(X, Y ) (Y − E(Y )) var(Y )

and is certainly Gaussian. This difference is zero-mean. Its variance is var(X) + [

cov(X, Y ) [cov(X, Y )]2 cov(X, Y ) 2 ] var(Y ) − 2 cov(X, Y ) = var(X) − . var(Y ) var(Y ) var(Y )


19

and P : F → [0, 1] is a σ-additive set function such that P (Ω) = 1. The idea is to specify the likelihood of various outcomes (elements of Ω). If one can specify the probability of individual outcomes (e.g., when Ω is countable), then one can choose F = 2Ω , so that all sets of outcomes are events. However, this is generally not possible as the example of the uniform distribution on [0, 1] shows. (See Appendix C.)

2.6.1

Stars and Bars Method

In many problems, we use a method for counting the number of ordered groupings of identical objects. This method is called the stars and bars method. Suppose we are given identical objects we call stars. Any ordered grouping of these stars can be obtained by separating them by bars. For example, || ∗ ∗ ∗ |∗ separates four stars into four groups of sizes 0, 0, 3, and 1. Suppose we wish to separate N stars into M ordered groups. We need M − 1 bars to form M groups. The number of orderings is the number of ways of placing the N identical ¡ ¢ −1 stars and M − 1 identical bars into N + M − 1 spaces, N +M . M Creating compound objects of stars and bars is useful when there are bounds on the sizes of the groups.

2.7

Solved Problems

Example 2.7.1. Describe the probability space {Ω, F, P } that corresponds to the random experiment “picking five cards without replacement from a perfectly shuffled 52-card deck.” 1. One can choose Ω to be all the permutations of A := {1, 2, . . . , 52}. The interpretation of ω ∈ Ω is then the shuffled deck. Each permutation is equally likely, so that pω = 1/(52!) for ω ∈ Ω. When we pick the five cards, these cards are (ω1 , ω2 , . . . , ω5 ), the top 5 cards of the deck.

20

CHAPTER 2. PROBABILITY SPACE

2. One can also choose Ω to be all the subsets of A with five elements. In this case, each ¡ ¢ subset is equally likely and, since there are N := 52 5 such subsets, one defines pω = 1/N for ω ∈ Ω. 3. One can choose Ω = {ω = (ω1 , ω2 , ω3 , ω4 , ω5 ) | ωn ∈ A and ωm 6= ωn , ∀m 6= n, m, n ∈ {1, 2, . . . , 5}}. In this case, the outcome specifies the order in which we pick the cards. Since there are M := 52!/(47!) such ordered lists of five cards without replacement, we define pω = 1/M for ω ∈ Ω. As this example shows, there are multiple ways of describing a random experiment. What matters is that Ω is large enough to specify completely the outcome of the experiment. Example 2.7.2. Pick three balls without replacement from an urn with fifteen balls that are identical except that ten are red and five are blue. Specify the probability space. One possibility is to specify the color of the three balls in the order they are picked. Then Ω = {R, B}3 , F = 2Ω , P ({RRR}) =

10 9 8 5 4 3 , . . . , P ({BBB}) = . 15 14 13 15 14 13

Example 2.7.3. You flip a fair coin until you get three consecutive ‘heads’. Specify the probability space. One possible choice is Ω = {H, T }∗ , the set of finite sequences of H and T . That is, n {H, T }∗ = ∪∞ n=1 {H, T } .

This set Ω is countable, so we can choose F = 2Ω . Here, P ({ω}) = 2−n where n := length of ω. This is another example of a probability space that is bigger than necessary, but easier to specify than the smallest probability space we need.


21

Example 2.7.4. Let Ω = {0, 1, 2, . . .}. Let F be the collection of subsets of Ω that are either finite or whose complement is finite. Is F a σ-field? No, F is not closed under countable set operations. For instance, {2n} ∈ F for each n ≥ 0 because {2n} is finite. However, A := ∪∞ n=0 {2n} is not in F because both A and Ac are infinite. Example 2.7.5. In a class with 24 students, what is the probability that no two students have the same birthday? Let N = 365 and n = 24. The probability is α :=

N N −1 N −2 N −n+1 × × × ··· × . N N N N

To estimate this quantity we proceed as follows. Note that Z n N −n+k N −n+x ln( ln(α) = )≈ ln( )dx N N 1 k=1 Z 1 = N ln(y)dy = N [yln(y) − y]1a n X

a

= −(N − n + 1)ln(

N −n+1 ) − (n − 1). N

(In this derivation we defined a = (N − n + 1)/N .) With n = 24 and N = 365 we find that α ≈ 0.48. Example 2.7.6. Let A, B, C be three events. Assume that P (A) = 0.6, P (B) = 0.6, P (C) = 0.7, P (A ∩ B) = 0.3, P (A ∩ C) = 0.4, P (B ∩ C) = 0.4, and P (A ∪ B ∪ C) = 1. Find P (A ∩ B ∩ C). We know that (draw a picture) P (A ∪ B ∪ C) = P (A) + P (B) + P (C) − P (A ∩ B) − P (A ∩ C) − P (B ∩ C) + P (A ∩ B ∩ C).

22


Substituting the known values, we find 1 = 0.6 + 0.6 + 0.7 − 0.3 − 0.4 − 0.4 + P (A ∩ B ∩ C), so that P (A ∩ B ∩ C) = 0.2. Example 2.7.7. Let Ω = {1, 2, 3, 4} and let F = 2Ω be the collection of all the subsets of Ω. Give an example of a collection A of subsets of Ω and probability measures P1 and P2 such that (i). P1 (A) = P2 (A), ∀A ∈ A. (ii). The σ-field generated by A is F. (This means that F is the smallest σ-field of Ω that contains A.) (iii). P1 and P2 are not the same. Let A= {{1, 2}, {2, 4}}. Assign probabilities P1 ({1}) = 18 , P1 ({2}) = 18 , P1 ({3}) = 38 , P1 ({4}) = 38 ; and P2 ({1}) = 1 12 , P2 ({2})

=

2 12 , P2 ({3})

=

5 12 , P2 ({4})

4 12 .

=

Note that P1 and P2 are not the same, thus satisfying (iii). 1 8

+

1 8

+

2 12

=

P1 ({1, 2}) = P1 ({1}) + P1 ({2}) = P2 ({1, 2}) = P2 ({1}) + P2 ({2}) =

1 12

=

1 4

1 4

Hence P1 ({1, 2}) = P2 ({1, 2}). P1 ({2, 4}) = P1 ({2}) + P1 ({4}) =

1 8

P2 ({2, 4}) = P2 ({2}) + P2 ({4}) =

2 12

+

3 8

+

= 4 12

1 2

=

1 2

Hence P1 ({2, 4}) = P2 ({2, 4}). Thus P1 (A) = P2 (A)∀A ∈ A, thus satisfying (i). To check (ii), we only need to check that ∀k ∈ Ω, {k} can be formed by set operations on sets in A ∪ φ∪ Ω. Then any other set in F can be formed by set operations on {k}. {1} = {1, 2} ∩ {2, 4}C


23

{2} = {1, 2} ∩ {2, 4} {3} = {1, 2}C ∩ {2, 4}C {4} = {1, 2}C ∩ {2, 4}. Example 2.7.8. Choose a number randomly between 1 and 999999 inclusive, all choices being equally likely. What is the probability that the digits sum up to 23? For example, the number 7646 is between 1 and 999999 and its digits sum up to 23 (7+6+4+6=23). Numbers between 1 and 999999 inclusive have 6 digits for which each digit has a value in {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. We are interested in finding the numbers x1 +x2 +x3 +x4 +x5 +x6 = 23 where xi represents the ith digit. First consider all nonnegative xi where each digit can range from 0 to 23, the number ¡ ¢ of ways to distribute 23 amongst the xi ’s is 28 5 . But we need to restrict the digits xi < 10. So we need to subtract the number of ways to distribute 23 amongst the xi ’s when xk ≥ 10 for some k. Specifically, when xk ≥ 10 we can express it as xk = 10 + yk . For all other j 6= k write yj = xj . The number of ways to arrange 23 amongst xi when some xk ≥ 10 is the same as the number of ways to arrange ¡ ¢ P yi so that 6i=1 yi = 23 − 10 is 18 5 . There are 6 possible ways for some xk ≥ 10 so there ¡ ¢ are a total of 6 18 5 ways for some digit to be greater than or equal to 10, as we can see by using the stars and bars method (see 2.6.1). However, the above counts events multiple times. For instance, x1 = x2 = 10 is counted both when x1 ≥ 10 and when x2 ≥ 10. We need to account for these events that are counted multiple times. We can consider when two digits are greater than or equal to 10: xj ≥ 10 and xk ≥ 10 when j 6= k. Let xj = 10 + yj and xk = 10 + yk and xi = yi ∀i 6= j, k. Then the number of ways to distribute 23 amongst xi when there are 2 greater than or equal to 10 is P equivalent to the number of ways to distribute yi when 6i=1 yi = 23 − 10 − 10 = 3. There ¡¢ ¡¢ are 85 ways to distribute these yi and there are 62 ways to choose the possible two digits that are greater than or equal to 10.

24


We are interested in when the sum of xi ’s is equal to 23. So we can have at most 2 xi ’s greater than or equal to 10. So we are done. ¡ ¢ ¡18¢ ¡6¢¡8¢ Thus there are 28 − 6 5 5 + 2 5 numbers between 1 through 999999 whose digits sum up to 23. The probability that a number randomly chosen has digits that sum up to 23 is

−6(18 + 6 8 (28 5) 5 ) (2)(5) 999999

.

P Example 2.7.9. Let A1 , A2 , . . . , An , n ≥ 2 be events. Prove that P (∪ni=1 Ai ) = i P (Ai ) − P P n+1 P (A ∩ A ∩ . . . ∩ A ). 1 2 n i
= P (∪ni=1 Ai ) + P (An+1 ) − P (∪ni=1 (Ai ∩ An+1 )) X X X P (Ai ∩ Aj ∩ Ak ) − . . . = P (Ai ) − P (Ai ∩ Aj ) + i

i
+ (−1) −

X

i
X P (A1 ∩ A2 ∩ . . . ∩ An ) + P (An+1 ) − ( P (Ai ∩ An+1 )

P (Ai ∩ Aj ∩ An+1 ) +

i
X

i

P (Ai ∩ Aj ∩ Ak ∩ An+1 ) − . . .

i
+ (−1)n+1 P (A1 ∩ A2 ∩ . . . ∩ An ∩ An+1 )) X X X P (Ai ∩ Aj ∩ Ak ) − . . . = P (Ai ) − P (Ai ∩ Aj ) + i

i
+ (−1)

i
P (A1 ∩ A2 ∩ . . . ∩ An+1 )

Example 2.7.10. Let {An , n ≥ 1} be a collection of events in some probability space P {Ω, F, P }. Assume that ∞ n=1 P (An ) < ∞. Show that the probability that infinitely many of those events occur is zero. This result is known as the Borel-Cantelli Lemma. To prove this result we must write the event “infinitely many of the events An occur”

1.4

Functions of a random variable

Recall that a random variable X on a probability space (Ω, F, P ) is a function mapping Ω to the real line R , satisfying the condition {ω : X(ω) ≤ a} ∈ F for all a ∈ R. Suppose g is a function mapping R to R that is not too bizarre. Specifically, suppose for any constant c that {x : g(x) ≤ c} is a Borel subset of R. Let Y (ω) = g(X(ω)). Then Y maps Ω to R and Y is a random variable. See Figure 1.6. We write Y = g(X). X

g

Ω X(ω)

g(X(ω))

Figure 1.6: A function of a random variable as a composition of mappings. Often we’d like to compute the distribution of Y from knowledge of g and the distribution of X. In case X is a continuous random variable with known distribution, the following three step procedure works well: (1) Examine the ranges of possible values of X and Y . Sketch the function g. (2) Find the CDF of Y , using FY (c) = P {Y ≤ c} = P {g(X) ≤ c}. The idea is to express the event {g(X) ≤ c} as {X ∈ A} for some set A depending on c. (3) If FY has a piecewise continuous derivative, and if the pmf fY is desired, differentiate FY . If instead X is a discrete random variable then step 1 should be followed. After that the pmf of Y can be found from the pmf of X using X pY (y) = P {g(X) = y} = pX (x) x:g(x)=y

Example 1.4 Suppose X is a N (µ = 2, σ 2 = 3) random variable (see Section 1.6 for the definition) and Y = X 2 . Let us describe the density of Y . Note that Y = g(X) where g(x) = x2 . The support of the distribution of X is the whole real line, and the range of g over this support is R+ . Next we find the CDF, FY . Since P {Y ≥ 0} = 1, FY (c) = 0 for c < 0. For c ≥ 0, √ √ FY (c) = P {X 2 ≤ c} = P {− c ≤ X ≤ c} √ √ − c−2 X −2 c−2 = P{ √ ≤ √ ≤ √ } 3 3 3 √ √ c−2 − c−2 = Φ( √ ) − Φ( √ ) 3 3 Differentiate with respect to c, using the chain rule and the fact, Φ0 (s) = ( fY (c) =

√ c−2 2 √ 1 {exp(−[ √ ] ) 24πc 6

√1 2π

2

exp(− s2 ) to obtain

√

+ exp(−[ − √c−2 ]2 )} 6

0 9

if y ≥ 0 if y < 0

(1.7)

Example 1.5 Suppose a vehicle is traveling in a straight line at speed a, and that a random direction is selected, subtending an angle Θ from the direction of travel which is uniformly distributed over the interval [0, π]. See Figure 1.7. Then the effective speed of the vehicle in the

B Θ a Figure 1.7: Direction of travel and a random direction. random direction is B = a cos(Θ). Let us find the pdf of B. The range of a cos(Θ) as θ ranges over [0, π] is the interval [−a, a]. Therefore, FB (c) = 0 for c ≤ −a and FB (c) = 1 for c ≥ a. Let now −a < c < a. Then, because cos is monotone nonincreasing on the interval [0, π], c } a c = P {Θ ≥ cos−1 ( )} a cos−1 ( ac ) = 1− π

FB (c) = P {a cos(Θ) ≤ c} = P {cos(Θ) ≤

1

Therefore, because cos−1 (y) has derivative, −(1 − y 2 )− 2 , ( fB (c) =

√ 1 π a2 −c2

0

| c |< a | c |> a

A sketch of the density is given in Figure 1.8.

fB

−a

0

a

Figure 1.8: The pdf of the effective speed in a uniformly distributed direction.

10

Θ

0

Y

Figure 1.9: A horizontal line, a fixed point at unit distance, and a line through the point with random direction. Example 1.6 Suppose Y = tan(Θ), as illustrated in Figure 1.9, where Θ is uniformly distributed over the interval (− π2 , π2 ) . Let us find the pdf of Y . The function tan(θ) increases from −∞ to ∞ as θ ranges over the interval (− π2 , π2 ). For any real c, FY (c) = P {Y ≤ c} = P {tan(Θ) ≤ c} = P {Θ ≤ tan−1 (c)} =

tan−1 (c) + π

π 2

Differentiating the CDF with respect to c yields that Y has the Cauchy pdf: fY (c) =

1 π(1 + c2 )

−∞
Example 1.7 Given an angle θ expressed in radians, let (θ mod 2π) denote the equivalent angle in the interval [0, 2π]. Thus, (θ mod 2π) is equal to θ + 2πn, where the integer n is such that 0 ≤ θ + 2πn < 2π. Let Θ be uniformly distributed over [0, 2π], let h be a constant, and let ˜ = (Θ + h mod 2π) Θ ˜ Let us find the distribution of Θ. ˜ takes values in the interval [0, 2π], so fix c with 0 ≤ c < 2π and seek to find Clearly Θ ˜ P {Θ S ≤ c}. Let A denote the interval [h, h + 2π]. Thus, Θ + h is uniformly distributed over A. Let ˜ ≤ c if and only if Θ + h ∈ B. Therefore, B = n [2πn, 2πn + c]. Thus Θ Z 1 ˜ P {Θ ≤ c} = T 2π dθ A B T By sketching the set B, it is easy to see that A B is either a single interval of length c, or the ˜ ≤ c} = c , so that Θ ˜ is itself union of two intervals with lengths adding to c. Therefore, P {Θ 2π uniformly distributed over [0, 2π] Example 1.8 Let X be an exponentially distributed random variable with parameter λ. Let Y = bXc, which is the integer part of X, and let R = X − bXc, which is the remainder. We shall describe the distributions of Y and R. 11

Proposition 1.10.1 Under the above assumptions, Y is a continuous type random vector and for y in the range of g: ∂x fX (x) fY (y) = ∂y = fX (x) (y) ∂y | (x) | ∂x

Example 1.10

Let U , V have the joint pdf: u + v 0 ≤ u, v ≤ 1 fU V (u, v) = 0 else

and let X = U 2 and Y = U (1 + V ). Let’s find the pdf fXY . The vector (U, V ) in the u − v plane is transformed into the vector (X, Y ) in the x − y plane under a mapping g that maps u, v to x = u2 and y = u(1 + v). The image in the x − y plane of the square [0, 1]2 in the u − v plane is the set A given by A = {(x, y) : 0 ≤ x ≤ 1, and

√

√ x ≤ y ≤ 2 x}

See Figure 1.12 The mapping from the square is one to one, for if (x, y) ∈ A then (u, v) can be y 2

v 1

1

u

x 1

Figure 1.12: Transformation from the u − v plane to the x − y plane. recovered by u =

√

x and v =

√y x

− 1. The Jacobian determinant is

∂x ∂u ∂y ∂u

∂x ∂v ∂y ∂v

0 2 = 2u 1 + v u = 2u

Therefore, using the transformation formula and expressing u and V in terms of x and y yields ( √ √y x+( x −1) if (x, y) ∈ A 2x fXY (x, y) = 0 else Example 1.11 Let U and V be independent continuous type random variables. Let X = U + V and Y = V . Let us find the joint density of X, Y and the marginal density of X. The mapping u x u+v g: → = v y v 24

is invertible, with inverse given by u = x − y and v = y. The absolute value of the Jacobian determinant is given by ∂x ∂x 1 1 ∂v ∂u ∂y ∂y = 0 1 = 1 ∂u ∂u Therefore fXY (x, y) = fU V (u, v) = fU (x − y)fV (y) The marginal density of X is given by Z ∞ Z fX (x) = fXY (x, y)dy = −∞

∞

fU (x − y)fV (y)dy

−∞

That is fX = fU ∗ fV .

Example 1.12 Let X1 and X2 be independent N (0, σ 2 ) random variables, and let X = (X1 , X2 )T denote the two-dimensional random vector with coordinates X1 and X2 . Any point of x ∈ R2 can 1 be represented in polar coordinates by the vector (r, θ)T such that r = kxk = (x21 + x22 ) 2 and θ = tan−1 ( xx21 ) with values r ≥ 0 and 0 ≤ θ < 2π. The inverse of this mapping is given by x1 = r cos(θ) x2 = r sin(θ) We endeavor to find the pdf of the random vector (R, Θ)T , the polar coordinates of X. The pdf of X is given by fX (x) = fX1 (x1 )fX2 (x2 ) =

1 − r22 e 2σ 2πσ 2

The range of the mapping is the set r > 0 and 0 < θ ≤ 2π. On the range, ∂x ∂x 1 ∂x1 cos(θ) −r sin(θ) ∂θ = r = ∂r ∂( ) ∂x2 ∂x2 sin(θ) r cos(θ) = r ∂r ∂θ θ Therefore for (r, θ)T in the range of the mapping, ∂x r − r22 fR,Θ (r, θ) = fX (x) r = e 2σ ∂( θ ) 2πσ 2 Of course fR,Θ (r, θ) = 0 off the range of the mapping. The joint density factors into a function of r and a function of θ, so R and Θ are independent. Moreover, R has the Rayleigh density with parameter σ 2 , and Θ is uniformly distributed on [0, 2π].

25

ELEG–636 Homework #1, Spring 2003

1. Show that if $%'& $*,+.-/ , ()" then !#"

and

Answer:

02134

? In the same way

)

Then

2.J @Express b $% the density a .

5687079%:;1< # 5687079%:;1<>=# 56

213 ()"

# (#"

Y X

)" (

@ ? IA2JLBDK CFE.G.HE I- JLAOK NCFE.G B CME.G N CFE.G ?QP%J R IA JLB K CFE.G.HE 1< P%J R I- JLA2K NCME.G 13 R R B CFE.G N CFE.G ?QP J R H IK JLB K CME.GSHE P J R IH JLK K NTCFE.G R R B CFE.G N CME.G ?VUSW & 8#" L$*,+* U.W & (#" X,+ ? 8#" L$%'& 8#" @$%,+.Z\[*]

$%

of the RV

in terms of

if (a)

Answer: (a)

"

@Xc

5d#eX 5df[^]8eX 5d g@eX

h'i

h i

X 7 d i 56 Xde0e9XQXdj i

560e9X

Xc

h i

X " 1 "

560e

)"

X X 1

X_7 i XVX_j i

X_7 i XVX_j i

[^$*_` $

; (b)

[^$*_


X_7 i h i K L J K J Hk B C G B C Gml X_j i H X_7 i h i K K J H BD C G H BTC G X_j i H H Xd7 hni i

Xo:p XQXdj i

(b)

"

@X`

56)eX 56f[^](eX 56 a Jq b ]reX

h i

X_7 i J q 56 a e XVX_j i X_7 hni 56j X s _ X j U.W X_7 hni USW XsX_j (#"

@X X

X` "

h i I JLK Jrtvu Hk BTC H h i Jgtvu A B C C G.G

Answer: (1) Let ~

':

and

and

X

i i

C fG Gml Xd7 i Xdj

i

i

g Since

i

Xd7 i Xdj

3. The RVs and densities L J y J{ 3b exponential

$%xw a b are $*>independent =z @Xx a with X _: q Find the densities of the following RVs: | ;}

i

w

J b a

are independent, we have

Y/<

<]8 < R 8213 J R w a J{ C J G a J 13 w b J a J{ Y w a 2


(2)

"

<

5d

R J R R J R R

ex56e/ ~

56eY8 0X X213X

$*213$* 1
! w:; 1

1@

w ] b < w:]2 Z 4. The RVs and are i and independent. Show that, if ~ = ¥ Y¦ o¢<£ ¤ ~ ¡ ~ ¡

Y/<

"

Answer: § 0 Q Let ~ , and . Since and are Gaussian, so

g by finding the mean and variance of .

& 6+¨

& & *+

©

So,

i

& + ~ + &ª & +@: & +

r

%+ & %+

£ ¤

& *+ & %+ J « ® a ¬

Thus,

+ & ¨ ~

R 213 J R Ja « ® 13 ¯ R £ ¤ £ ¤ 3

r

, then

is also Gaussian. We can find


& + ~ is already obtained, which is

& +* ~

5. Use the moment generating function, show that the linear transformation of a Gaussian random vector is also Gaussian. Proof: Let be a °± real random Gaussian vector, then the density function is

J(¶ qJ·¹¸ GSº» ¸3¼ ¶ C q
]

Let ½ be a °± real vector, then the moment generating function of

¾ ` ½

J ¶ qJ· ¸ G º » ¸ ¼ ¶ C q

Let À

is

& a¿ º q + a ¿º q

be a linear transform of

0 Á

Then

The moment generating function of

À

´ q À6Â À

is

¾ ½

&a ¿ º } & a ¿ º Ã & a C Ãº +

q + ¿Gº q +

Using the moment generating function of , we have

¾ ½

a · ¸ º C Ãº ¿ G ¶ C Ãº ¿ G º » ¸ C Ãº ¿ G a ·¹Ä º ¿ ¶ ¿ º» Ä ¿

¾

½ . which has the same form of So, is also Gaussian. w $ 6. Let - ° ¡Å-ÇÆ I be four IID random variables with exponential distribution with = 1. X - ÈÉ °

Æ I

É $ > = e_eÊ °

X

(a) Determine and plot the pdf of ° XËY (b) Determine and plot the pdf of X ° (c) Determine and plot the pdf of Å ° X (d) Compare the pdf of ° with that of the Gaussian density.

Å

4


$% a J b $*

Answer: Let The characteristic function of

$%

Ì

is

fÍr

Í ÏÎ (

$ >=ÐÐÐT=$ - Since I ° ° are i.i.d.,

z Ñ ² X X]!ÐÐÐ/r X C G Evaluating both sides by the characteristic functions, we have Ì

Ñ ² fÍr & >a ÒÓ C G So,

Ì

Ì

Ñ ² Õ ² C G +* É Ô C G Æ I

zÑ ² fÍr\Ö ÍØ× C G ÏÎ ! X whose inverse Fourier transform yields the pdf of - ° X - J%I a J b

Ñ ² X X >Ù C G 9 6@=zÚ@=Ê ¦ This expression holds for any positive integer , including 7. The mean and covariance of a Gaussian random vector

are given by, respectively,

ß Û ÜÞÝ à á

and

ÜÞÝ I

I

ß

Plot the 1 , 2 , and 3 concentration ellipses representing the contours of the density function in $ =$^3 å $ 7ç the plane. âäã,° : The radius of an ellipse with major axis a (along ) and minor axis æ $* (along ) is given by è ç

ç å

æ

ëÇì å Çê e ê eí çî ¤ . Compute the 1 ellipse specified by £ ï I and æ £ ï and then rotate and where i C '& $ CFI ðSG $ CFðSG + mC ñMG óò q ñ : Û q translate each point ã using the transformation . Answer: ô Ë J%I Ý ÅË ß Ë ÅË ãé°

Çê :

5

æ


So,

ô

q ]c Let

I ³ ¤ Ú £ a J ÷ kC Ê ¤

[^$ I = $ '$ I 9 The linear transform

is a rotation of

Ê<ú/û

$ I $ ß Ý

a J ¶ C qJõ BG º ö B ¼ ¶ C qJõ BG J I G J C ¶ % J I G C J G C J G l ¶% $ I Ç$

Ýäù ù

ù

ß

Ý

3

I ß

of the original axes.

I : Ë

[^$ I = $ So,

ç So, the radius of the ellipse is

è

The concentration ellipse of (

or

ù

3ø:$

$ I 9

è

æ

Ú

ç æ þÇÿ3üz ê ç %üý W ê : % è æ ) is thus

I : Ë $ I Ç$ 9

3ø:$

3

[*$ I = $ When the function is chosen differently, the figure will be different. But the orientation of the ellipses are the same.

6


x2

3 2 1

2

x1

1

7

ELEG–636 Test #1, March 25, 1999

NAME:

1. (35 pts) Let y = minfjx1 j; x2 g where x1 and x2 are i.i.d. inputs with cdf and pdf Fx () and fx (), respectively. For simplicity, assume fx () is symmetric about 0, i.e., fx (x) = fx (,x). Determine the cdf and pdf of y in terms of the distribution of the inputs. Plot the pdf of y for fx () uniform on [,1; 1]. Note that Fx (x) , Fx (,x) for x 0 Fjxj (x) = 0 otherwise

(

Also

Fminfx1 ;x2 g (x)

Thus,

=

(

(

= 1 , P fx1 xgP fx2 xg = 1 , (1 , Fx (x))(1 , Fx (x)) 1

2

= 1 , (1 , Fjx j (y))(1 , Fx (y)) 1 , (1 , Fx (y) + Fx (,y))(1 , Fx (y)) for y 0 1 , (1 , Fx (y)) otherwise Fy (y )

2

1

2

2Fx (y) , Fx (,y) , Fx2 (y) + Fx (y)Fx (,y) for y 0 otherwise Fx (y ) If fx () is symmetric about 0, then fx (x) = fx (,x) and Fx (x) = 1 , Fx (,x), giving =

Fy (y ) =

(

2Fx (y) , (1 , Fx (y)) , Fx2 (y) + Fx (y)(1 , Fx (y)) Fx (y ) =

(

Taking the derivative, fy (y )

=

=

4Fx (y) , 2Fx2 (y) , 1 Fx (y )

(

(

for y 0 otherwise

4fx (y) , 4fx (y)Fx (y) fx (y )

4fx(y)(1 , Fx (y)) fx (y )

1

for y 0 otherwise

for y 0 otherwise

for y 0 otherwise

ELEG–636 Test #1, March 25, 1999

NAME:

2. (35 pts) Consider the observed samples yi

= + xi

for i = 1; 2; : : : ; N . We wish to estimate the location parameter using a maximum likelihood estimator operating on the observations y1 ; y2 ; : : : ; yN . Consider two cases:

N (0; 2 ), for i = 1; 2; : : : ; N . (10 pts) The xi terms are independent with distribution xi N (0; i2 ), for i = 1; 2; : : : ; N . (10 pts) The xi terms are i.i.d. with distribution xi

(15 pts) Are the estimates unbiased? What is the variance of the estimates? Are they consistent?

j =

fyj (y )

N Y p1

2

Thus, M L

and taking the derivative,

= arg max ,

N X (yi , M L ) i=1

For the case of changing variances, N X (yi , M L) i=1

1 , (yi2,2) = 22 22 i=1 e

i2

2

N=2 P ,

,

N (yi )2 i=1 2 2

e

N X (yi , )2 i=1

22

N X = 0 ) M L = N1 yi i=1

PN y PN w y i=1 i i = 0 ) M L = PN 1 M L = Pi=1 N w i

2

i

i=1 2 i

i=1

i

which is a normalized filter, where wi = 12 for i = 1; 2; : : : ; N . For each estimate E fM L g = , and they are thus unbiased.

8 PN 8 PN !2 9 !2 9 = < = < w x w y , w i i i i i i=1 i=1 = E var(M L )[N ] = E f(M L , )2 g = E P P N ; : Ni=1 wi ; : i=1 wi P PN w x x w g PN w2 2 PN w Ef N 1 i=1 j =1 i i j j i=1 i i i=1 i = = = = P P P P N N N N ( w )2 ( w )2 ( w )2 w i=1

i

i=1

i

i=1

i

i=1

i

Since wi > 0, we have var(M L )[N + 1] < var(M L )[N ]. This, combined with the fact that the estimator is unbiased means the estimate is consistent.

3

ELEG{636 Test #1, March 23, 2000

NAME:

1. (30 pts) The random variables x and y are independent and uniformly distributed on p the interval [0,1]. Determine the conditional distribution frjA(rjA) where r = x2 + y2 and A = fr 1g. Answer: Examine the joint density fx;y (x; y) in the xy plane. Since x and y are independent,

fx;y (x; y) = fx(x)fy (y) = 1 for 0 x; y 1 This de nes a uniform density over the region 0 x; y 1 in the rst quadrant of the xy plane. p Note that r = x2 + y2 de nes an arc in the rst quadrant. Also, if 0 r 1 the area under the uniform density up to radius r is simply given by

Z q 2 2 Fr (r) = P r[ x + y r] = p 2 2 fx;y (x; y)dxdy x +y r Z 2 r for 0 r 1 = p 1dxdy = 2 2 x +y

4

r

Then for A = fr 1g.

FrjA (rjA) =

Fr;A (r; A) Fr (r) r2 4 = = = r2 for 0 r 1 P r[A] Fr (1) 4

Thus, frjA(rjA) = 2r for 0 r 1 and 0 elsewhere.

1

ELEG{636 Test #1, April 5, 2001

NAME:

1. (35 pts) Probability questions:

(10 pts) Let x be a random variable and set y = x2 . Derive a simpli ed expression for f (y jx 0).

(15 pts) Suppose now that y = a sin(x + ), where and a > 0 are constants. Determine f (y ). y

(10 pts) Suppose further that x is uniformly distributed over [ ; ]. Determine f (y) for this special case. y

Answer: Clearly, F (yjx 0) = 0 for y < 0. Then for y 0,

p j 0) = P r(PY r(Xy;X0) 0) = F (1 y)F

F (y x

Fx (0)

x

x

Thus

j 0) = 2p

f (y x

(0)

U (y ) :

p

fx ( y ) U (y ): y(1 Fx (0))

Now for y = g(x) = a sin(x + ) we have, assuming jyj a, in nitely many solutions xn n = 0;

1; 2; : : :. Also,

Note that

g 2 (x

n

)+g

0

2 (x

= arcsin(y=a)

g 0 (xn ) n

)=

a2 cos2 (x g 0 (xn )

Thus fy (y ) =

=

n

= a cos(x + ) n

+ ) + a2 sin2 (x

n

q

g 2 (xn ) =

a2

X f (x ) 1 =p 2 g (x ) a x 0

i

a2

X

n

n

q

y2

+ ) = a2 . Or, y2 :

fx (xn );

i

If x U ( ; ) then there is only a single solution, and fy (y ) =

p

1

2 a2

1

y2

;

jyj a

jyj a

ELEG–636 Test #1, April 14, 2003

NAME:

1. (30 pts) Probability questions:

(15 pts) Let x be a random variables with density fx (x) given below. Let y the shown function. Determine fy (y ) and Fy (y ).

= g (x)

be

(15 pts) Let x and y be independent, zero mean, unit variance Gaussian random variables. Define w = x2 + y 2 and z = x2 : Determine fw;z (w; z ). Are w and z independent?

Answer: Note that

(

fx (x) =

Thus Fx (x) =

p Since x = y for 0 y 1,

1 x 4

+

1 (x 2

0

8 > <

0

1 2 x 8 > :

8 > <

Fy (y ) =

> :

+

1

p

1

x<2

otherwise x<0

, 0:5)

0

x<2 2x

y<1 1y

py , 0:5)

y<0

0 +

1 u( 2

y<1 1y

0

1

8 > <

1 y 8 > :

0

y<0

0

1 y 8 > :

=

1 u(x 2

Fx ( y ) 0

8 > <

=

, 0:5)

0 +

1 u(y 2

1

, 0:25)

y<0

0

y<1 1y

Taking the derivitive yields (

fy (y ) =

1 8

+

1 (y 2

, 0:25) + 0

1

3 (y 8

, 1)

0

y1

otherwise

ELEG–636 Test #1, April 14, 2003

NAME:

Tha Jabobian of the transformation is

d(x2 +y2 ) d(x2 +y2 ) dx dy d(x2 ) d(x2 ) dx dy

The reverse transformation is easily seen to be x w z . Thus,

=

J (x; y ) =

fw;z (w; z ) =

fx;y (x; y )

j j

4 xy

+

j j

4 xy

x= y=

2x

2y

2x

0

j j

= 4 xy

p pz and y = w , x

2

p x= z p y = w,z

fx;y (x; y )

=

fx;y (x; y )

j j

+

pz , pw , z

4 xy

fx;y (x; y )

j j

4 xy

pw , z ,

pz ,pw , z p x=, z p y =, w,z

x= y=

+

=

(1) Since x and y are independent, fx;y (x; y ) =

1 2

e

,(x2 +y2 ) 2

Thus

1 p p e,w= u(w )u(z )u(w , z ) 2 z w , z where the last three terms indicate w; z 0 and w z . 2

fw;z (w; z ) =

2

ELEG–636 Midterm, April 7, 2009

NAME:

1. [30 pts] Probability: (a) [15 pts] Prove the Bienayme inequality, which is a generalization of the Tchebycheff inequality, E{|X − a|n } P r{|X − a| ≥ } ≤ n for arbitrary a and distribution of X. (b) [15 pts] Consider the uniform distribution over [−1, 1]. i. [10 pts] Determine the moment generating function for this distribution. ii. [5 pts] Use the moment generating function to generate a simple expression for 0 the k th moment, mk . Answer: (a) n

Z

∞

Z

n

|x − a| fx (x)dx ≥

E{|x − a| } = −∞

n

Z

|x − a| fx (x)dx ≥ x−a|≥

n fx (x)dx

x−a|≥

E{|X − a|n } =n P r{|x − a| ≥ } ⇒ P r{|X − a| ≥ } ≤ n (b) 1

1 s −s 2s (e − e ) s 6= 0 e dx = 1 s=0 −1 k d Φ(s) ⇒ E{xk } = dk s s=0 dΦ(s) 1 s 1 s −s −s E{x} = = (e + e ) − (e − e ) ds s=0 2s 2s2 s=0 1 s 1 s −s −s =0 = (e − e ) − (e − e ) 2 4 s=0

1 Φ(s) = 2

Z

sx

Repeat the differentiation, limit (l’Hpital’s rule) process. The analytical solution is simpler: Z 1 1 k 1 − (−1)k+1 0 k = 1, 3, 5, . . . k E{x } = x dx = = 1 k = 0, 2, 4, . . . 2 −1 2(k + 1) k+1

1

ELEG–636 Midterm, April 7, 2009

NAME:

2 ) 3. [35 pts] Let Z = X +N , where X and N are independent with distributions N ∼ N (0, σN 1 1 and fX (x) = 2 δ(x − 2) + 2 δ(x + 2).

(a) [15 pts] Determine the MAP, MS, MAE, and ML estimates for X in terms of Z. (b) [10 pts] Determine the bias of each estimate, i.e., determine whether or not each estimate is biased. (c) [10 pts] Determine the variances of the estimates. Answer: 2 ) + 1 N (2, σ 2 ). (a) Since X and N are independent, fZ (z) = fX (z) ∗ fN (z) = 12 N (−2, σN N 2 Also 2 fZ|X (z|x) =N (x, σN )

x ˆM L = arg max fZ|X (z|x) = z x

2 )(δ(x − 2) + δ(x + 2)) fZ|X (z|x)fX (x) N (x, σN = fX|Z (x|z) = fZ (z) 2fZ (z) 2 z>0 x ˆM AP = arg max fX|Z (x|z) = x −2 z < 0 Z ∞ Z ∞ 1 x ˆM S = xfX|Z (x|z)dx = xf (z|x)fX (x)dx fZ (z) −∞ Z|X −∞ 2 2 )| 2N (2, σN x=z − 2N (−2, σN )|x=z = 2fZ (z) 2 2 )| N (2, σN )|x=z − N (−2, σN x=z =2 2 )| 2 )| + N (−2, σ N (2, σN x=z N x=z Z xˆM AE Z xˆM AE 1 1 = fX|Z (x|z)dx = fZ|X (z|x)fX (x)dx 2 fZ (z) −∞ −∞ Z xˆM AE 1 2 2 fZ|X (z|x)fX (x)dx = N (2, σN ⇒ )|x=z + N (−2, σN )|x=z 4 −∞ Z xˆM AE 1 2 2 2 N (x, σN )(δ(x − 2)+δ(x + 2))dx = N (2, σN )|x=z + N (−2, σN )|x=z ⇒ 2 −∞

Note the LHS is not continuous ⇒ x ˆM AE not well defined. (b) Note fZ (z) is symmetric about 0 ⇒ E{ˆ xM L } = E{z} = 0 ⇒ x ˆM L is unbiased (E{x} = 0). Similarly, E{ˆ xM AP } = 2P r{z > 0} − 2P r{z < 0} = 0 ⇒ x ˆM AP is unbiased. Also, x ˆM S is an odd function (about 0) of z ⇒ E{ˆ xM S } = 0 ⇒ x ˆM S is unbiased. 2 2 2 2 2 2 (c) σM ˆM AP = ±2). Determining L = σZ = σX + σN = 4 + σN . Also, σM AP = 4 (since x 2 σM S is not trivial, and will not be considered.

3


1. A token is placed at the origin on a piece of graph paper. A coin biased to heads is given, P (H) = 2/3. If the result of a toss is heads, the token is moved one unit to the right, and if it is a tail the token is moved one unit to the left. Repeating this 1200 times, what is a probability that the token is on a unit N , where 350 ≤ N ≤ 450? Simulate the system and plot the histogram using 10,000 realizations. Solution: Let x = # of heads. Then 350 ≤ x − (1200 − x) ≤ 450 ⇒ 775 ≤ x ≤ 825 and i 1200−i 825 X 2 1 1200 P r(775x ≤ 825) = i 3 3 i=775

which can be approximated using the DeMoivre–Laplace approximation ! ! i2 X i − np i − np n 1 2 −Φ p (p)i (1 − p)n−i ≈ Φ p i np(1 − p) np(1 − p) i=i1

where Φ(x) =

Rx

1 −x2 /2 dx −∞ 2π e

2. Random variable X is characterized by cdf FX (x) = (1 − e−x )U (x) and event C is defined by C = {0.5 < X ≤ 1}. Determine and plot FX (x|C) and fX (x|C). Solution: Evaluating P r(X ≤ x, 0.5 < X ≤ 1) for the allowable three cases x < 0.5

P r(X ≤ x, 0.5 < X ≤ 1) = 0

0.5 ≤ x ≤ 1

P r(X ≤ x, 0.5 < X ≤ 1) = FX (x) − FX (0.5) = e−0.5 − e−x

x>1

P r(X ≤ x, 0.5 < X ≤ 1) = FX (1) − FX (0.5) = e−0.5 − e−1 = 0.2386

Also, P r(C) = FX (1) − FX (0.5) = e−0.5 − e−1 = 0.2386. Thus  0 x < 0.5 P r(X ≤ x, 0.5 < X ≤ 1)  −0.5 −x (e − e )/0.2386 0.5 ≤ x ≤ 1 fX (x|C) = =  P r(0.5 < X ≤ 1) 1 x>1 3. Prove that the characteristic function for the univariate Gaussian distribution, N (η, σ 2 ), is ω2σ2 φ(ω) = exp jωη − 2 Next determine the moment generating function and determine the first four moments.

1


Solution: ∞

1 (x − η)2 √ φ(ω) = ejωx dx exp 2 2σ 2πσ −∞ 2 Z ∞ 1 (x − 2ηx + η 2 − 2jωxσ 2 ) √ = exp dx 2σ 2 2πσ −∞ Z ∞ 1 (−η 2 + (η 2 + jωσ 2 η)2 (x − (ηx + jωσ 2 )2 √ exp dx = exp 2σ 2 2σ 2 2πσ −∞ Z ∞ (−η 2 + (η 2 + jωσ 2 η)2 1 (x − (ηx + jωσ 2 )2 √ = exp exp dx 2σ 2 2σ 2 2πσ −∞ (−η 2 + (η 2 + jωσ 2 η)2 = exp 2σ 2 2 2 which reduces to φ(ω) = exp jωη − ω 2σ . The moment generating function is simple Z

s2 σ 2 Φ(s) = exp sη + 2 and mk =

dk Φ(s) | , dk s s=0

which yields

m1 = η

m2 = σ 2 + η 2

m3 = 3ησ 2 + η 3

m4 = 3σ 4 + 6σ 2 η 2 + η 4

4. Let Y = X 2 . Determine fY (y) for: (a) fX (x) = 0.5 exp{−|x|} (b) fX (x) = exp{−|x|}U (X) √ Solution: Y = X 2 ⇒ X = ± y and dY /dX = 2X. Thus fX (x) fX (x) + fY (y) = |2x| x=√y |2x| x=−√y Substituting and simplifying fX (x) = 0.5 exp{−|x|}

⇒

fX (x) = exp{−|x|}U (x)

⇒

√ 1 fY (y) = √ e− y U (y) 2 y √ 1 fY (y) = √ e− y U (y) 2 y

5. Given the joint pdf fXY (x, y) fXY (x, y) =

8xy, 0 < y < 1, 0 < x < y 0, otherwise

Determine (a) fx (x), (b) fY (y), (c) fY (y|x), and (d) E[Y |x]. Solution: 2


(a) fX (x) =

R∞

(b) fY (y) =

R∞

=

R1

−∞ fXY (x, y)dx =

Ry

−∞ fXY (x, y)dy

(c) fY (y|x) = (d) E[Y |x] =

fXY (x,y) fX (x)

=

x

−∞ yfY (y|x)dy =

4y 3 0 < y < 1 0 otherwise

x
0

R∞

4x − 4x3 0 < x < 1 0 otherwise

8xydy = 8xydx =

o

2y 1−x2

2y 2 x 1−x2 dy

R1

=

2 3

1−x3 1−x2

=

2 3

1+x+x2 1+x

6. Let W and Z be RVs defined by W = X2 + Y 2

and

Z = X2

where X and Y are independent; X, Y ∼ N (0, 1). (a) Determine the joint pdf fW Z (w, z). (b) Are W and Z independent? Solution: Given the system of equations 2x 2y w z = 4|xy| J = x y 2x 0 Note we must have w, z ≥ 0 and w ≥ z. Thus the inverse system (roots) are √ √ y = ± w − z. x = ± z, Thus fXY (x, y) √ fW Z (w, z) = 4|xy| x = ±√ z y =± w−z

(∗)

Note also that, since X, Y ∼ N (0, 1), fXY (x, y) =

1 − x2 +y2 2 e 2π

(∗∗)

Substituting (∗∗) into (∗) [which has four terms] and simplifying yields fW Z (w, z) =

ew/2 p U (w − z)U (z) 2π z(w − z)

(∗ ∗ ∗)

Note W and Z are not independent. Counter example proof: Suppose W and Z are independent. Then fW (w)fZ (z) > 0 for all w, z > 0. But this violates (∗ ∗ ∗), as fW Z (w, z) > 0 only for w ≥ z.

3


1. Let

R=

2 −2 −2 5

Express R as R = QΩQH , where Ω is diagonal. Solution: 2 − λ −2 −2 5 − λ

= λ2 − 7λ + 6 = 0

Than solving Rqi = λi qi gives q1 = where

√1 [1, −2]T 5

⇒

and q2 =

Q = [q1 , q2 ]

λ1 = 6, λ2 = 1

and

Ω=

√1 [2, 1]T . 5

6 0 0 1

Thus R = QΩQH

2. The two-dimensional covariance matrix can be expressed as: σ12 ρσ1 σ2 C= ρ∗ σ1 σ2 σ22 (a) Find the simplest expression for the eigenvalues of C. (b) Specialize the results to the case σ 2 = σ22 = σ22 . (c) What are the eigenvectors in the special case (b) when ρ is real? Solution: (a) 2 σ1 − λ ρσ1 σ2 ρ∗ σ1 σ2 σ 2 − λ 2

= λ2 − (σ12 + σ22 )λ + (1 − |p|2 )σ12 σ22 = 0 p (σ12 + σ22 ) ± σ14 + σ24 − 2σ12 σ22 + 4|p|2 σ12 σ22 ⇒λ= 2

(b) For σ 2 = σ22 = σ22 λ=

2σ 2 ±

p 4|p|2 σ 4 = (1 ± |p|)σ 2 2

3. Let x[n] = Aejω0 n where the complex amplitude A is a RV with random magnitude and phase A = |A|ejφ . Show that a sufficient condition for the random process to be stationary is that the amplitude and phase are independent and that the phase is uniformly distributed over [−π, π]. Solution: First note E{x[n]} = E{A}ejω0 n and E{A} = E{|A|}E{ejφ } = 0 1


by independence and uniform distribution of φ. Thus it has a fixed mean. Next note E{x[n]x∗ [n − k]} = E{|A|2 }ejω0 k which is strictly a function of k ⇒ WSS. 4. Let Xi be i.i.d. RVs uniformly distributed on [0, 1] and define Y =

20 X

Xi .

i=1

Utilize Tchebycheff’s inequality to determine a bound for P r{8 < Y < 12}. Solution: Note ηx = inequality

1 2

and σx2 =

P r{|Y − ηy | ≥ 2} ≤

1 12 .

σ 2 y

2

=

Thus ηy = 10 and σy2 = 5 12

⇒

20 12

= 53 . Utilize Tchebycheff’s

P r{8 < Y < 12} ≥ 1 −

5 7 = 12 12

5. Let X ∼ N (0, 2σ 2 ) and Y ∼ N (1, σ 2 ) be independent RVs. Also, define Z = XY . Find the Bays estimate of X from observation Z: (a) Using the squared error criteria. (b) Using the absolute error criteria. 6. Let X and Y be independent RVs characterized by fX (x) = ae−ax U (x) and fY (y) = ae−ay U (y). Also, define Z = XY . Find the Bays estimate of X from observation Z using the uniform cost function. Solution: Fz|x (z|x) = P r(xy ≤ z|x) = P r(y ≤ z/x) = Fy (z/x)

⇒

fz|x (z|x) =

1 fy (z/x) x

1 x ˆ = arg max fz|x (z|x)fx (x) = arg max fy (z/x)fx (x) x 1 −az/x −ax −1 = arg max ae ae U (x)U (z) = arg max a2 x−1 e−a(zx +x) U (x)U (z) x −1 −1 ⇒ 0 = − a2 x−2 e−a(zx +x) + (a2 x−1 e−a(zx +x) )(−a(1 − zx−2 )) 0 = − x−1 − a(1 − zx−2 ) ⇒ ax2 + x − z = 0 √ −1 ± 1 + 4az ⇒x ˆ= 2a 7. Random processes x[n] and y[n] are defined by x[n] = v1 [n] + 3v2 [n − 1] y[n] = v2 [n + 1] + 3v2 [n − 1] where v1 [n] and v2 [n] are independent white noise processes, each with variance 0.5. 2


1. Let fx (t) be symmetric about 0. Prove that µ is the expected value of a sample distributed according to fx−µ (t). Solution. Since fx (t) is symmetric about 0, fx (t) is even. Z +∞

E[(x − µ)] =

−∞ Z +∞

= −∞

tfx−µ (t)dt tfx (t − µ)dt

Let u = t − µ, E[(x − µ)] =

Z +∞

u + µfx (u)du

−∞

Z +∞

Z +∞

ufx (u) du +

= −∞

| {z }

−∞

µfx (u)du

odd

Z +∞

= 0+µ −∞

fx (u)du

= µ

2. The complimentary cumulative distribution function is defined as Qx (x) = 1 − Fx (x), or more explicitly in the zero mean, unit variance Gaussian distribution case as Z ∞ 1 1 √ exp − t2 dt. Qx (x) = 2 2π x Show that

1 1 Qx (x) ≈ √ exp − x2 . 2 2πx

1 t exp − 12 t2 dt. Also Hint: use integration by parts on Qx (x) = x∞ √2πt explain why the approximation improves x as increases.

R

Solution. Rb Rb 0 0 b Recall integrationby parts: a f (t)g (t)dt = f (t)g(t)|a − a f (t)g(t)dt. 1 Let g 0 (t) = t exp − 21 t2 and f (t) = √2πt Z ∞

Qx (x) = x

1 1 √ t exp − t2 dt 2 2πt

1


=

≈

Z ∞ ∞ 1 1 1 1 √ exp − t2 − exp − t2 dt 2 2 2 2πt 2πt x x | {z } →0 as x→∞ 1 1 √ exp − x2 2 2πx

−√

1 1 2 Since x∞ √2πt dt goes to zero as x goes to infinity, the ap2 exp − 2 t proximation improves x as increase.

R

3. The probability density function for a two dimensional random vector is defined by (

fx (x) =

Ax21 x2 x1 , x2 ≥ 0 and x1 + x2 ≤ 1 0 otherwise

(a) Determine Fx (x) and the value of A. (b) Determine the marginal density fx2 (x). (c) Are fx1 (x) and fx2 (x) independent? Show why or why not. Solution. (a) Fx1 ,x2 (∞, ∞) =

Z 1 Z 1−x1 0

x22 x2 2 0 0 Z 1 (1 − x1 )2 Ax21 dx1 2 0 Z A 1 4 (x − 2x31 + x21 )dx1 2 0 1 A 60 1 Z 1

= = = = =

0

Ax21 x2 dx2 dx1

Ax21

(1)

Therefore, A = 60. Defining Fx1 ,x2 (u, v) = P r(x1 ≤ u, x2 ≤ v), we have • x1 < 0 or x2 < 0, then F (x1 , x2 ) = 0.

2


• x1 , x2 ≥ 0 and x1 + x2 ≤ 1, then Z x1 Z x2

F (x1 , x2 ) =

60u2 vdvdu

0

0

= 10x31 x22 • 0 ≤ x1 , x2 ≤ 1 and x1 + x2 ≥ 1, then

F (x1 , x2 ) = 1 −

Z 1−x2 Z 1−u 0

= 10x22 −

x2 3 20x2 +

2

60u vdvdu −

15x42 − 4x52 +

Z 1 Z 1−u

x1 10x31

60u2 vdvdu

0

− 15x41 + 6x51 − 1

• 0 ≤ x1 ≤ 1 and x2 ≥ 1, then

F (x1 , x2 ) = 1 −

Z 1 Z 1−u x1

60u2 vdvdu

0

= 10x31 − 15x41 + 6x51 • 0 ≤ x2 ≤ 1 and x1 ≥ 1, then

F (x1 , x2 ) = 1 −

Z 1−x2 Z 1−u 0

= 10x22 −

x2 3 20x2 +

60u2 vdvdu

15x42 − 4x52

• x1 , x2 ≥ 1, then F (x1 , x2 ) = 1. So  0 x1 < 0 or x2 < 0    3 2   10x1 x2 x1 , x2 ≥ 0, x1 + x2 ≤ 1    10x2 − 20x3 + 15x4 − 4x5 + 10x3 − 15x4 + 6x5 − 1 0 ≤ x , x ≤ 1, x + x ≥ 1 1 2 1 2 2 2 2 2 1 1 1 F (x1 , x2 ) =  10x31 − 15x41 + 6x51 0 ≤ x1 ≤ 1, x2 ≥ 1     10x22 − 20x32 + 15x42 − 4x52 0 ≤ x2 ≤ 1, x1 ≥ 1   

x1 , x2 ≥ 1

1

(b) Z 1−x2

fx2 (x2 ) =

0

60x21 x2 dx1

= 20x2 (1 − x2 )3 3


(c) Since Z 1−x1

fx1 (x1 ) =

0

60x21 x2 dx2

= 30x21 (1 − x1 )2 , fx1 ,x2 (x1 , x2 ) 6= fx1 (x1 )fx2 (x2 ). Therefore, fx1 (x1 ) and fx2 (x2 ) are NOT independent. 4. Consider the two independent marginal distributions (

1 0 ≤ x1 ≤ 1 0 otherwise

fx1 (x) = (

2x 0 ≤ x2 ≤ 1 0 otherwise

fx2 (x) = Let A be the event x1 ≤ x2 . (a) Find and sketch fx (x). (b) Determine P r{A}.

(c) Determine fx|A (x|A). Are the components independent, i.e., are fx1 |A (x|A) and fx2 |A (x|A) independent? Solution. (a) Since two marginal distributions are independent, fX (X) = fx1 (x1 )fx2 (x2 ) (

=

2x2 0 ≤ x1 , x2 ≤ 1 0 otherwise

(b) Z 1 Z 1−x2

P r(A) =

2x2 dx1 dx2 0

Z 1

= 0

= =

0

2x22 dx2

2x32 1 3 0 2 3 4

(2)


(c) fX|A (X|A) =

fX (X) P r(A) (

=

fx1 |A (x1 |A) = = =

3x2 0 ≤ x1 < x2 ≤ 1 0 otherwise

Z 1

3x2 dx2 x1 3x22 1 2 x1 3(1 − x1 )2 2

fx2 |A (x2 |A) =

,

0 ≤ x1 ≤ 1

Z x2

2x2 dx1 0

= 2x22 ,

0 ≤ x2 ≤ 1

fX|A (X|A) 6= fx1 |A (x1 |A)fx2 |A (x2 |A). Therefore, fx1 |A (x1 |A) and fx2 |A (x2 |A) are NOT independent. 5. The entropy H for a random vector is defined as −E{ln fx (x)}. Show that for the complex Gaussian case H = N (1 + ln π) + ln |Cx |. Determine the corresponding expression when the vector is real. Solution. The complex Gaussian p.d.f. is fx (x) =

1 π N |Cx |

exp[−(x − mx )H C−1 x (x − mx )]

Then, H = −E{ln fx (x)} = E[(x − mx )H C−1 x (x − mx )] + N ln π + ln |Cx |

5


Note H −1 E[(x − mx )H C−1 x (x − mx )] = E[trace((x − mx ) Cx (x − mx ))] H = trace(C−1 x E[(x − mx )(x − mx ) ])

= trace(C−1 x Cx ) = trace(I) = N Therefore H = N + N ln π + ln |Cx | = N (1 + ln π) + ln |Cx | Similarly, when the vector is real H =

1 1 N (1 + ln(2π)) + ln |Cx | 2 2

6. Let x = 3u − 4v y = 2u + v where u and v are unit mean, unit variance, uncorrelated Gaussian random variables. (a) Determine the means and variances of x and y. (b) Determine the joint density of x and y. (c) Determine the conditional density of y given x. Solution. (a) E(x) = E(3u − 4v) = 3E(u) − 4E(v) = 3−4 = −1 E(y) = E(2u + v) = 2E(u) + E(v) = 2+1 = 3 6


σx2 = E(x2 ) − E 2 (x) = E[(3u − 4v)2 ] − 1 = 25 σy2 = E(y 2 ) − E 2 (y) = E[(2u + v)2 ] − 9 = 5 (b) Note "

x y

#

"

3 −4 2 1

|

{z

=

Thus A

−1

1 = 11

A

"

#"

u v

#

}

1 4 −2 3

#

and fx,y (x, y) = = =

fu,v (A−1 [x, y]T ) abs |A| 1 fu,v ((x + 4y)/11, (−2x + 3y)/11) 11 1 1 x + 4y −2x + 3y exp(− [( − 1)2 + ( − 1)2 ]) 22π 2 11 11

(c) Note x is Gaussian fx (x) = √

1 1 exp − (x + 1)2 2 × 25 2π × 5

Thus fy|x (y|x) = = =

fx,y (x, y) fx (x) √ 2π × 5 1 x + 4y −2x + 3y 1 exp − [( − 1)2 + ( − 1)2 ] + (x + 1)2 22π 2 11 11 2 × 25 r 5 2 1 x + 4y −2x + 3y 1 2 2 2 exp − [( − 1) + ( − 1) − (x + 1) ] 22 π 2 11 11 25

7


7. Consider the orthogonal transformation of the correlated zero mean random variables x1 and x2 "

y1 y2

#

"

=

cos θ sin θ − sin θ cos θ

#"

x1 x2

#

Note E{x21 } = σ12 , E{x22 } = σ22 , and E{x1 x2 } = ρσ1 σ2 . Determine the angle θ such that y1 and y2 are uncorrelated. Solution.

(

y1 = x1 cos θ + x2 sin θ y2 = −x1 sin θ + x2 cos θ

E(y1 y2 ) = E[(x1 cos θ + x2 sin θ)(−x1 sin θ + x2 cos θ)] = sin θ cos θE[x22 ] + (cos2 θ − sin2 θ)E[x1 x2 ] − sin θ cos θE[x21 ] = sin θ cos θ(σ22 − σ12 ) + (cos2 θ − sin2 θ)ρσ1 σ2 (σ 2 − σ12 ) = sin 2θ · 2 + cos 2θ · ρσ1 σ2 2 If y1 and y2 are uncorrelated, E(y1 y2 ) = 0. For −π/2 ≤ θ < π/2, θ=

2ρσ1 σ2 1 arctan 2 2 σ2 − σ12

8. The covariance matrix and mean vector for a real Gaussian density are "

Cx =

1 0.5 0.5 1

and

"

mx =

1 0

#

#

(a) Determine the eigenvalues and eigenvectors. (b) Generate a mesh plot of the distribution using MATLAB. (c) Change the off-diagonal values to −0.5 and repeat (a) and (b).

8


Solution. (a) Solve |Cx − λI| = 0. (1 − λ)2 − 0.25 = (λ − 0.5)(λ − 1.5) = 0 Hence, eigenvalues are 0.5 and 1.5. For λ = 0.5, the corresponding eigenvector is [1, −1]T . For λ = 1.5, the corresponding eigenvector is [1, 1]T . (c) Eigenvalues are 0.5 and 1.5. For λ = 0.5, the corresponding eigenvector is [1, 1]T . For λ = 1.5, the corresponding eigenvector is [1, −1]T . 9. Let {xk (n)}K k=1 be i.i.d. zero mean, unit variance uniformly distributed random variables and set yK (n) =

K X

xk (n).

k=1

(a) Determine and plot the pdf of yK (n) for K = 2, 3, 4. (b) Compare the pdf’s to the Gaussian density. (c) Perform the comparison experimentally using MATLAB. That is, generate K sequences of n = 1, 2, . . . , N uniformly distributed samples. Add the sequences and plot the resulting distribution (histogram). Fit the results to a Gaussian distribution for various K and N . Solution. (a) {xk (n)}K k=1 are i.i.d. zero mean, unit variance uniformly distributed random variables. ( 1/2a xk ∈ [−a, a] fxk (xk ) = 0 otherwise Since E[x2k ] = 1, 1 a 2 x dx 2a −a x3 a = 6a −a a2 = 3 = 1

E[x2k ] =

9

Z


⇒a= That is

(

fxk (xk ) =

√

3

√ √ xk ∈ [− 3, 3] otherwise

1 √ 2 3

0

For K=2, y2 (n) = x1 (n) + x2 (n). fy2 (x) = fx1 (x) ∗ fx2 (x) =

 x 1 √   12 + 2 3  

x − 12 + 0

1 √ 2 3

√ −2 3 ≤ x < 0 √ 0
For K=3, y3 (n) = x1 (n) + x2 (n) + x3 (n) = y2 (n) + x3 (n). fy3 (x) = fy2 (x) ∗ fx3 (x)

=

√  (x+3√ 3)2    48 3   2  3−x √ 8 3√

 (x−3√ 3)2    48 3  

0

10

√ √ −3 3 ≤ x < − 3 √ √ − 3≤x< 3 √ √ 3≤x≤3 3 otherwise

SolvedProblems[1]

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