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Probabilidade, um curso moderno com aplicações - Sheldon Ross - Solutions/chap-9-11pt.pdf
147 
Chapter 9 
 
Problems and Theoretical Exercises 
 
1. (a) P(2 arrivals in (0, s) ⏐2 arrivals in (0, 1)} 
 
 =P{2 in (0, s), 0 in (s, 1)}/e−λλ2/2) 
 = [e−λs(λs)2/2][e−(1−s)λ]/(e−λλ2/2) = s2 = 1/9 when s = 1/3 
 
 (b) 1 − P{both in last 40 minutes) = 1 − (2/3)2 = 5/9 
 
2. e−3s/60 
 
3. e−3s/60 + (s/20)e−3s/60 
 
8. The equations for the limiting probabilities are: 
 
 ∏c = .7∏c + .4∏s + .2∏g 
 ∏s = .2∏x + .3∏s + .4∏g 
 ∏g = .1∏c + .3∏s + .4∏g 
 ∏c + ∏s + ∏g = 1 
 
 and the solution is: ∏c = 30/59, ∏s = 16/59, ∏g = 13/59. Hence, Buffy is cheerful 3000/59 
percent of the time. 
 
9. The Markov chain requires 4 states: 
 
 0 = RR = Rain today and rain yesterday 
 
 1 = RD = Dry today, rain yesterday 
 
 2 = DR = Rain today, dry yesterday 
 
 3 = DD = Dry today and dry yesterday 
 
 with transition probability matrix 
 
 P = 
.8 .2 0 0
0 0 .3 .7
.4 .6 0 0
0 0 .2 .8
 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
148 Chapter 9 
 
 The equations for the limiting probabilities are: 
 
 ∏0 = .8∏0 + .4∏2 
 ∏1 = .2∏0 + .6∏2 
 ∏2 = .3∏1 + .2∏3 
 ∏3 = .7∏1 + .8∏3 
 ∏0 + ∏1 + ∏2 + ∏3 = 1 
 
 which gives 
 
 ∏0 = 4/15, ∏1 = ∏2 = 2/15, ∏3 = 7/15. 
 
 Since it rains today when the state is either 0 or 2 the probability is 2/5. 
 
10. Let the state be the number of pairs of shoes at the door he leaves from in the morning. 
Suppose the present state is i, where i > 0. Now after his return it is equally likely that one 
door will have i and the other 5 − i pairs as it is that one will have i − 1 ant the other 6 − i. 
Hence, since he is equally likely to choose either door when he leaves tomorrow it follows 
that 
 
 Pi,i = Pi,5−i = Pi,i−1 = Pi,6−i = 1/4 
 
 provided all the states i, 5 − i, i − 1, 6 − i are distinct. If they are not then the probabilities are 
added. From this it is easy to see that the transition matrix Pij, i, j = 0, 1, …, 5 is as follows: 
 
 P = 
1/ 2 0 0 0 0 1/ 2
1/ 4 1/ 4 0 0 1/ 4 1/ 4
0 1/ 4 1/ 4 1/ 4 1/ 4 0
0 0 1/ 2 1/ 2 0 0
0 1/ 4 1/ 4 1/ 4 1/ 4 0
1/ 4 1/ 4 0 0 1/ 4 1/ 4
 
 
 Since this chain is doubly stochastic (the column sums as well as the row sums all equal to 
one) it follows that ∏i = 1/6, i = 0, …, 5, and thus he runs barefooted one-sixth of the time. 
 
11. (b) 1/2 
 
 (c) Intuitively, they should be independent. 
 
 (d) From (b) and (c) the (limiting) number of molecules in urn 1 should have a binomial 
distribution with parameters (M, 1/2). 
 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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Probabilidade, um curso moderno com aplicações - Sheldon Ross - Solutions/chap-8-11pt.pdf
140 
Chapter 8 
 
Problems 
 
1. P{0 ≤ X ≤ 40} = 1 − P{⏐X − 20⏐ > 20} ≥ 1 − 20/400 = 19/20 
 
2. (a) P{X ≥ 85} ≤ E[X]/85 = 15/17 
 
 (b) P{65 ≤ X ≤ 85) = 1 − P{⏐X − 75⏐ > 10} ≥ 1 − 25/100 
 
 (c) 
1
25
/ 75 5
25
n
i
i
P X n
n=
⎧ ⎫⎪ ⎪− > ≤⎨ ⎬
⎪ ⎪⎩ ⎭
∑ so need n = 10 
 
3. Let Z be a standard normal random variable. Then, 
 
 
1
/ 75 5
n
i
i
P X n
=
⎧ ⎫⎪ ⎪− >⎨ ⎬
⎪ ⎪⎩ ⎭
∑ ≈ P{⏐Z⏐ > n } ≤ .1 when n = 3 
 
4. (a) 
20
1
15 20 /15i
i
P X
=
⎧ ⎫
> ≤⎨ ⎬
⎩ ⎭
∑ 
 
 (b) 
20 20
1 1
15 15.5i i
i i
P X P X
= =
⎧ ⎫ ⎧ ⎫
> = >⎨ ⎬ ⎨ ⎬
⎩ ⎭ ⎩ ⎭
∑ ∑ 
 ≈ 15.5 20
20
P Z
⎧ ⎫−
>⎨ ⎬
⎩ ⎭
 
 = P{Z > −1.006} 
 ≈ .8428 
 
5. Letting Xi denote the i
th roundoff error it follows that 
50
1
i
i
E X
=
⎡ ⎤
⎢ ⎥
⎣ ⎦
∑ = 0, 
 Var
50
1
i
i
X
=
⎛ ⎞
⎜ ⎟⎝ ⎠∑ = 50 Var(X1) = 50/12, where the last equality uses that .5 + X is uniform (0, 1) 
and so Var(X) = Var(.5 + X) = 1/12. Hence, 
 
 { }3iP X >∑ ≈ P{⏐N(0, 1)⏐ > 3(12/50)1/2} by the central limit theorem 
 = 2P{N(0, 1) > 1.47 = .1416 
 
6. If Xi is the outcome of the i
th roll then E[Xi] = 7/2 Var(Xi) = 35/12 and so 
 
79 79
1 1
300 300.5i i
i i
P X P X
= =
⎧ ⎫ ⎧ ⎫
≤ = ≤⎨ ⎬ ⎨ ⎬
⎩ ⎭ ⎩ ⎭
∑ ∑ 
 ≈ 
1/ 2
300.5 79(7 / 2)
(0,1) { (0,1) 1.58}
(79 35/12)
P N P N
⎧ ⎫−≤ = ≤⎨ ⎬
×⎩ ⎭
 = .9429 
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter 8 141 
 
7. 
100
1
525 500
525 (0,1) { (0,1) .5}
(100 25)
i
i
P X P N P N
=
⎧ ⎫⎧ ⎫ −⎪ ⎪> ≈ > = >⎨ ⎬ ⎨ ⎬
×⎪ ⎪⎩ ⎭ ⎩ ⎭
∑ = .3085 
 
 where the above uses that an exponential with mean 5 has variance 25. 
 
8. If we let Xi denote the life of bulb i and let Ri be the time to replace bulb i then the desired 
probability is 
100 99
1 1
550i i
i i
P X R
= =
⎧ ⎫
+ ≤⎨ ⎬
⎩ ⎭
∑ ∑ . Since i iX R+∑ ∑ has mean 100 × 5 + 99 × .25 = 
524.75 and variance 2500 + 99/48 = 2502 it follows that the desired probability is 
approximately equal to P{N(0, 1) ≤ [550 − 524.75]/(2502)1/2} = P{N(0, 1) ≤ .505} = .693 
 It should be noted that the above used that 
 
 Var(Ri) = Var
1
Unif[0,1]
2
⎛ ⎞
⎜ ⎟⎝ ⎠
 = 1/48 
 
9. Use the fact that a gamma (n, 1) random variable is the sum of n independent exponentials 
with rate 1 and thus has mean and variance equal to n, to obtain: 
 
 .01
X n
P
n
⎧ ⎫−
>⎨ ⎬
⎩ ⎭
 = { }/ .01P X n n n− > 
 ≈ { }(0,1) .01P N n> 
 = { }2 (0,1) .01P N n> 
 
 Now P{N(0, 1) > 2.58} = .005 and so n = (258)2. 
 
10. If Wn is the total weight of n cars and A is the amount of weight that the bridge can withstand 
then Wn − A is normal with mean 3n − 400 and variance .09n + 1600. Hence, the probability 
of structural damage is 
 
 P{Wn − A ≥ 0} ≈ { }(400 3 ) / .09 1600P Z n n≥ − + 
 
 Since P{Z ≥ 1.28} = .1 the probability of damage will exceed .1 when n is such that 
 
 400 − 3n ≤ 1.28 .09 1600n + 
 
 The above will be satisfied whenever n ≥ 117. 
 
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
142 Chapter 8 
 
12. Let Li denote the life of component i. 
 
 
100
1
1
1000 50(101)
10ii
E L
=
⎡ ⎤
= +⎢ ⎥
⎣ ⎦
∑ = 1505 
 Var
2100 100 100
2 2
1 1 1
1
10 (100) (100)(101)
10 100ii i i
i
L i
= = =
⎛ ⎞ ⎛ ⎞= + = + +⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠∑ ∑ ∑ 
 
 Now apply the central limit theorem to approximate. 
 
13. (a) 
74
{ 80} 15/ 7
14 /5
X
P X P
⎧ ⎫−
> = >⎨ ⎬
⎩ ⎭
 ≈ PPZ > 2.14} ≈ .0162 
 
 (b) 
74
{ 80} 24 / 7
14 /8
Y
P Y P
⎧ ⎫−
> = >⎨ ⎬
⎩ ⎭
 ≈ P{Z > 3.43} ≈ .0003 
 
 (c) Using that ( ) 196 / 64 196 / 25SD Y X− = + ≈ 3.30 we have 
 
 { 2.2}P Y X− > = { }/3.30 2.2 / 3.30}P Y X− > 
 ≈ P{Z > .67} ≈ .2514 
 
 (d) same as in (c) 
 
14. Suppose n components are in stock. The probability they will last for at least 2000 hours is 
 
 p = 
1
2000 100
2000
30
n
i
i
n
P X P Z
n=
⎧ ⎫ ⎧ ⎫−≥ ≈ ≥⎨ ⎬ ⎨ ⎬
⎩ ⎭⎩ ⎭
∑ 
 
 where Z is a standard normal random variable. Since 
 
 .95 = P{Z ≥ −1.64} it follows that p ≥ .95 if 
 
 
2000 100
30
n
n
−
 ≤ −1.64 
 
 or, equivalently, 
 
 (2000 − 100n)/ n ≤ −49.2 
 
 and this will be the case if n ≥ 23. 
 
15. 
10,000
1
2,700,000i
i
P X
=
⎧ ⎫⎪ ⎪>⎨ ⎬
⎪ ⎪⎩ ⎭
∑ ≈ P{Z ≥ (2,700,000 − 2,400,000)/(800 ⋅ 100)} = P{Z ≥ 3.75} ≈ 0 
 
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter 8 143 
 
16. (a) Number AJ’s jobs, let Xi be the time it takes to do job i, and let XA = 
20
1 ii
X
=∑ be the time 
that it takes AJ to finish all 20 jobs. Because 
 
[ ] 20(50) 1000,AE X = = Var(XA) = 20(100) = 2000 
 
 the central limit theorem gives that 
 
1000 900 1000
{ 900}
2000 2000
A
A
X
P X P
−⎧ ⎫−
≤ = ≤⎨ ⎬
⎩ ⎭
 
 
{ 2.236}
1 (2.236) .013
P Z
Φ
≈ ≤ −
= − =
 
 
 (b) Similarly, if we let XM be the time that it takes MJ to finish all of her 20 jobs, then by the 
central limit theorem XM is approximately normal with mean and variance 
 
[ ] 20(52) 1040,ME X = = Var(XM) = 20(225) = 4500 
 
 Thus 
 
1040 900 1040
{ 900}
4500 4500
M
M
X
P X P
−⎧ ⎫−
≤ = ≤⎨ ⎬
⎩ ⎭
 
 
{ 2.087}
1 (2.087) 0.18
P Z
Φ
≈ ≤ −
= − =
 
 
 (c) Because the sum of independent normal random variables is also normal, M AD X X≡ − 
is approximately normal with mean and variance 
 
E[D] = 1040 − 1000 = 40, Var(D) = 4500 + 2000 = 6500 
 
 Hence, 
 
40 40
{ 0}
6500 6500
D
P D P
⎧ ⎫− −
> = ≥⎨ ⎬
⎩ ⎭
 
 
{ .4961}
(.4961) .691
P Z
Φ
≈ ≥ −
= =
 
 
 Thus even though AJ is more likely than not to finish earlier than MJ, MJ has the better 
chance to finish within 900 minutes. 
 
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
144 Chapter 8 
 
19. Let Yi denote the additional number of fish that need to be caught to obtain a new type when 
there are at present i distinct types. Then Yi is geometric with parameter 
4
4
i−
. 
 
 E[Y] = 
3
0
4 4 25
1 4
3 2 3ii
E Y
=
⎡ ⎤
= + + + =⎢ ⎥
⎣ ⎦
∑ 
 
 Var[Y] = Var
3
0
4 130
2 12
9 9ii
Y
=
⎛ ⎞
= + + =⎜ ⎟⎝ ⎠∑ 
 
 Hence, 
 
 
25 25 1300 1
3 3 9 10
P Y
⎧ ⎫⎪ ⎪− > ≤⎨ ⎬
⎪ ⎪⎩ ⎭
 
 
 and so we can take a = 
25 1300
3
−
, b = 
25 1300
3
+
. 
 
 Also, 
 
2
25 130 1
3 10130 9
P Y a
a
⎧ ⎫− > ≤ =⎨ ⎬
+⎩ ⎭
 when a = 
1170
3
. 
 
 Hence 
25 1170
3
P Y
⎧ ⎫+⎪ ⎪>⎨ ⎬
⎪ ⎪⎩ ⎭
 ≤ .1. 
 
21. g(x) = xn(n−1) is convex. Hence, by Jensen’s Inequality 
 
 E[Y n/(n−1)] ≥ E[Y])n/(n−1) Now set Y = X n−1 and so 
 E[X n] ≥ (E[X n−1])n/(n−1) or (E[X n])1/n ≥ (E[X n−1])1/(n−1) 
 
22. No 
 
23. (a) 20/26 ≈ .769 
 
 (b) 20/(20 + 36) = 5/14 ≈ .357 
 
 (d) p ≈ P{Z ≥ (25.5 − 20)/ 20 } ≈ P{Z ≥ 1.23} ≈ .1093 
 
 (e) p = .112184 
 
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter 8 145 
 
Theoretical Exercises 
 
1. This follows immediately from Chebyshev’s inequality. 
 
2. P{D >α} = P{⏐X − μ⏐ > αμ} ≤ 
2
2 2 2 2
1
r
ς
α μ α
= 
 
3. (a) 
λ λ
λ
= 
 
 (b) /(1 )
(1 )
np
np p
np p
= −
−
 
 
 (c) answer = 1 
 
 (d) 
1/ 2
3
1/12
= 
 
 (e) answer = 1 
 
 (d) answer = /μ σ 
 
4. For ε > 0, let δ > 0 be such that ⏐g(x) − g(c) ⏐ < ε whenever ⏐x − c⏐ ≤ δ. Also, let B be such 
that ⏐g(x) ⏐ < B. Then, 
 
 E[g(Zn)] = ( ) ( ) ( ) ( )n n
x c x c
g x dF x g x dF x
δ δ− ≤ − >
+∫ ∫ 
 ≤ (ε + g(c))P{⏐Zn − c⏐ ≤ δ} + BP{⏐Zn − c⏐ > δ} 
 
 In addition, the same equality yields that 
 
 E[g(Zn)] ≥ (g(c) − ε)P{⏐Zn − c⏐ ≤ δ} − BP{⏐Zn − c⏐ > δ} 
 
 Upon letting n → ∞ , we obtain that 
 
 lim sup E[g(Zn)] ≤ g(c) + ε 
 lim inf E[g(Zn)] ≥ g(c) − ε 
 
 The result now follows since ε is arbitrary. 
 
5. Use the notation of the hint. The weak law of large numbers yields that 
 
 1lim {( ... ) / } 0n
n
P X X n c ε
→∞
+ + − > = 
 
 Since X1 + … + Xn is binomial with parameters n, x, we have 
 
 1
1
...
( / ) (1 )
n
k n kn
k
nX X
E f f k n x x
kn
−
=
⎡ ⎤ ⎛ ⎞+ +⎛ ⎞ = −⎜ ⎟⎢ ⎥ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦
∑ 
 
 The result now follows from Exercise 4. 
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
146 Chapter 8 
 
6. E[X] = 
1 1
{ } { }
k
i i k
i P X i i P X i
∞
= = +
= + =∑ ∑ 
 ≥ 
1
{ }
k
i
i P X k
=
=∑ 
 = P{X = k}k(k + 1)/2 
 ≥ 
2
{ }
2
k
P X k= 
 
7. Take logs and apply the central limit theorem 
 
8. It is the distribution of the sum of t independent exponentials each having rate λ. 
 
9. 1/2 
 
10. Use the Chernoff bound: e−tiM(t) = ( 1)
te tieλ − − will obtain its minimal value when t is chosen 
to satisfy 
 
 λet = i, and this value of t is negative provided i < λ. 
 
 Hence, the Chernoff bound gives 
 
 P{X ≤ i} ≤ ei−λ(λ/i)i 
 
11. e−tiM(t) = (pet + q)ne−ti and differentiation shows that the value of t that minimizes it is such 
that 
 npet = i(pet + q) or et = 
( )
iq
n i p−
 
 
 Using this value of t, the Chernoff bound gives that 
 
 P{X ≥ i} ≤ ( ) /( )
n
i i iiq q n i p iq
n i
⎛ ⎞+ −⎜ ⎟⎝ ⎠−
 
 = 
( ) ( )
( )
n i i
i i n
nq n i p
i q n i
−
−
 
 
12. 1 = E[eθX] ≥ eθE[X] by Jensen’s inequality. 
 
 Hence, θE[X] ≤ 0 and thus θ > 0. 
 
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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Probabilidade, um curso moderno com aplicações - Sheldon Ross - Solutions/chap-7-11pt.pdf
104 
Chapter 7 
 
Problems 
 
1. Let X = 1 if the coin toss lands heads, and let it equal 0 otherwise. Also, let Y denote the 
value that shows up on the die. Then, with p(i, j) = P{X = i, Y = j} 
 
 E[return] = 
6 6
1 1
2 (1, ) (0, )
2j j
j
jp j p j
= =
+∑ ∑ 
 = 
1
(42 10.5)
12
+ = 52.5/12 
 
2. (a) 6 ⋅ 6 ⋅ 9 = 324 
 
 (b) X = (6 − S)(6 − W)(9 − R) 
 
 (c) E[X] = 6(6)(6)P{S = 0, W = 0, R = 3} + 6(3)(9)P{S = 0, W = 3, R = 0} 
 + 3(6)(9)P{S = 3, W = 0, R = 0} + 6(5)(7)P{S = 0, W = 1, R = 2} 
 + 5(6)(7)P{S = 1, W = 0, R = 2} + 6(4)(8)P{S = 0, W = 2, R = 1} 
 + 4(6)(8)P{S = 2, W = 0, R = 1} + 5(4)(9)P{S = 1, W = 2, R = 0} 
 + 4(5)(9)P{S = 2, W = 1, R = 0} + 5(5)(8)P{S = 1, W = 1, R = 1} 
 
 = 
9 6 9 6 61
216 324 420 6 384 9 360 6 200(6)(6)(9)
21 3 3 2 2 2
3
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
+ + ⋅ + + +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎣ ⎦
⎜ ⎟⎝ ⎠
 
 ≈ 198.8 
 
3. If the first win is on trial N, then the winnings is W = 1 − (N − 1) = 2 − N. Thus, 
 
 (a) P(W > 0) = P(N = 1) = 1/2 
 (b) P(W < 0) = P(N > 2) = 1/4 
 (c) E[W] = 2 − E[N] = 0 
 
4. 
1 1
2
0 0 0
1 1
0 0 0
1 1
0 0 0
1
[ ] / 2 1/ 6
1
[ ] / 2 1/ 4
1
[ ] 1/ 2
y
y
y
E XY xy dxdy y dy
y
E X x dxdy y dy
y
E Y y dxdy ydy
y
= = =
= = =
= = =
∫ ∫ ∫
∫ ∫ ∫
∫ ∫ ∫
 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter 7 105 
 
5. The joint density of the point (X, Y) at which the accident occurs is 
 
 f(x, y) = 
1
9
, −3/2 < x, y < 3/2 
 = f(x) f(y) 
 
 where 
 
 f(a) = 1/3, −3/2 < a < 3/2. 
 
 Hence we may conclude that X and Y are independent and uniformly distributed on 
 (−3/2, 3/2) Therefore, 
 
 E[⏐X⏐ + ⏐Y⏐] = 
3/ 2 3/ 2
3/ 2 0
1 4
2 3/ 2
3 3
x dx xdx
−
= =∫ ∫ . 
 
6. 
10 10
1 1
[ ]i i
i i
E X E X
= =
⎡ ⎤
=⎢ ⎥
⎣ ⎦
∑ ∑ = 10(7/2) = 35. 
 
8. E[number of occupied tables] = 
1 1
[ ]
N N
i i
i i
E X E X
= =
⎡ ⎤
=⎢ ⎥
⎣ ⎦
∑ ∑ 
 Now, 
 
 E[Xi] = P{i
th arrival is not friends with any of first i − 1} 
 = (1 − p)i−1 
 
 and so 
 
 E[number of occupied tables] = 1
1
(1 )
N
i
i
p −
=
−∑ 
 
7. Let Xi equal 1 if both choose item i and let it be 0 otherwise; let Yi equal 1 if neither A nor B 
chooses item i and let it be 0 otherwise. Also, let Wi equal 1 if exactly one of A and B choose 
item i and let it be 0 otherwise. Let 
 
 X = 
10
1
i
i
X
=
∑ , Y = 
10
1
i
i
Y
=
∑ , W = 
10
1
i
i
W
=
∑ 
 
 (a) E[X] = 
10
1
[ ]i
i
E X
=
∑ = 10(3/10)2 = .9 
 
 (b) E[Y] = 
10
1
[ ]i
i
E Y
=
∑ = 10(7/10)2 = 4.9 
 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
106 Chapter 7 
 
 (c) Since X + Y + W = 10, we obtain from parts (a) and (b) that 
 
 E[W] = 10 − .9 − 4.9 = 4.2 
 
 Of course, we could have obtained E[W] from 
 
 E[W] = 
10
1
[ ]i
i
E W
=
∑ = 10(2)(3/10)(7/10) = 4.2 
 
9. Let Xj equal 1 if urn j is empty and 0 otherwise. Then 
 E[Xj] = P{ball i is not in urn j, i ≥ j} = (1 1/ )
n
i j
i
=
−∏ 
 Hence, 
 (a) E[number of empty urns] = 
1
(1 1/ )
n n
j i j
i
= =
−∑∑ 
 (b) P{none are empty} = P{ball j is in urn j, for all j} 
 = 
1
1/
n
j
j
=
∏ 
 
10. Let Xi equal 1 if trial i is a success and 0 otherwise. 
 
 (a) .6. This occurs when P{X1 = X2 = X3} = 1. It is the largest possible since 
 1.8 = { 1} 3 { 1}i iP X P X= = =∑ . Hence, P{Xi = 1} = .6 and so 
 
 P{X = 3} = P{X1 = X2 = X3 = 1} ≤ P{Xi = 1} = .6. 
 
 (b) 0. Letting 
 X1 = 
1 if .6
0 otherwise
U ≤
 , X2 = 
1 if .4
0 otherwise
U ≤
 , X3 = 
1 if .3
0 otherwise
U ≤
 
 
 Hence, it is not possible for all Xi to equal 1. 
11. Let Xi equal 1 if a changeover occurs on the i
th flip and 0 otherwise. Then 
 
 E[Xi] = P{i − 1 is H, i is T} + P{i − 1 is T, i is H} 
 = 2(1 − p)p, i ≥ 2. 
 
 E[number of changeovers] = 
1
[ ]
n
i i
i
E X E X
=
⎡ ⎤ =⎣ ⎦∑ ∑ = 2(n − 1)(1 − p) 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter 7 107 
 
12. (a) Let Xi equal 1 if the person in position i is a man who has a woman next to him, and let it 
equal 0 otherwise. Then 
 
 E[Xi] = 
1
, if 1, 2n
2 2 1
1 ( 1)( 2)
1 , otherwise
2 (2 1)(2 2)
n
i
n
n n
n n
⎧ =⎪ −⎪
⎨ ⎡ ⎤− −⎪ −⎢ ⎥− −⎪ ⎣ ⎦⎩
 
 
 Therefore, 
 
 
1
n
i
i
E X
=
⎡ ⎤
⎢ ⎥
⎣ ⎦
∑ = 
2
1
[ ]
n
i
i
E X
=
∑ 
 = 
1 2 3
(2 2)
2 2 1 4 2
n n
n
n n
⎛ ⎞+ −⎜ ⎟⎝ ⎠− −
 
 = 
23
4 2
n n
n
−
−
 
 
 (b) In the case of a round table there are no end positions and so the same argument as in part 
(a) gives the result 
 
 
2( 1)( 2) 3
1
(2 1)(2 2) 4 2
n n n
n
n n n
⎡ ⎤− −
− =⎢ ⎥− − −⎣ ⎦
 
 
 where the right side equality assumes that n > 1. 
 
13. Let Xi be the indicator for the event that person i is given a card whose number matches his 
age. Because only one of the cards matches the age of the person i 
 
 
1000 1000
1 1
[ ]i i
i i
E X E X
= =
⎡ ⎤
=⎢ ⎥
⎣ ⎦
∑ ∑ = 1 
 
14. The number of stages is a negative binomial random variable with parameters m and 1 − p. 
Hence, its expected value is m/(1 − p). 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
108 Chapter 7 
 
15. Let Xi,j , i ≠ j equal 1 if i and j form a matched pair, and let it be 0 otherwise. 
 
 Then 
 E[Xi,j] = P{i, j is a matched pair} = 
1
( 1)n n −
 
 Hence, the expected number of matched pairs is 
 
 , ,
1 1
[ ]
2 ( 1) 2i j i ji j i j
n
E X E X
n n< <
⎡ ⎤ ⎛ ⎞
= = =⎢ ⎥ ⎜ ⎟ −⎝ ⎠⎢ ⎥⎣ ⎦
∑ ∑ 
 
16. E[X] = 
2
2
/ 2
/ 21
2 2
x
y
y x
e
y e dy
π π
−
−
>
=∫ 
 
17. Let Ii equal 1 if guess i is correct and 0 otherwise. 
 
 (a) Since any guess will be correct with probability 1/n it follows that 
 
 E[N] = 
1
[ ] / 1
n
i
i
E I n n
=
= =∑ 
 
 (b) The best strategy in this case is to always guess a card which has not yet appeared. For 
this strategy, the ith guess will be correct with probability 1/(n − i + 1) and so 
 
 E[N] = 
1
1/( 1)
n
i
n i
=
− +∑ 
 
 (c) Suppose you will guess in the order 1, 2, …, n.
That is, you will continually guess card 1 
until it appears, and then card 2 until it appears, and so on. Let Ji denote the indicator 
variable for the event that you will eventually be correct when guessing card i; and note 
that this event will occur if among cards 1 thru i, card 1 is first , card 2 is second, …, and 
card i is the last among these i cards. Since all i! orderings among these cards are equally 
likely it follows that 
 
 E[Ji] = 1/i! and thus E[N] = 
1 1
1/ !
n n
i
i i
E J i
= =
⎡ ⎤
=⎢ ⎥
⎣ ⎦
∑ ∑ 
 
18. E[number of matches] = 
52
1
1 match on card
,
0 ---i i
i
E I I
⎡ ⎤ ⎧
= ⎨⎢ ⎥
⎩⎣ ⎦
∑ 
 = 
1
52 4
13
= since E[Ii] = 1/13 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter 7 109 
 
19. (a) E[time of first type 1 catch] − 1 =
1
1
1
p
− using the formula for the mean of a geometric 
random variable. 
 
 (b) Let 
 
 Xj = 
1 a type is caught before a type 1
0 otherwise.
j⎧
⎨
⎩
 
 
 Then 
 
 
1 1
[ ]j j
j j
E X E X
≠ ≠
⎡ ⎤
=⎢ ⎥
⎢ ⎥⎣ ⎦
∑ ∑ 
 
1
{type before type 1}
j
P j
≠
=∑ 
 1
1
/( )j j
j
P P P
≠
= +∑ , 
 
 where the last equality follows upon conditioning on the first time either a type 1 or type j 
is caught to give. 
 
 P{type j before type 1} = P{j⏐j or 1} = 
1
j
j
P
P P+
 
 
20. Similar to (b) of 19. Let 
 
 Xj = 
1 ball removed before ball 1
0 ---
j⎧
⎨
⎩
 
 
 
1 1 1
[ ] {ball before ball 1}j j
j j j
E X E X P j
≠ ≠ ≠
⎡ ⎤
= =⎢ ⎥
⎢ ⎥⎣ ⎦
∑ ∑ ∑ 
 
1
{ or 1}
j
P j j
≠
=∑ 
 
1
( ) / (1) ( )
j
W j W W j
≠
= +∑ 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
110 Chapter 7 
 
21. (a) 
3 97100 1 364
365
3 365 365
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠
 
 
 (b) Let Xj = 
1 if day is someones birthday
0 ---
j⎧
⎨
⎩
 
 
100365 365
1 1
364
[ ] 365 1
365j j
E X E X
⎡ ⎤⎡ ⎤ ⎛ ⎞= = −⎢ ⎥⎢ ⎥ ⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦ ⎣ ⎦
∑ ∑ 
 
22. From Example 3g, 1 + 
6 6 6 6
6
5 4 3 2
+ + + + 
 
23. 
5 8 5 8
1 1 1 1
[ ] ( )i i i iE X Y E X E Y
⎡ ⎤
+ = +⎢ ⎥
⎣ ⎦
∑ ∑ ∑ ∑ 
 
2 3 3 147
5 8
11 20 120 110
= + = 
 
24. Number the small pills, and let Xi equal 1 if small pill i is still in the bottle after the last large 
pill has been chosen and let it be 0 otherwise, i = 1, …, n. Also, let Yi, i = 1, …, m equal 1 if 
the ith small pill created is still in the bottle after the last large pill has been chosen and its 
smaller half returned. 
 
 Note that X = 
1 1
n m
i i
i i
X Y
= =
+∑ ∑ . Now, 
 
 E[Xi] = P{small pill i is chosen after all m large pills} 
 = 1/(m + 1) 
 
 E[Yi] = P{ith created small pill is chosen after m − i existing large pills} 
 = 1/(m − i + 1) 
 
 Thus, 
 
 (a) E[X] = n/(m + 1) + 
1
1/( 1)
m
i
m i
=
− +∑ 
 
 (b) Y = n + 2m − X and thus 
 
 E[Y] = n + 2m − E[X] 
 
25. P{N ≥ n} P{X1 ≥ X2 ≥ … ≥ Xn} = 
1
!n
 
 E[N] = 
1 1
1
{ }
!n n
P N n e
n
∞ ∞
= =
≥ = =∑ ∑ 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter 7 111 
 
26. (a) E[max] = 
1
0
{max }P t dt>∫ 
 = 
1
0
(1 {max )}P t dt− ≤∫ 
 = 
1
0
(1 /
1
n nt dt
n
− =
+∫ 
 
 (b) E[min] = 
1
0
{min }4p t t>∫ 
 = 
1
0
1
(1 )
1
nt dt
n
− =
+∫ 
 
27. Let X denote the number of items in a randomly chosen box. Then, with Xi equal to 1 if item 
i is in the randomly chosen box 
 
 E[X] = 
101 101
1 1
101
[ ] 10
10i ii i
E X E X
= =
⎡ ⎤
= = >⎢ ⎥
⎣ ⎦
∑ ∑ 
 
 Hence, X can exceed 10, showing that at least one of the boxes must contain more than 10 items. 
 
28. We must show that for any ordering of the 47 components there is a block of 12 consecutive 
components that contain at least 3 failures. So consider any ordering, and randomly choose a 
component in such a manner that each of the 47 components is equally likely to be chosen. 
Now, consider that component along with the next 11 when moving in a clockwise manner 
and let X denote the number of failures in that group of 12. To determine E[X], arbitrarily 
number the 8 failed components and let, for i = 1, …, 8, 
 
 Xi = 
1, if failed component is among the group of 12 components
0, otherwise
i⎧
⎨
⎩
 
 
 Then, 
 
 X = 
8
1
i
i
X
=
∑ 
 
 and so 
 
 E[X] = 
8
1
[ ]i
i
E X
=
∑ 
 
 Because Xi will equal 1 if the randomly selected component is either failed component 
number i or any of its 11 neighboring components in the counterclockwise direction, it 
follows that E[Xi] = 12/47. Hence, 
 
 E[X] = 8(12/47) = 96/47 
 
 Because E[X] > 2 it follows that there is at least one possible set of 12 consecutive 
components that contain at least 3 failures. 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
112 Chapter 7 
 
29. Let Xii be the number of coupons one needs to collect to obtain a type i. Then 
 
 
)
1 2
3 4
1 2
3, 4
1 2 3 4
[ ] 8, 1,2
{ ] 8 / 3, 3,4
[min( , )] 4
[min( , )] 2, 1,2, 3,4
[min( , )] 4 / 3
[min( , , )] 8 / 5, 3,4
[min( , )] 8 / 7, 1,2
[min( , , , ] 1
i
i
i j
j
i
E X i
E X i
E X X
E X X i j
E X X
E X X X j
E X X X i
E X X X X
= =
= =
=
= = =
=
= =
= =
=
 
 
 (a) E[max Xi] = 2 ⋅ 8 + 2 ⋅ 8/3 − (4 + 4 ⋅ 2 + 4/3) + (2 ⋅ 8/5 + 2 ⋅ 8/7) − 1 = 
437
35
 
 
 (b) E[max(X1, X2)] = 8 + 8 − 4 = 12 
 
 (c) E[max(X3, X4)] = 8/3 + 8/3 − 4/3 = 4 
 
 (d) Let Y1 = max(X1, X2), Y2 = max(X3, X4). Then 
 
 E[max(Y1, Y2)] = E[Y1] + E[Y2] − E[min(Y1, Y2)] 
 
 giving that 
 
 E[min(Y1, Y2)] = 12 + 4 − 
437 123
35 35
= 
 
30. E[(X − Y)]2 = Var(X − Y) = Var(X) + Var(−Y) = 2σ2 
 
31. 
10
1
1
Var 10 Var( )i
i
X X
=
⎛ ⎞
=⎜ ⎟⎝ ⎠∑ . Now 
 Var(X1) = 
2 2
1[ ] (7 / 2)E X − 
 = [1 + 4 + 9 + 16 + 25 + 36]/6 − 49/4 
 = 35/12 
 
 and so 
10
1
Var i
i
X
=
⎛ ⎞
⎜ ⎟⎝ ⎠∑ = 350/12. 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter 7 113 
 
32. Use the notation in Problem 9, 
 
 X = 
1
n
j
j
X
=
∑ 
 
 where Xj is 1 if box j is empty and 0 otherwise. Now, with 
 
 E[Xj] = P{Xj = 1} = (1 1/ )
n
i j
i
=
−∏ , we have that 
 Var(Xj) = E[Xj](1 − E[Xj]). 
 
 Also, for j < k 
 E[XjXk] = 
1
(1 1/ ) (1 2 / )
k n
i j i k
i i
−
= =
− −∏ ∏ 
 Hence, for j < k, 
 
 Cov(Xj, Xk) = 
1
(1 1/ ) (1 2 / ) (1 1/ ) (1 1/ )
k n n n
i j i k i j i k
i i i i
−
= = = =
− − − − −∏ ∏ ∏ ∏ 
 Var(X ) = 
1
[ ](1 [ ]) 2Cov( , )
n
j j j k
j
E X E X X X
=
− +∑ 
 
33. (a) E[X2 + 4X + 4] = E[X2] + 4E[X] + 4 = Var(X) + E2[X] + 4E[X] + 4 = 14 
 
 (b) Var(4 + 3X) = Var(3X) = 9Var(X) = 45 
 
34. Let Xj = 
1 if couple are seated next to each other
0 otherwise
j⎧
⎨
⎩
 
 
 (a) 
10
1
2 20
10
19 19j
E X
⎡ ⎤
= =⎢ ⎥
⎣ ⎦
∑ ; P{Xj = 1} = 219 since there are 2 people seated next to wife j 
and so the probability that one of them is her husband is 
2
19
. 
 
 (b) For i ≠ j, E[XiXj] = P{Xi = 1, Xj = 1} 
 = P{Xi = 1}P{Xj = 1⏐Xi = 1} 
 = 
2 2
19 18
 since given Xi = 1 we can regard couple i as a single entity. 
 
 Var
210
1
2 2 2 2 2
10 1 10 9
19 19 19 18 19jj
X
=
⎛ ⎞ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + ⋅ −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎢ ⎥⎣ ⎦
∑ 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
114 Chapter 7 
 
35. (a) Let X1 denote the number of nonspades preceding the first ace and X2 the number of 
nonspades between the first 2 aces. It is easy to see that 
 
 P{X1 = i, X2 = j} = P{X1 = j, X2 = i} 
 
 and so X1 and X2 have the same distribution. Now E[X1] = 
48
5
 by the results of Example 
3j and so E[2 + X1 + X2] = 
106
5
. 
 
 (b) Same
method as used in (a) yields the answer 
39 265
5 1
14 14
⎛ ⎞+ =⎜ ⎟⎝ ⎠
. 
 
 (c) Starting from the end of the deck the expected position of the first (from the end) heart is, 
from Example 3j, 
53
14
. Hence, to obtain all 13 hearts we would expect to turn over 
 52 − 53
14
 + 1 = 
13
14
(53). 
 
36. Let Xi = 
1 roll lands on 1
0 otherwise
i⎧
⎨
⎩
, Yi = 
1 roll lands on 2
0 otherwise
i⎧
⎨
⎩
 
 
 Cov(Xi, Yj) = E[Xi Yj] − E[Xi]E[Yj] 
 = 
i
1
 (since X 0 when 
36
1 1
0
36 36
ji j Y i j
i j
⎧− = = =⎪⎪
⎨
⎪ − = ≠⎪⎩
 
 
 Cov , Cov( , )i j i j
i j i j
X Y X Y=∑ ∑ ∑∑ 
 
36
n= − 
 
37. Let Wi, i = 1, 2, denote the i
th outcome. 
 
 Cov(X, Y) = Cov(W1 + W2 , W1 − W2) 
 = Cov(W1, W1) − Cov(W2, W2) 
 = Var(W1) − Var(W2) = 0 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter 7 115 
 
38. E[XY] = 2
0 0
2
x
xy e dydx
∞
−∫ ∫ 
 = 2 2 2
0 0
1 (3) 1
8 8 4
x yx e dx y e dy
Γ∞ ∞− −= = =∫ ∫ 
 
 E[X] = 
2
0 0
2
( ) , ( )
x x
x x
e
xf x dx f x dy
x
∞ −
=∫ ∫ = 2e−2x 
 = 
1
2
 
 
 E[Y] = 
2
0 0
2
( ) , ( )
x
Y Y
e
yf y dy f y dx
x
∞ ∞ −
=∫ ∫ 
 = 
2
0
2 x
y
e
y dxdy
x
∞ ∞ −
∫ ∫ 
 = 
2
0 0
2
x xe
y dydx
x
∞ −
∫ ∫ 
 = 2 2
0
1 (2) 1
4 4 4
xxe dx ye dy
Γ∞ − −= = =∫ ∫ 
 
 Cov(X, Y) = 
1 1 1 1
4 2 4 8
− = 
 
39. Cov(Yn, Yn) = Var(Yn) = 3σ2 
 Cov(Yn, Yn+1) = Cov(Xn + Xn+1 + Xn+2, Xn+1 + Xn+2 + Xn+3) 
 = Cov(Xn+1 + Xn+2, Xn+1 + Xn+2) = Var(Xn+1 + Xn+2) = 2σ2 
 Cov(Yn, Yn+2) = Cov(Xn+2, Xn+2) = σ2 
 Cov(Yn, Yn+j) = 0 when j ≥ 3 
 
40. fY(y) = e
−y
/1 x y ye dx e
y
− −=∫ . In addition, the conditional distribution of X given that Y = y is 
exponential with mean y. Hence, 
 
 E[Y] = 1, E[X] = E[E[X⏐Y]] = E[Y] = 1 
 
 Since, E[XY] = E[E[XY⏐Y]] = E[YE[X⏐Y]] = E[Y2] = 2 (since Y is exponential with mean 1, it 
follows that E[Y2] = 2). Hence, Cov(X, Y) = 2 − 1 = 1. 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
116 Chapter 7 
 
41. The number of carp is a hypergeometric random variable. 
 
 E[X] = 
60
10
 = 6 
 
 Var(X) = 
20(80) 3 7 336
99 10 10 99
= from Example 5c. 
 
42. (a) Let Xi = 
1 pair consists of a man and a woman
0 otherwise
i⎧
⎨
⎩
 
 
 E[Xi] = P{Xi = 1} = 
10
19
 
 E[XiXj] = P{Xi = 1, Xj = 1} = P{Xi = 1}P{Xj = 1⏐X2 = 1} 
 = 
10 9
19 17
, i ≠ j 
 
 
10
1
100
19i
E X
⎡ ⎤
=⎢ ⎥
⎣ ⎦
∑ 
 
 Var 
210
2
1
10 10 10 9 10 900 18
10 1 10 9
19 19 19 17 19 17(19)i
X
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + ⋅ − =⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎢ ⎥⎣ ⎦
∑ 
 
 (b) Xi = 
1 pair consists of a married couple
0 otherwise
i⎧
⎨
⎩
 
 
 E[Xi] = 
1
19
, E[XiXj] = P{Xi = 1}P{Xj = 1⏐Xi = 1} = 
1 1
19 17
, i ≠ j 
 
 
10
1
10
19i
E X
⎡ ⎤
=⎢ ⎥
⎣ ⎦
∑ 
 Var
210
2
1
1 15 1 1 1 180 18
10 10 9
19 19 19 17 19 17(19)i
X
⎡ ⎤⎛ ⎞ ⎛ ⎞= + ⋅ − =⎢ ⎥⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎢ ⎥⎣ ⎦
∑ 
 
43. E[R] = n(n + m + 1)/2 
 
 Var(R) = 
2
2
1 1
1 2
n m
i
i
nm n m
n m n m
+
=
⎡ ⎤
⎢ ⎥+ +⎛ ⎞⎢ ⎥− ⎜ ⎟⎢ ⎥⎝ ⎠+ − +
⎢ ⎥
⎣ ⎦
∑
 
 The above follows from Example 3d since when F = G, all orderings are equally likely and 
the problem reduces to randomly sampling n of the n + m values 1, 2, …, n + m. 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter 7 117 
 
44. From Example 8l 
n nm
n m n m
+
+ +
. Using the representation of Example 2l the variance can be 
computed by using 
 
 E[I1Il+j] = 
0
1
1 2
n m n
n m n m n m
⎧
⎪
−⎨
⎪ + + − + −⎩
 
, 1
, 1 1
j
n j
=
− ≤ <
 
 
 E[IiIi+j] = 
0
( 1)( 1)
( )( 1)( 2)( 3)
mn m n
n m n m n m n m
⎧
⎪
− −⎨
⎪ + + − + − + −⎩
 
, 1
, 1 1
j
n j
=
− ≤ <
 
 
45. (a) 1 2 2 3
1 2 2 3
Cov( , ) 1
2Var( ) Var( )
X X X X
X X X X
+ +
=
+ +
 
 (b) 0 
 
46. E[I1I2] = 
12
1 2
2
[ bank rolls ]P{bank rolls }
i
E I I i i
=
∑ 
 = 2( {roll is greater than }) {bank rolls }
i
P i P i∑ 
 = 21[ ]E I 
 ≥ (E[I1])2 
 = E[I1] E[I2] 
 
47. (a) It is binomial with parameters n − 1 and p. 
 
 (b) Let xi,j equal 1 if there is an edge between vertices i and j, and let it be 0 otherwise. Then, 
Di = ,k i i kX≠∑ , and so, for i ≠ j 
 
 Cov(Di, Dj) = Cov , ,,i k r j
k i r j
X X
≠ ≠
⎛ ⎞
⎜ ⎟
⎝ ⎠
∑ ∑ 
 = , ,( , )i k r j
k i r j
Cov X X
≠ ≠
∑∑ 
 = Cov(Xi, j , Xi, j) 
 = Var(Xi, j) 
 = p(1 − p) 
 
 where the third equality uses the fact that except when k = j and r = i, Xi ,k and Xr, j are 
independent and thus have covariance equal to 0. Hence, from part (a) and the preceding 
we obtain that for i ≠ j, 
 
 ρ(Di, Dj) = 
(1 ) 1
( 1) (1 ) 1
p p
n p p n
− =
− − −
 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
118 Chapter 7 
 
48. (a) E[X] = 6 
 
 (b) E[X⏐Y = 1] = 1 + 6 = 7 
 
 (c) 
2 3 4
1 4 1 4 1 4 1 4
1 2 3 4 (5 6)
5 5 5 5 5 5 5 5
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
 
 
49. Let Ci be the event that coin i is being flipped (where coin 1 is the one having head 
probability .4), and let T be the event that 2 of the first 3 flips land on heads. Then 
 
 P(C1⏐T) = 1 1
1 1 2 2
( ) ( )
( ) ( ) ( ) ( )
P T C P C
P T C P C P T C P C+
 
 = 
2
2 2
3(.4) (.6)
3(.4) (.6) 3(.7) (.3)+
 = .395 
 
 Now, with Nj equal to the number of heads in the final j flips, we have 
 
 E[N10⏐T] = 2 + E[N7⏐T] 
 
 Conditioning on which coin is being used, gives 
 
 E[N7⏐T] = E[N7⏐TC1]P(C1T) + E[N7TC2]P(C2⏐T) = 2.8(.395) + 4.9(.605) = 4.0705 
 
 Thus, E[N10⏐T] = 6.0705. 
 
50. fX⏐Y(x⏐y) = 
/
/
/
0
/ 1
/
x y y
x y
x y y
e e y
e
y
e e y dx
− −
−
∞
− −
=
∫
, 0 < x < ∞ 
 
 Hence, given Y = y, X is exponential with mean y, and so 
 
 E[X 2⏐Y = y] = 2y2 
 
51. fX⏐Y(x⏐y) = 
0
/ 1
,
/
y
y
y
e y
y
e y dx
−
−
=
∫
 0 < x < y 
 E[X 3⏐Y = y] = 3 3
0
1
/ 4
y
x dx y
y
=∫ 
 
52. The average weight, call it E[W], of a randomly chosen person is equal to average weight of 
all the members of the population. Conditioning on the subgroup of that person gives 
 
 E[W] = 
1 1
{ member of subgroup ]
r r
i i i
i i
E W i p w p
= =
=∑ ∑ 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter 7 119 
 
53. Let X denote the number of days until the prisoner is free, and let I denote the initial door 
chosen. Then 
 
 E[X] = E[X⏐I = 1](.5) + E[X⏐I = 2](.3) + E[X⏐I = 3](.2) 
 = (2 + E[X])(.5) + (4 + E[X])(.3) + .2 
 
 Therefore, 
 E[X] = 12 
 
54. Let Ri denote the return from the policy that stops the first time a value at least as large as i 
appears. Also, let X be the first sum, and let pi = P{X = i}. Conditioning on X yields 
 
 E[R5] = 
12
5
2
[ } i
i
E R X i p
=
=∑ 
 = E[R5)(p2 + p3 + p4) + 
12
5
i
i
ip
=
∑ − 7p7 
 = 5
6
[ ]
36
E R + 5(4/36) + 6(5/36) + 8(5/36) + 9(4/36) + 10(3/36) + 11(2/36) + 12(1/36) 
 = 5
6
[ ]
36
E R + 190/36 
 Hence, E[R5] = 19/3 ≈ 6.33. In the same fashion, we obtain that 
 
 E[R6] = 6
10 1
[ ]
36 36
E R + [30 + 40 + 36 + 30 + 22 + 12] 
 
 implying that 
 
 E[R6] = 170/26 ≈ 6.54 
 
 Also, 
 
 E[R8] = 8
15 1
[ ] (140)
36 36
E R + 
 
 or, 
 
 E[R8] = 140/21 ≈ 6.67 
 
 In addition, 
 
 E[R9] = 9
20 1
[ ] (100)
26 36
E R + 
 
 or 
 
 E[R9] = 100/16 = 6.25 
 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
120 Chapter 7 
 
 And 
 
 E[R10] = 10
24 1
[ ] (64)
36 36
E R + 
 or 
 
 E[R10] = 64/12 ≈ 5.33 
 
 The maximum expected return is E[R8]. 
 
55. Let N denote the number of ducks.
Given N = n, let I1, …, In be such that 
 Ii = 
1 if duck is hit
0 otherwise
 i⎧
⎨
⎩
 
 
 E[Number hit⏐N = n] = 
1
n
i
i
E I
=
⎡ ⎤
⎢ ⎥
⎣ ⎦
∑ 
 = 
10
1
.6
[ ] 1 1 ,
n
i
i
E I n
n=
⎡ ⎤⎛ ⎞= − −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
∑ since given 
 
 N = n, each hunter will independently hit duck i with probability .6/n. 
 
 E[Number hit] = 
10
6
0
.6
1 1 6 / !n
n
n e n
n
∞
−
=
⎡ ⎤⎛ ⎞− −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
∑ 
 
56. Let Ii = 
1 elevator stops at floor 
0 otherwise
i⎧
⎨
⎩
. Let X be the number that enter on the ground floor. 
 
 
1 1
1
[ ] 1
kN N
i i
i i
N
E I X k E I X k N
N= =
⎡ ⎤⎡ ⎤ −⎛ ⎞= = = = −⎢ ⎥⎢ ⎥ ⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦ ⎣ ⎦
∑ ∑ 
 10
1 0
1 (10)
!
kN k
i
i k
N
E I N N e
N k
∞
−
= =
⎡ ⎤ −⎛ ⎞= −⎢ ⎥ ⎜ ⎟⎝ ⎠⎣ ⎦
∑ ∑ 
 = N − Ne−10/N = N(1 − e−10/N) 
 
57. 
1
[ ] [ ]
N
i
i
E X E N E X
=
⎡ ⎤
=⎢ ⎥
⎣ ⎦
∑ = 12.5 
 
58. Let X denote the number of flips required. Condition on the outcome of the first flip to 
obtain. 
 
 E[X] = E[X⏐heads]p + E[x⏐tails](1 − p) 
 = [1 + 1/(1 − p)]p + [1 + 1/p](1 − p) 
 = 1 + p/(1 − p) + (1 − p)/p 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter 7 121 
 
59. (a) E[total prize shared] = P{someone wins} = 1 − (1 − p)n+1 
 
 (b) Let Xi be the prize to player i. By part (a) 
 
 
1
1
1
1 (1 )
n
n
i
i
E X p
+
+
=
⎡ ⎤
= − −⎢ ⎥
⎣ ⎦
∑ 
 
 But, by symmetry all E[Xi] are equal and so 
 
 E[X] = [1 − (1 − p)n+1]/(n + 1) 
 
 (c) E[X] = p E[1/(1 + B)] where B, which is binomial with parameters n and p, represents the 
number of other winners. 
 
60. (a) Since the sum of their number of correct predictions is n (one for each coin) it follows 
that one of them will have more than n/2 correct predictions. Now if N is the number of 
correct predictions of a specified member of the syndicate, then the probability mass 
function of the number of correct predictions of the member of the syndicate having more 
than n/2 correct predictions is 
 
 P{i correct} = P{N = i} + P(N = n − i} i > n/2 
 = 2P{N = i} 
 = P{N = i⏐N > n/2} 
 
 (b) X is binomial with parameters m, 1/2. 
 
 (c) Since all of the X + 1 players (including one from the syndicate) that have more than n/2 
correct predictions have the same expected return we see that 
 
 (X + 1) ⋅ Payoff to syndicate = m + 2 
 
 implying that 
 
 E[Payoff to syndicate] = (m + 2) E[(X + 1)−1] 
 
 (d) This follows from part (b) above and (c) of Problem 56. 
 
61. (a) P(M ≤ x) = 1
1 1
( )
( ) ( ) ( ) (1 )
1 (1 ) ( )
n n
n n
pF x
P M x N n P N n F x p p
p F x
∞ ∞
−
= =
≤ = = = − =
− −∑ ∑ 
 
 (b) P(M ≤ x⏐N = 1) = F(x) 
 
 (c) P(M ≤ x⏐N > 1) = F(x)P(M ≤ x) 
 
 (d) P(M ≤ x) = P(M ≤ x⏐N = 1)P(N = 1) + P(M ≤ x⏐N > 1)P(N > 1) 
 = F(x)p + F(x)P(M ≤ x)(1 − p) 
 
 again giving the result 
 
 P(M ≤ x) = ( )
1 (1 ) ( )
pF x
p F x− −
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
122 Chapter 7 
 
62. The result is true when n = 0, so assume that 
 
 P{N(x) ≥ n} = xn/(n − 1)! 
 
 Now, 
 
 P{N(x) ≥ n + 1} = 
1
1
0
{ ( ) 1 }P N x n U y dy≥ + =∫ 
 = 
0
{ ( ) }
x
P N x y n dy− ≥∫ 
 = 
0
{ ( ) }
x
P N u n du≥∫ 
 = 1
0
/( 1)!
x
nu n− −∫ du by the induction hypothesis 
 = xn/n! 
 
 which completes the proof. 
 
 (b) E[N(x)] = 
0 0 0
{ ( ) { ( ) 1} / !n x
n n n
P N x n P N x n x n e
∞ ∞ ∞
= = =
> = ≥ + = =∑ ∑ ∑ 
 
63. (a) Number the red balls and the blue balls and let Xi equal 1 if the i
th red ball is selected and 
let it by 0 otherwise. Similarly, let Yj equal 1 if the j
th blue ball is selected and let it be 0 
otherwise. 
 
 Cov , Cov( , )i j i j
i j i j
X Y X Y
⎛ ⎞
=⎜ ⎟
⎝ ⎠
∑ ∑ ∑∑ 
 Now, 
 E[Xi] = E[Yj] = 12/30 
 
 E[XiYj] = P{red ball i and blue ball j are selected} = 
28 30
10 12
⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
 
 Thus, 
 Cov(X, Y) = 2
28 30
80 (12 / 30)
10 12
⎡ ⎤⎛ ⎞ ⎛ ⎞
−⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦
 = −96/145 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter 7 123 
 
 (b) E[XY⏐X] = XE[Y⏐X] = X(12 − X)8/20 
 
 where the above follows since given X, there are 12-X additional balls to be selected from 
among 8 blue and 12 non-blue balls. Now, since X is a hypergeometric random variable 
it follows that 
 
 E[X] = 12(10/30) = 4 and E[X 2] = 12(18)(1/3)(2/3)/29 + 42 = 512/29 
 
 As E[Y] = 8(12/30) = 16/5, we obtain 
 
 E[XY] = 
2
(48 512 / 29)
5
− = 352/29, 
 and 
 Cov(X, Y) = 352/29 − 4(16/5) = −96/145 
 
64. (a) E[X] = E[X⏐type 1]p + E[X⏐type 2](1 − p) = pμ1 + (1 − p)μ2 
 
 (b) Let I be the type. 
 E[X⏐I] = μI , Var(X⏐I) = 2Iσ 
 Var(X) = 2[ ] Var( )I IE σ μ+ 
 = 2 2 2 2 21 2 1 2 1 2(1 ) (1 ) [ (1 ) ]p p p p p pσ σ μ μ μ μ+ − + + − − + − 
 
65. Let X be the number of storms, and let G(B) be the events that it is a good (bad) year. Then 
 
 E[X] = E[X⏐G]P(G) + E[X⏐B]P(B) = 3(.4) + 5(.6) = 4.2 
 
 If Y is Poisson with mean λ, then E[Y 2] = λ + λ2. Therefore, 
 
 E[X 2] = E[X 2⏐G]P(G) + E[X 2⏐B]P(B) = 12(.4) + 30(.6) = 22.8 
 
 Consequently, 
 
 Var(X) = 22.8 − (4.2)2 = 5.16 
 
66. E[X 2] = 2 2 2
1
{ [ 1] [ 2] [ 3]}
3
E X Y E X Y E X Y= + = + = 
 = 2 2
1
{9 [(5 ) ] [(7 ) ]}
3
E X E X+ + + + 
 = 2
1
{83 24 [ ] 2 [ ]}
3
E X E X+ + 
 = 2
1
{443 2 [ ]}
3
E X+ since E[X] = 15 
 
 Hence, 
 
 Var(X) = 443 − (15)2 = 218. 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
124 Chapter 7 
 
67. Let Fn denote the fortune after n gambles. 
 
 E[Fn] = E[E[Fn⏐Fn−1]] = E[2(2p − 1)Fn−1p + Fn−1 − (2p − 1)Fn−1] 
 = (1 + (2p − 1)2)E[Fn− 1] 
 = [1 + (2p − 1)2]2E[Fn−2] 
 
 = [1 + (2p − 1)2]nE[F0] 
 
68. (a) .6e−2 + .4e−3 
 
 (b) .6e−2
3 3
32 3.4
3! 3!
e−+ 
 
 (c) P{3⏐0} = 
3 3
2 2 3 3
2 3
2 3
.6 .4
{3,0} 3! 3!
{0} .6 .4
e e e e
P
P e e
− − − −
− −
+
=
+
 
 
69. (a) 
0
1
2
x xe e dx
∞
− − =∫ 
 (b) 
3
3
0 0
1 (4) 1
3! 96 96 16
x x yxe e dx e y dy
Γ∞ ∞− − −= = =∫ ∫ 
 
 (c) 
3
0
4
0
3! 2 2
813
x x x
x x
x
e e e dx
e e dx
∞
− − −
∞
− −
= =
∫
∫
 
 
70. (a) 
1
0
1/ 2pdp =∫ 
 
 (b) 
1
2
0
1/ 3p dp =∫ 
 
71. P{X = i} = 
1
0
{ }P X i p dp=∫ = 
1
0
(1 )i n i
n
p p dp
i
−⎛ ⎞ −⎜ ⎟⎝ ⎠∫ 
 = 
!( )!
1/( 1)
( 1)!
n i n i
n
i n
⎛ ⎞ −
= +⎜ ⎟ +⎝ ⎠
 
 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter 7 125 
 
72. (a) P{N ≥ i} = 
1 1
1
0 0
{ } (1 ) 1/iP N i p dp p dp i−≥ = − =∫ ∫ 
 (b) P{N = i} = P{N ≥ i} − P{N ≥ i + 1} = 1
( 1)i i +
 
 
 (c) E[N] = 
1 1
{ } 1/
i i
P N i i
∞ ∞
= =
≥ = = ∞∑ ∑ . 
 
73. (a) E[R] = E[E[R⏐S]] = E[S] = μ 
 
 (b) Var(R⏐S) = 1, E[R⏐S] = S 
 Var(R) = 1 + Var(S) = 1 + σ2 
 
 (c) fR(r) = ( ) ( )S R Sf s F r s ds∫ 
 = 
2 2 2( ) / 2 ( ) / 2s r sC e e dsμ σ− − − −∫ 
 = 
2 2
2 2exp 21 1
r
K S
μ σ σ
σ σ
⎧ ⎫⎛ ⎞ ⎛ ⎞+⎪ ⎪− −⎨ ⎬⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠⎪ ⎪⎩ ⎭
∫ ds exp {−(ar2 + br)} 
 
 Hence, R is normal. 
 
 (d) E[RS] = E[E[RS⏐S]] = E[SE[R⏐S]] = E[S 2] = μ2 + σ2 
 
 Cov (R, S) = μ2 + σ2 − μ2 = σ2 
 
75. X is Poisson with mean λ = 2 and Y is Binomial with parameters 10, 3/4. Hence 
 
 (a) P{X + Y = 2} = P{X = 0)P{Y = 2} + P{X = 1}P{Y = 1} + P{X = 2}P{Y = 0} 
 = 2 2 8 2 9 2 10
10 10
(3/ 4) (1/ 4) 2 (3/ 4)(1/ 4) 2 (1/ 4)
2 1
e e e− − −
⎛ ⎞ ⎛ ⎞
+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
 
 
 (b) P{XY = 0} = P{X = 0} + P{Y = 0} − P{X = Y = 0} 
 = e−2 + (1/4)10 − e−2(1/4)10 
 
 (c) E[XY] = E[X]E[Y] = 2 ⋅ 10 ⋅ 3
4
 = 15 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
126 Chapter 7 
 
77. The joint moment generating function, E[etX+sY] can be obtained either by using 
 
 E[etX+sY] = ( , )tX sYe f x y dy dx+∫ ∫ 
 
 or by noting that Y is exponential with rate 1 and, given Y, X is normal with mean Y and 
variance 1. Hence, using this we obtain 
 
 E[etX+sY⏐Y] = esYE[EtX⏐Y] = 
2 / 2sY Yt te e + 
 
 and so 
 
 E[etX+sY] = 
2 / 2 ( )[ ]t s t Ye E e + 
 = 
2 / 2 1(1 )te s t −− − , s + t < 1 
 
 Setting first s and then t equal to 0 gives 
 
 E[etX] = 
2 / 2 1(1 ) ,te t −− t < 1 
 E[esY] = (1 − s)−1, s < 1 
 
78. Conditioning on the amount of the initial check gives 
 
 E[Return] = E[Return⏐A]/2 + E[Return⏐B]/2 
 = {AF(A) + B[1 − F(A)]}/2 + {BF(B) + A[1 − F(B)]}/2 
 = {A + B + [B − A][F(B) − F(A)]}/2 
 > (A + B)/2 
 
 where the inequality follows since [B − A] and [F(B) − F(A) both have the same sign. 
 
 (b) If x < A then the strategy will accept the first value seen: if x > B then it will reject the 
first one seen; and if x lies between A and B then it will always yield return B. Hence, 
 
 E[Return of x-strategy] = 
if 
( ) / 2 otherwise
B A x B
A B
< <
+
 
 
 (c) This follows from (b) since there is a positive probability that X will lie between A and B. 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter 7 127 
 
79. Let Xi denote sales in week i. Then 
 
 E[X1 + X2] = 80 
 Var(X1 + X2) = Var(X1) + Var(X2) + 2 Cov(X1, X2) 
 = 72 + 2[.6(6)(6)] = 93.6 
 
 (a) With Z being a standard normal 
 
 P(X1 + X2 > 90) = 
90 80
93.6
P Z
⎛ ⎞−
>⎜ ⎟⎝ ⎠
 
 = P(Z > 1.034) ≈ .150 
 
 (b) Because the mean of the normal X1 + X2 is less than 90 the probability that it exceeds 90 
is increased as the variance of X1 + X2 increases. Thus, this probability is smaller when 
the correlation is .2. 
 
 (c) In this case, 
 P(X1 + X2 > 90) = 
90 80
72 2[.2(6)(6)]
P Z
⎛ ⎞−
>⎜ ⎟+⎝ ⎠
 
 = P(Z > 1.076) ≈ .141 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
128 Chapter 7 
 
Theoretical Exercises 
 
1. Let μ = E[X]. Then for any a 
 
 E[(X − a)2 = E[(X − μ + μ − a)2] 
 = E[(X − μ)2] + (μ − a)2 + 2E[(x − μ)(μ − a)] 
 = E[(X − μ)2] + (μ − a)2 + 2(μ − a)E[(X − μ)] 
 = E[(X − μ)2 + (μ − a)2 
 
2. E[⏐X − a⏐ = ( ) ( ) ( ) ( )
x a x a
a x f x dx x a f x dx
< >
− + −∫ ∫ 
 = aF(a) − ( ) ( ) [1 ( )]
x a x a
xf x dx xf x dx a F a
< >
+ − −∫ ∫ 
 
 Differentiating the above yields 
 derivative = 2af(a) + 2F(a) − af(a) − af(a) − 1 
 Setting equal to 0 yields that 2F(a) = 1 which establishes the result. 
 
3. E[g(X, Y)] = 
0
{ ( , ) }P g X Y a da
∞
>∫ 
 = 
( , )
0 , : 0
( , )
( , ) ( , )
g x y
x y
g x y a
f x y dydxda daf x y dydx
∞
>
=∫ ∫ ∫ ∫ ∫ ∫ 
 = ( , )g x y dydx∫ ∫ 
 
4. g(X) = g(μ) + g′(μ)(X − μ) + g′′(μ) 
2( )
2
X μ−
 + … 
 ≈ g(μ) + g′(μ)(X − μ) + g′′(μ) 
2( )
2
X μ−
 
 
 Now take expectations of both sides. 
 
5. If we let Xk equal 1 if Ak occurs and 0 otherwise then 
 
 X = 
1
n
k
k
X
=
∑ 
 Hence, 
 E[X] = 
1 1
[ ] ( )
n n
k k
k k
E X P A
= =
=∑ ∑ 
 But 
 E[X] = 
1 1
{ } ( )
n n
k
k k
P X k P C
= =
≥ =∑ ∑ . 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter 7 129 
 
6. X = 
0
( )X t dt
∞
∫ and taking expectations gives 
 E[X] = 
0 0
[ ( )] { }E X t dt P X t dt
∞ ∞
= >∫ ∫ 
 
7. (a) Use Exercise 6 to obtain that 
 
 E[X] = 
0 0
{ } { }P X t dt P Y t dt
∞ ∞
> ≥ >∫ ∫ = E[Y] 
 
 (b) It is easy to verify that 
 
 X+ ≥st Y+ and Y− ≥ st X− 
 
 Now use part (a). 
 
8. Suppose X ≥st Y and f is increasing. Then 
 P{f(X) > a} = P{X > f −1(a)} 
 ≥ P{Y > f −1(a)} since x ≥st Y 
 = P{f(Y) > a} 
 
 Therefore, f(X) ≥st f(Y) and so, from Exercise 7, 
 E[f(X)] ≥ E[f(Y)]. 
 
 On the other hand, if E[f(X)] ≥ E[f(Y)] for all increasing functions f, then by letting f be the 
increasing function 
 
 f(x) = 
1 if
0 otherwise
x t>
 
 
 then 
 
 P{X > t} = E[f(X)] ≥ E[f(Y)] = P{Y > t} 
 
 and so X >st Y. 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
130 Chapter 7 
 
9. Let 
 
 Ij = 
th1 if a run of size begins at the flip
0 otherwise
k j⎧
⎨
⎩
 
 
 Then 
 
 Number of runs of size k = 
1
1
n k
j
j
I
− +
=
∑ 
 E[Number of runs of size k = 
1
1
n k
j
j
E I
− +
=
⎡ ⎤
⎢ ⎥
⎢ ⎥⎣ ⎦
∑ 
 = P(I1 = 1) + 1
2
( 1) ( 1)
n k
j n k
j
P I P I
−
− +
=
= + =∑ 
 = pk(1 − p) + (n − k − 1)pk(1 − p)2 + pk(1 − p) 
 
10. 1 = 1
1 1 1 1 1
n n n n n
i i i i iE X X E X X nE X X
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
∑ ∑ ∑ ∑ ∑ 
 
 Hence, 
 
 
1 1
/
k n
i iE X X k n
⎡ ⎤
=⎢ ⎥
⎣ ⎦
∑ ∑ 
 
11. Let 
 Ij = 
1 outcome never occurs
0 otherwise
j⎧
⎨
⎩
 
 
 Then X = 
1
r
jI∑ and E[X] = 
1
(1 )
r
n
j
j
p
=
−∫ 
 
12. For X having the Cantor distribution, E[X] = 1/2, Var(X) = 1/8 
 
13. Let 
 Ij = 
1 record at 
0 otherwise
j⎧
⎨
⎩
 
 
 1
1 1 1 1
[ ] { is largest of ,..., } 1/
n n n n
j j j jE I E I P X X X j
⎡ ⎤
= = =⎢ ⎥
⎣ ⎦
∑ ∑ ∑ ∑ 
 
 Var
1 1 1
1 1
Var( ) 1
n n n
j jI I j j
⎛ ⎞ ⎛ ⎞
= = −⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠∑ ∑ ∑ 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter 7 131 
 
15. μ = 
1
n
i
i
p
=
∑ by letting Number = 
1
n
i
i
X
=
∑ where 1 is success0 ---i
i
X
⎧
= ⎨
⎩
 
 
 Var(Number) = 
1
(1 )
n
i i
i
p p
=
−∑ 
 
 maximization of variance occur when pi ≡ μ/n 
 
 minimization of variance when pi = 1, i = 1, …, [μ], p[μ]+1 = μ − [μ] 
 
 To prove the maximization result, suppose that 2 of the pi are unequal—say pi ≠ pj. Consider 
a new p-vector with all other pk, k ≠ i, j, as before and with 
2
i j
i j
p p
p p
+
= = . Then in the 
variance formula, we must show 
 
 2 1
2 2
i j i jp p p p+ +⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
 ≥ pi(1 − pi) + pj(1 − pj) 
 
 or equivalently, 
 
 2 2 22 ( )i j i j i jp p p p p p+ − = − ≥ 0. 
 
 The maximization is similar. 
 
16. Suppose that each element is, independently, equally likely to be colored red or blue. If we 
let Xi equal 1 if all the elements of Ai are similarly colored, and let it be 0 otherwise, then 
1
r
ii
X
=∑ is the number of subsets whose elements all have the same color. Because 
 
 [ ]
1 1 1
2(1/ 2) i
r r r
A
i i
i i i
E X E X
= = =
⎡ ⎤
= =⎢ ⎥
⎣ ⎦
∑ ∑ ∑ 
 
 it follows that for at least one coloring the number of monocolored subsets is less than or 
equal to 1
1
(1/ 2) i
r A
i
−
=∑ 
 
17. ( ) 2 2 2 21 2 1 2Var (1 ) (1 )X Xλ λ λ σ λ σ+ − = + − 
 
2
2 2 2
1 2 2 2
1 2
( ) 2 2(1 ) 0
d
d
σλσ λ σ λ
λ σ σ
= − − = ⇒ =
+
 
 As Var(λX1 + (1 − λ)X2) = 21 2( (1 ) )E X Xλ λ μ⎡ ⎤+ − −⎣ ⎦ we want this value to be small. 
 
 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
132 Chapter 7 
 
18. (a. Binomial with parameters m and Pi + Pj. 
 
 (b) Using (a) we have that Var(Ni + Nj) = m(Pi + Pj)(1 − Pi − Pj) and thus 
 
 m(Pi + Pj)(1 − Pi − Pj) = mPi(1 − Pi) + mPj(1 − Pj) + 2 Cov(Ni, Nj) 
 
 Simplifying the above shows that 
 
 Cov(Ni, Nj) = −mPiPj. 
 
19. Cov(X + Y, X − Y) = Cov(X, X) + Cov(X, −Y) + Cov(Y, X) + Cov(Y, −Y) 
 = Var(X) − Cov(X, Y) + Cov(Y, X) − Var(Y) 
 = Var(X) − Var(Y) = 0. 
 
20. (a) Cov(X, Y⏐Z) 
 = E[XY − E[X⏐Z]Y − XE[Y⏐Z] + E[X⏐Z]E[Y⏐Z] [Z] 
 = E[XY⏐Z] − E[X⏐Z] E[Y⏐Z] − E[X⏐Z]E[Y⏐Z] + E[X⏐Z]E[Y⏐Z] 
 = E[XY⏐Z] − E[X⏐Z]E[Y⏐Z] 
 
 where the next to last equality uses the fact that given Z, E[X⏐Z] and E[Y⏐Z] can be 
treated as constants. 
 
 (b) From (a) 
 
 E[Cov(X, Y⏐Z)] = E[XY] − E[E[X⏐Z]E[Y⏐Z]] 
 
 On the other hand, 
 
 Cov(E[X⏐Z], E[Y⏐Z] = E[E[X⏐Z]E[Y⏐Z]]