Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers
Edition 3 Roy D. Yates and David J. Goodman
Yates ates and Goodman Goodman 3e Proble Problem/ m/Sol Soluti ution on Set: Set: and 9.3.1 Problem 3.6.2
3.6.2, 3.6 .2, 9.1.2, 9.1.2, 6.5.3, 6.5.3, 9.2 9.2.3, .3,
•
Given the random variable X X in in Problem 3.4.2, let V = g( g (X ) = |X | X |. (a) Find P V V (v ). (b) Find F V V (v ). (c) Find E[V E[V ]. ]. Problem 3.6.2 Solution
From the solution to Problem 3.4.2, the PMF of X is
P X X (x) =
(a) The PMF of V = |X | X | satisfies
0.2 0.5 0.3 0
x = − = −11, x = 0, x = 1, otherwise.
P V (v ) = P [|X | = v = v]] = P X V (v X (v ) + P X X (−v ) .
(1)
(2)
In particular, P V = P X 0.5, V (0) = P X (0) = 0.
(3)
P V = P X 0.5. V (1) = P X (−1) + P X X (1) = 0.
(4)
1
The complete expression for the PMF of V is
P V (v) =
0.5 v = 0, 0.5 v = 1, 0 otherwise.
(5)
(b) From the PMF, we can construct the staircase CDF of V .
F V (v) =
0 v < 0, 0.5 0 ≤ v < 1, 1 v ≥ 1.
(6)
(c) From the PMF P V (v), the expected value of V is E [V ] =
P V (v) = 0(1/2) + 1(1/2) = 1/2.
(7)
v
You can also compute E[V ] directly by using Theorem 3.10. Problem 9.1.2
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Flip a biased coin 100 times. On each flip, P[H ] = p. Let X i denote the number of heads that occur on flip i. What is P X (x)? Are X 1 and X 2 independent? Define Y = X 1 + X 2 + · · · + X 100 . Describe Y in words. What is P Y (y)? Find E[Y ] and Var[Y ]. 33
Problem 9.1.2 Solution
The random variable X 33 is a Bernoulli random variable that indicates the result of flip 33. The PMF of X 33 is
P X
33
1 − p x = 0, (x) = p x = 1, 0 otherwise.
2
(1)
Note that each X i has expected value E[X ] = p and variance Var[X ] = p(1 − p). The random variable Y = X 1 + · · · + X 100 is the number of heads in 100 coin flips. Hence, Y has the binomial PMF
100
P Y (y) =
y
py (1 − p)100
y
y = 0, 1, . . . , 100,
−
0
otherwise.
(2)
Since the X i are independent, by Theorems 9.1 and 9.3, the mean and variance of Y are E [Y ] = 100 E [X ] = 100 p, Var[Y ] = 100 Var[X ] = 100 p(1 − p). Problem 6.5.3
(3)
Find the PDF of W = X + Y when X and Y have the joint PDF f X,Y (x, y) =
2 0 ≤ x ≤ y ≤ 1, 0 otherwise.
Problem 6.5.3 Solution
The joint PDF of X and Y is f X,Y (x, y) =
2 0 ≤ x ≤ y ≤ 1, 0 otherwise.
(1)
We wish to find the PDF of W where W = X + Y . First we find the CDF of W , F W (w), but we must realize that the CDF will require different integrations for different values of w. Y
Y=X
For values of 0 ≤ w ≤ 1 we look to integrate the shaded area in the figure to the right.
w
Area of Integration
w
w −x
2
F W (w) =
0
X+Y=w X w
3
x
w2 2 dy dx = . 2
(2)
For values of w in the region 1 ≤ w ≤ 2 we look to integrate over the shaded region in the graph to the right. From the graph we see that we can integrate with respect to x first, ranging y from 0 to w/2, thereby covering the lower right triangle of the shaded region and leaving the upper trapezoid, which is accounted for in the second term of the following expression:
Y w Y=X
Area of Integration
w
X+Y=w
y
2
F W (w) =
X
0
w
w −y
1
2 dxdy +
0
w2 = 2w − 1 − . 2
w
2
2 dxdy
0
(3)
Putting all the parts together gives the CDF
0
F W (w) =
w2 2
2w − 1 − 1
w2 2
w < 0, 0 ≤ w ≤ 1, 1 ≤ w ≤ 2, w > 2,
(4)
and (by taking the derivative) the PDF f W (w) =
Problem 9.2.3
w 2 − w 0
0 ≤ w ≤ 1, 1 ≤ w ≤ 2, otherwise.
(5)
X is the continuous uniform (a,b) random variable. Find the MGF φX (s). Use the MGF to calculate the first and second moments of X . Problem 9.2.3 Solution
We find the MGF by calculating E[esX ] from the PDF f X(x). b
φX (s) = E e
sX
=
e
sX
a
4
1 ebs − eas dx = . b−a s(b − a)
(1)
Now to find the first moment, we evaluate the derivative of φX (s) at s = 0. dφ X (s) E [X ] = ds
s bebs − aeas − ebs − eas = (b − a)s2
s=0
.
(2)
s=0
Direct evaluation of the above expression at s = 0 yields 0/0 so we must apply l’Hˆ opital’s rule and differentiate the numerator and denominator. bs
as
2
bs
2
as
+s b e −a e s 0 2(b − a)s b2 ebs − a2 eas b+a = lim = . s 0 2(b − a) 2
E [X ] = lim
be − ae
→
bs
as
− be − ae
(3)
→
To find the second moment of X , we first find that the second derivative of φX (s) is 2
s d φX (s) = ds2
2
2
bs
2
as
b e −a e
bs
as
− 2s be − ae (b − a)s3
bs
as
+ 2 be − ae
.
(4)
Substituting s = 0 will yield 0/0 so once again we apply l’Hˆopital’s rule and differentiate the numerator and denominator. 2
2
3
bs
3
as
s b e −a e d φX (s) E X = lim = lim s 0 s 0 ds2 3(b − a)s2 b3 − a3 = = (b2 + ab + a2 )/3. 3(b − a) 2
→
→
(5)
In this case, it is probably simpler to find these moments without using the MGF. Problem 9.3.1
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N is the binomial (100, 0.4) random variable. M is the binomial (50, 0.4) random variable. M and N are independent. What is the PMF of L = M + N ? Problem 9.3.1 Solution
N is a binomial (n = 100, p = 0.4) random variable. M is a binomial (n = 50, p = 0.4) random variable. Thus N is the sum of 100 independent Bernoulli ( p = 0.4) and M is the sum of 50 independent Bernoulli ( p = 0.4) random variables. Since 5
M and N are independent, L = M + N is the sum of 150 independent Bernoulli ( p = 0.4) random variables. Hence L is a binomial n = 150, p = 0.4) and has PMF P L (l) =
150 (0.4)l (0.6)150 l . l
6
−
(1)
