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需要这本书的习题解答(solution manual) Stochastic Processes Theory for Applications by R.G.Gallager
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徐*浩 徐*浩 2018-11-19 17:22:47

亲,这里为您找到

Stochastic Processes,

Theory for Applications

Solutions to Selected Exercises

R.G.Gallager

December 2, 2013

这一本书的练习参考资料,您可选择查看下是否是您需要的内容(以下是正文部分内容,共计130页)

2 APPENDIX A. SOLUTIONS TO EXERCISES

A.1 Solutions for Chapter 1

Exercise 1.2: This exercise derives the probability of an arbitrary (non-disjoint) union of events, derives

the union bound, and derives some useful limit expressions.

a) For 2 arbitrary events A 1 and A 2 , show that

A 1

[

A 2 = A 1

[

(A 2 ?A 1 ), (A.1)

where A 2 ?A 1 = A 2 A c 1 . Show that A 1 and A 2 ? A 1 are disjoint. Hint: This is what Venn diagrams were

invented for.

Solution: Note that each sample point ! is in A 1 or A c

1 , but not both. Thus each ! is in

exactly one of A 1 , A c

1 A 2

or A c

1 A c 2 . In the first two cases, ! is in both sides of (A.1) and in

the last case it is in neither. Thus the two sides of (A.1) are identical. Also, as pointed out

above, A 1 and A 2 ? A 1 are disjoint. These results are intuitively obvious from the Venn

diagram,

?

?

6

⇥ ⇥ ?

B

B

B M

6

?

A

A U

?

? ↵

A 1 A 2

A 1 A 2 A 2 A c 1 = A 2 ?A 1 A 1 A c 2

b) For each n ? 2 and arbitrary events A 1 ,... ,A n , define B n = A n ?

S n?1

i=1

A i . Show that B 1 ,B 2 ,... are

disjoint events and show that for each n ? 2,

S n

i=1 A i

=

S n

i=1 B i . Hint: Use induction.

Solution: Let B 1 = A 1 . From (a), B 1 and B 2 are disjoint and (from (A.1)), A 1

S

A 2 =

B 1

S

B 2 . Let C n =

S n

i=1 A i . We use induction to prove that C n

=

S n

i=1 B i

and that the

B n are disjoint. We have seen that C 2 = B 1

S

B 2 , which forms the basis for the induction.

We assume that C n?1 =

S n?1

i=1

B i and prove that C n =

S n

i=1 B i .

C n = C n?1

[

A n = C n?1

[

A n C c

n?1

= C n?1

[

B n =

[ n

i?1

B i .

In the second equality, we used (A.1), letting C n?1 play the role of A 1 and A n play the role

of A 2 . From this same application of (A.1), we also see that C n?1 and B n = A n ?C n?1 are

disjoint. Since C n?1 =

S n?1

i=1

B i , this also shows that B n is disjoint from B 1 ,... ,B n?1 .

c) Show that

Pr

n [ 1

n=1

A n

o

= Pr

n [ 1

n=1

B n

o

=

X 1

n=1

Pr{B n }.

Solution: If ! 2

S 1

n=1 A n , then it is in A n

for some n ? 1. Thus ! 2

S n

i=1 B i , and thus

! 2

S 1

n=1 B n . The same argument works the other way, so

S 1

n=1 A n

=

S 1

n=1 B n . This

establishes the first equality above, and the second is the third axiom of probability.

d) Show that for each n, Pr{B n }  Pr{A n }. Use this to show that

Pr

n [ 1

n=1

A n

o

X 1

n=1

Pr{A n }.

A.1. SOLUTIONS FOR CHAPTER 1 3

Solution: Since B n = A n ?

S n?1

i=1

A i , we see that ! 2 B n implies that ! 2 A n . From (1.5),

this implies that Pr{B n }  Pr{A n } for each n. Thus

Pr

n [ 1

n=1

A n

o

=

X 1

n=1

Pr{B n } 

X 1

n=1

Pr{A n }.

e) Show that Pr

?S

1

n=1 A n

= lim n!1 Pr

?S

n

i=1 A i

. Hint: Use (c). Note that this says that the probability

of a limit is equal to the limit of the probabilities. This might well appear to be obvious without a proof,

but you will see situations later where similar appearing interchanges cannot be made.

Solution: From (c),

Pr

n [ 1

n=1

A n

o

=

X 1

n=1

Pr{B n } = lim

k!1

X k

n=1

Pr{B n }.

From (b), however,

X k

n=1

Pr{B n } = Pr

(

k

[

n=1

B n

)

= Pr

(

k

[

n=1

A n

)

.

f) Show that Pr

?T

1

n=1 A n

= lim n!1 Pr

?T

n

i=1 A i

. Hint: Remember De Morgan’s equalities.

Solution: Using De Morgans equalities,

Pr

(

1

\

n=1

A n

)

= 1 ? Pr

(

1

[

n=1

A c

n

)

= 1 ? lim

k!1

Pr

(

k

[

n=1

A c

n

)

= lim

k!1

Pr

(

k

\

n=1

A n

)

.

Exercise 1.4: Consider a sample space of 8 equiprobable sample points and let A 1 ,A 2 ,A 3 be three

events each of probability 1/2 such that Pr{A 1 A 2 A 3 } = Pr{A 1 }Pr{A 2 }Pr{A 3 }.

a) Create an example where Pr{A 1 A 2 } = Pr{A 1 A 3 } =

1

4

but Pr{A 2 A 3 } =

1

8 . Hint: Make a table with a

row for each sample point and a column for each of the above 3 events and try di↵erent ways of assigning

sample points to events (the answer is not unique).

Solution: Note that exactly one sample point must be in A 1 ,A 2 , and A 3 in order to make

Pr{A 1 A 2 A 3 } = 1/8. In order to make Pr{A 1 A 2 } = 1/4, there must be an additional sample

point that contains A 1 and A 2 but not A 3 . Similarly, there must be a sample point that

contains A 1 and A 3 but not A 2 . These give rise to the first three rows in the table below.

The other sample points can each be in at most 1 of the events A 1 ,A 2 , and A 3 , and in order

to make each event have probability 1/2, each of these sample points must be in exactly

one of the three events. This uniquely specifies the table below except for which sample

point lies in each event.

4 APPENDIX A. SOLUTIONS TO EXERCISES

Sample points A 1 A 2 A 3

1 1 1 1

2 1 1 0

3 1 0 1

4 1 0 0

5 0 1 0

6 0 1 0

7 0 0 1

8 0 0 1

b) Show that, for your example, A 2 and A 3 are not independent. Note that the definition of statistical inde-

pendence would be very strange if it allowed A 1 ,A 2 ,A 3 to be independent while A 2 and A 3 are dependent.

This illustrates why the definition of independence requires (1.14) rather than just (1.15).

Solution: Note that Pr{A 2 A 3 } = 1/8 6= Pr{A 2 }Pr{A 3 }, so A 2 and A 3 are dependent.

We also note that Pr{A c

1 A c 2 A c 3 } = 0 6= Pr{A c 1 }Pr{A c 2 }Pr{A c 3 }, further reinforcing the

conclusion that A 1 ,A 2 ,A 3 are not statistical independence. Although the definition in

(1.14) of statistical independence of more than 2 events looks strange, it is clear from this

example that (1.15) is insu?cient in the sense that it only specifies part of the above table.

Exercise 1.9: (Proof of Theorem 1.4.1) The bounds on the binomial in this theorem are based on

the Stirling bounds. These say that for all n ? 1, n! is upper and lower bounded by

p 2⇡n ⇣ n

e

⌘ n

< n! <

p 2⇡n ⇣ n

e

⌘ n

e 1/12n . (A.2)

The ratio,

p 2⇡n(n/e) n /n!, of the first two terms is monotonically increasing with n toward the limit 1,

and the ratio

p 2⇡n(n/e) n

exp(1/12n)/n! is monotonically decreasing toward 1. The upper bound is more

accurate, but the lower bound is simpler and known as the Stirling approximation. See [8] for proofs and

further discussion of the above facts.

a) Show from (A.2) and from the above monotone property that

n

k

!

<

r

n

2⇡k(n ? k)

n n

k k (n?k) n?k

.

Hint: First show that n!/k! <

p n/kn n k ?k e ?n+k

for k < n.

Solution: Since the ratio of the first two terms of (A.2) is increasing in n, we have

p 2⇡k(k/e) k /k! < p 2⇡n(n/e) n /n!.

Rearranging terms, we have the result in the hint. Applying the first inequality of (A.2) to

n ? k and combining this with the result on n!/k! yields the desired result.

b) Use the result of (a) to upper bound p S n (k) by

p S n (k) <

r

n

2⇡k(n ? k)

p k (1 ? p) n?k n n

k k (n?k) n?k

.

Show that this is equivalent to the upper bound in Theorem 1.4.1.

Solution: Using the binomial equation and then (a),

p S n (k) =

✓ n

k

p k (1 ? p) n?k <

r

n

2⇡k(n ? k)

n n

k k (n?k) n?k

p k (1 ? p) n?k .

A.1. SOLUTIONS FOR CHAPTER 1 5

This is the the desired bound on p S n (k). Letting ˜ p = k/n, this becomes

p S n (˜ pn) <

s

1

2⇡n˜ p(1 ? ˜ p)

p ˜ pn (1 ? p) n(1?˜ p)

˜ p ˜ pn (1 ? ˜ p) n(1?˜ p)

=

s

1

2⇡n˜ p(1 ? ˜ p)

exp

n

˜ pln

p

˜ p

+ ˜ pln

1 ? p

1 ? ˜ p

?◆

,

which is the same as the upper bound in Theorem 1.4.1.

c) Show that

n

k

!

>

r

n

2⇡k(n ? k)

n n

k k (n?k) n?k

1 ?

n

12k(n ? k)

?

.

Solution: Use the factorial lower bound on n! and the upper bound on k and (n ? k)!.

This yields

✓ n

k

>

r

n

2⇡k(n ? k)

n n

k k (n?k) n?k

exp

?

1

12k

?

1

12(n ? k)

>

r

n

2⇡k(n ? k)

n n

k k (n?k) n?k

1 ?

n

12k(n ? k)

?

,

where the latter equation comes from combining the two terms in the exponent and then

using the bound e ?x > 1 ? x.

d) Derive the lower bound in Theorem 1.4.1.

Solution: This follows by substituting ˜ pn for k in the solution to c) and substituting this

in the binomial formula.

e) Show that ?(p, ˜ p) = ˜ pln(

˜ p

p ) + (1 ? ˜ p)ln(

1?˜ p

1?p ) is 0 at ˜ p = p and nonnegative elsewhere.

Solution: Taking the first two derivatives of ?(p, ˜ p) with respect to ˜ p,

@?(p, ˜ p)

@˜ p

= ?ln

✓ p(1 ? ˜ p)

˜ p(1 ? p)

@f 2 (p, ˜ p)

@˜ p 2

=

1

˜ p(1 ? ˜ p) .

Since the second derivative is positive for 0 < ˜ p < 1, the minimum of ?(p, ˜ p) with respect to

˜ p is 0, is achieved where the first derivative is 0, i.e., at ˜ p = p. Thus ?(p, ˜ p) > 0 for ˜ p 6= p.

Furthermore, ?(p, ˜ p) increases as ˜ p moves in either direction away from p.

Exercise 1.11: a) For any given rv Y , express E[|Y |] in terms of

R

y<0 F Y (y)dy and

R

y?0 F

c

Y (y)dy. Hint:

Review the argument in Figure 1.4.

Solution: We have seen in (1.34) that

E[Y ] = ?

Z

y<0

F Y (y)dy +

Z

y?0

F c

Y

(y)dy.

Since all negative values of Y become positive in |Y |,

E[|Y |] = +

Z

y<0

F Y (y)dy +

Z

y?0

F c

Y

(y)dy.

6 APPENDIX A. SOLUTIONS TO EXERCISES

To spell this out in greater detail, let Y = Y

+

+ Y

?

where Y

+

= max{0,Y } and Y

?

=

min{Y,0}. Then Y = Y

+

+ Y

?

and |Y | = Y

+

? Y

?

= Y

+

+ |Y

? |.

Since E[Y

+ ] =

R

y?0 F

c

Y

(y)dy and E[Y

? ] = ? R

y<0 F Y (y)dy, the above results follow.

b) For some given rv X with E[|X|] < 1, let Y = X ? ↵. Using (a), show that

E[|X ? ↵|] =

Z

?1

F X (x)dx +

Z

1

F c X (x)dx.

Solution: This follows by changing the variable of integration in (a). That is,

E[|X ? ↵|] = E[|Y |] = +

Z

y<0

F Y (y)dy +

Z

y?0

F c

Y

(y)dy

=

Z

?1

F X (x)dx +

Z

1

F c

X (x)dx,

where in the last step, we have changed the variable of integration from y to x ? ↵.

c) Show that E[|X ? ↵|] is minimized over ↵ by choosing ↵ to be a median of X. Hint: Both the easy way

and the most instructive way to do this is to use a graphical argument illustrating the above two integrals

Be careful to show that when the median is an interval, all points in this interval achieve the minimum.

Solution: As illustrated in the picture, we are minimizing an integral for which the inte-

grand changes from F X (x) to F c

X (x) at x = ↵. If F X (x) is strictly increasing in x, then

F c

X

= 1 ? F X is strictly decreasing. We then minimize the integrand over all x by choosing

↵ to be the point where the curves cross, i.e., where F X (x) = .5. Since the integrand has

been minimized at each point, the integral must also be minimized.

0.5

1

0

F X (x)

F c X (x)

If F X is continuous but not strictly increasing, then there might be an interval over which

F X (x) = .5; all points on this interval are medians and also minimize the integral; Exercise

1.10 (c) gives an example where F X (x) = 0.5 over the interval [1,2). Finally, if F X (↵) ? 0.5

and F X (↵ ? ✏) < 0.5 for some ↵ and all ✏ > 0 (as in parts (a) and (b) of Exercise 1.10),

then the integral is minimized at that ↵ and that ↵ is also the median.

Exercise 1.12: Let X be a rv with CDF F X (x). Find the CDF of the following rv’s.

a) The maximum of n IID rv’s, each with CDF F X (x).

Solution: Let M + be the maximum of the n rv’s X 1 ,... ,X n . Note that for any real x,

M + is less than or equal to x if and only if X j  x for each j, 1  j  n. Thus

Pr{M +  x} = Pr{X 1  x,X 2  x,... ,X n  x} =

n

Y

j=1

Pr{X j  x},

where we have used the independence of the X j ’s. Finally, since Pr{X j  x} = F X (x) for

each j, we have F M + (x) = Pr{M +  x} =

? F

X (x)

? n .

A.1. SOLUTIONS FOR CHAPTER 1 7

b) The minimum of n IID rv’s, each with CDF F X (x).

Solution: Let M ? be the minimum of X 1 ,... ,X n . Then, in the same way as in ((a),

M ? > y if and only if X j > y for 1  j  n and for all choice of y. We could make the

same statement using greater than or equal in place of strictly greater than, but the strict

inequality is what is needed for the CDF. Thus,

Pr{M ? > y} = Pr{X 1 > y,X 2 > y,... ,X n > y} =

n

Y

j=1

Pr{X j > y},

It follows that 1 ? F M ? (y) =

1 ? F X (y)

⌘ n

.

c) The di↵erence of the rv’s defined in a) and b); assume X has a density f X (x).

Solution: There are many di?cult ways to do this, but also a simple way, based on first

conditioning on the event that X 1 = x. Then X 1 = M + if and only if X j  x for 2  j  n.

Also, given X 1 = M + = x, we have R = M + ? M ?  r if and only if X j > x ? r for

2  j  n. Thus, since the X j are IID,

Pr{M + =X 1 ,R  r | X 1 = x} =

n

Y

j=2

Pr{x?r < X j  x}

= [Pr{x?r < X  x}] n?1 = [F X (x) ? F X (x ? r)] n?1 .

We can now remove the conditioning by averaging over X 1 = x. Assuming that X has the

density f X (x),

Pr{X 1 = M + , R  r} =

Z

1

?1

f X (x)[F X (x) ? F X (x ? r)] n?1 dx.

Finally, we note that the probability that two of the X j are the same is 0 so the events

X j = M + are disjoint except with zero probability. Also we could condition on X j = x

instead of X 1 with the same argument (i.e., by using symmetry), so Pr{X j = M + , R  r} =

Pr{X 1 = M + R  r} It follows that

Pr{R  r} =

Z

1

?1

nf X (x)[F X (x) ? F X (x ? r)] n?1 dx.

The only place we really needed the assumption that X has a PDF was in asserting that

the probability that two or more of the X j ’s are jointly equal to the maximum is 0. The

formula can be extended to arbitrary CDF’s by being careful about this possibility.

These expressions have a simple form if X is exponential with the PDF ?e ??x for x ? 0.

Then

Pr{M ? ? y} = e ?n?y ; Pr{M +  y} =

? 1 ? e ??y ? n ;

Pr{R  y} =

? 1 ? e ??y ? n?1 .

We will see how to derive Pr{R  y} when we study Poisson processes.

Exercise 1.13: Let X and Y be rv’s in some sample space ⌦ and let Z = X + Y , i.e., for each

! 2 ⌦, Z(!) = X(!)+Y (!). The purpose of this exercise is to show that Z is a rv. This is a mathematical

fine point that many readers may prefer to simply accept without proof.

8 APPENDIX A. SOLUTIONS TO EXERCISES

a) Show that the set of ! for which Z(!) is infinite or undefined has probability 0.

Solution: Note that Z can be infinite (either ±1) or undefined only when either X or

Y are infinite or undefined. Since these are events of zero probability, Z can be infinite or

undefined only with probability 0.

b) We must show that {! 2 ⌦ : Z(!)  ↵} is an event for each real ↵, and we start by approximating

that event. To show that Z = X + Y is a rv, we must show that for each real number ↵, the set {! 2 ⌦ :

X(!)+Y (!)  ↵} is an event. Let B(n,k) = {! : X(!)  k/n}

T {Y (!)  ↵+(1?k)/n} for integer k > 0.

Let D(n) =

S

k B(n,k), and show that D(n) is an event.

Solution: We are trying to show that {Z  ↵} is an event for arbitrary ↵ and doing this

by first quantizing X and Y into intervals of size 1/n where k is used to number these

quantized elements. Part (c) will make sense of how this is related to {Z  ↵, but for

now we simply treat the sets as defined. Each set B(n,k) is an intersection of two events,

namely the event {! : X(!)  k/n} and the event {! : Y (!)  ↵ + (1?k)/n}; these must

be events since X and Y are rv’s. For each n, D(n) is a countable union (over k) of the

sets B(n,k), and thus D(n) is an event for each n and each ↵

c) On a 2 dimensional sketch for a given ↵, show the values of X(!) and Y (!) for which ! 2 D(n). Hint:

This set of values should be bounded by a staircase function.

Solution:

x

y

@

@

@

@

@

@

@

@

@

@

@

? 1

n

0

1

n

2

n

3

n

↵?1/n

↵?2/n

The region D(n) is sketched for ↵n = 5; it is the region below the staircase function above.

The kth step of the staircase, extended horizontally to the left and vertically down is the

set B(n,k). Thus we see that D(n) is an upper bound to the set {Z  ↵}, which is the

straight line of slope -1 below the staircase.

d) Show that

{! : X(!) + Y (!)  ↵} =

\

n?1

D(n). (A.3)

Explain why this shows that Z = X + Y is a rv.

Solution: The region {! : X(!) + Y (!)  ↵} is the region below the diagonal line of

slope -1 that passes through the point (0,↵). This region is thus contained in D(n) for

each n ? 1 and is thus contained in

T

n?1 D(n). On the other hand, each point ! for which

X(!)+Y (!) > ↵ is not contained in D(n) for su?ciently large n. This verifies (A.3). Since

D(n) is an event, the countable intersection is also an event, so {! : X(!) + Y (!)  ↵} is

an event. This applies for all ↵. This, in conjunction with (a), shows that Z is a rv.

e) Explain why this implies that if X 1 ,X 2 ,... ,X n are rv’s, then Y = X 1 + X 2 + ··· + X n is a rv. Hint:

A.1. SOLUTIONS FOR CHAPTER 1 9

Only one or two lines of explanation are needed.

Solution: We have shown that X 1 + X 2 is a rv, so (X 1 + X 2 ) + X 3 is a rv, etc.

Exercise 1.15: (Stieltjes integration) a) Let h(x) = u(x) and F X (x) = u(x) where u(x) is the unit

step, i.e., u(x) = 0 for ?1 < x < 0 and u(x) = 1 for x ? 0. Using the definition of the Stieltjes integral

in Footnote 19, show that

R 1

?1 h(x)dF X (x) does not exist. Hint: Look at the term in the Riemann sum

including x = 0 and look at the range of choices for h(x) in that interval. Intuitively, it might help initially

to view dF X (x) as a unit impulse at x = 0.

Solution: The Riemann sum for this Stieltjes integral is

P

n h(x n )[F(y n )?F(y n?1 )] where

y n?1 < x n  y n . For any partition {y n ; n ? 1}, consider the k such that y k?1 < 0  y k and

consider choosing either x n < 0 or x n ? 0. In the first case h(x n )[F(y n )?F(y n?1 )] = 0 and

in the second h(x n )[F(y n ) ? F(y n?1 )] = 1. All other terms are 0 and this can be done for

all partitions as ? ! 0, so the integral is undefined.

b) Let h(x) = u(x ? a) and F X (x) = u(x ? b) where a and b are in (?1, +1). Show that

R 1

?1 h(x)dF X (x)

exists if and only if a 6= b. Show that the integral has the value 1 for a < b and the value 0 for a > b. Argue

that this result is still valid in the limit of integration over (?1, 1).

Solution: Using the same argument as in (a) for any given partition {y n ; n ? 1}, consider

the k such that y k?1 < b  y k . If a = b, x k can be chosen to make h(x k ) either 0 or 1,

causing the integral to be undefined as in (a). If a < b, then for a su?ciently fine partion,

h(x k ) = 1 for all x k such that y k?1 < x k  y k . Thus that term in the Riemann sum is

1. For all other n, F X (y n ) ? F X (y n?1 ) = 0, so the Riemann sum is 1. For a > b and k

as before, h(x k ) = 0 for a su?ciently fine partition, and the integral is 0. The argument

does not involve the finite limits of integration, so the integral remains the same for infinite

limits.

c) Let X and Y be independent discrete rv’s, each with a finite set of possible values. Show that

R 1

?1 F X (z?

y)dF Y (y), defined as a Stieltjes integral, is equal to the distribution of Z = X + Y at each z other than the

possible sample values of Z, and is undefined at each sample value of Z. Hint: Express F X and F Y as sums

of unit steps. Note: This failure of Stieltjes integration is not a serious problem; F Z (z) is a step function,

and the integral is undefined at its points of discontinuity. We automatically define F Z (z) at those step

values so that F Z is a CDF (i.e., is continuous from the right). This problem does not arise if either X or

Y is continuous.

Solution: Let X have the PMF {p(x 1 ),... ,p(x K )} and Y have the PMF {p Y (y 1 ),... ,p Y (y J )}.

Then F X (x) =

P K

k=1 p(x k )u(x ? x k ) and F Y (y) =

P J

j=1 q(y j )u(y ? y j ). Then

Z

1

?1

F X (z ? y)dF Y (y) =

K

X

k=1

J

X

j=1

Z

1

?1

p(x k )q(y j )u(z ? y j ? x k )du(y ? y j ).

From (b), the integral above for a given k,j exists unless z = x k + y j . In other words, the

Stieltjes integral gives the CDF of X + Y except at those z equal to x k + y j for some k,j,

i.e., equal to the values of Z at which F Z (z) (as found by discrete convolution) has step

discontinuities.

To give a more intuitive explanation, F X (x) = Pr{X  x} for any discrete rv X has jumps

at the sample values of X and the value of F X (x k ) at any such x k includes p(x k ), i.e., F X


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