March 13 Homework Solutions
Math 151, Winter 2012
Chapter 7 Problems (pages 373-379)
Problem 30
If X and Y are independent and identically distributed with mean µ and variance σ 2 , find
E[(X − Y )]2 .
E[(X − Y )2 ] = E[X 2 − 2XY + Y 2 ]
= E[X 2 ] − 2E[XY ] + E[Y 2 ]
= (Var(X) + E[X]2 ) − 2(Cov(X, Y ) + E[X]E[Y ]) + (Var(Y ) + E[Y ]2 )
= (σ 2 + µ2 ) − 2(0 + µ2 ) + (σ 2 + µ2 )
= 2σ 2 .
Problem 34
If 10 married couples are randomly seated at a round table, compute
(a) the expected number of the number of wives who are seated next to their husbands.
Label the wives 1 through 10 and let Xi =P1 if the i-th wife is seated next to her
husband and Xi = 0 otherwise. Let X = 10
i=1 Xi be the number of wives seated
next to their husbands. Then
E[X] =
10
X
E[Xi ] =
i=1
10
X
P {Xi = 1}.
i=1
Given where the i-th wife sits, her husband may sit in one of the 19 remaining seats,
two of which are next to his wife. So P {Xi = 1} = 2/19 and hence
E[X] =
10
X
P {Xi = 1} = 10 ·
i=1
2
20
= .
19
19
(b) the variance of the number of wives who are seated next to their husbands.
We want to compute
Var(X) =
10
X
Var(Xi ) + 2
i=1
X
Cov(Xi , Xj ).
i<j
Since Xi only takes values 0 or 1, we have
Var(Xi ) = E[Xi2 ] − E[Xi ]2
= P {Xi2 = 1} − P {Xi = 1}2
= P {Xi = 1} − P {Xi = 1}2
2
2
2
=
−
19
19
34
=
.
361
1
Also, for i = 1, ..., 10 and j 6= i we have
Cov(Xi , Xj ) = E[Xi Xj ] − E[Xi ]E[Xj ] = P {Xi = Xj = 1} − P {Xi = 1}P {Xj = 1}.
We know P {Xi = Xj = 1} = P ({Xj = 1}|{Xi = 1})P {Xi = 1}. Given that the
i-th wife and her husband sit next to each other, the j-th wife can sit in any of the
18 remaining seats. If the j-th wife sits next to the i-th wife or her husband, the
j-th wife’s husband may sit in one of the remaining 17 seats, only one of which is
next to his wife. If the j-th wife does not sit next to the i-th wife or her husband,
then the j-th wife’s husband may sit in one of the remaining 17 seats, two of which
is next to his wife. Thus
16 2
1
2 1
·
+
·
= .
P ({Xi = 1}|{Xj = 1}) =
18 17 18 17
9
So P {Xi = Xj = 1} = 1/9 · 2/19 = 2/171. Hence
2
2
2
2
Cov(Xi , Xj ) =
−
=
.
171
19
3249
Hence
Var(X) =
10
X
i=1
Var(Xi ) + 2
X
i<j
10
2
360
34
+2·
·
=
.
Cov(Xi , Xj ) = 10 ·
361
2
3249
361
Problem 37
A die is rolled twice. Let X equal the sum of the outcomes, and let Y equal the first
outcome minus the second. Compute Cov(X, Y ).
Let Z1 be the outcome of the first role and let Z2 be the outcome of the second roll.
So X = Z1 + Z2 and Y = Z1 − Z2 . Then
Cov(X, Y ) = Cov(Z1 + Z2 , Z1 − Z2 )
= Cov(Z1 , Z1 ) − Cov(Z1 , Z2 ) + Cov(Z2 , Z1 ) − Cov(Z2 , Z2 )
= Var(Z1 ) − Var(Z2 ) since Cov(Z1 , Z2 ) − Cov(Z2 , Z1 )
= 0 since Var(Z1 ) = Var(Z2 ).
Note that Var(Z1 ) = Var(Z2 ) because Z1 and Z2 are identically distributed.
Problem 41
A pond contains 100 fish, of which 30 are carp. If 20 fish are caught, what are the mean
and variance of the number of carp among the 20? What assumptions are you making?
Let X denote the number of carp caught of the 20 fish caught. We assume that each
of the 100
ways to catch the 20 fish are equally likely. Then X satisfies a hypergeometric
20
distribution with parameters n = 20, N = 100, m = 30 (see Section 4.8.3 on page 160 in
the text). By the formulas on page 162 and 163 for the expected value and variance of
the hypergeometric series, we have
E[X] =
nm
20 · 30
=
=6
N
100
2
and
nm (n − 1)(m − 1)
nm
20 · 30 19 · 29
20 · 30
112
Var(X) =
+1−
=
+1−
=
.
N
N −1
N
100
99
100
33
Problem 45
If X1 , X2 , X3 , and X4 are (pairwise) uncorrelated random variables, each having mean
0 and variance 1, compute the correlations of
(a) X1 + X2 and X2 + X3 .
We have
Cov(X1 + X2 , X2 + X3 ) = Cov(X1 , X2 ) + Cov(X1 , X3 ) + Cov(X2 , X2 ) + Cov(X2 , X3 )
= 0 + 0 + Var(X2 ) + 0
= 1.
Note
Var(X1 + X2 ) = Var(X1 ) + Var(X2 ) + 2 Cov(X1 , X2 ) = 1 + 1 + 2 · 0 = 2.
Similarly Var(X2 + X3 ) = 0. So the correlation between X1 + X2 and X2 + X3 is
1
Cov(X1 + X2 , X2 + X3 )
1
= .
ρ(X1 + X2 , X2 + X3 ) = p
=√
2
2·2
Var(X1 + X2 ) Var(X2 + X3 )
(b) X1 + X2 and X3 + X4 .
Since
Cov(X1 + X2 , X3 + X4 ) = Cov(X1 , X3 ) + Cov(X1 , X4 ) + Cov(X2 , X3 ) + Cov(X2 , X4 )
=0+0+0+0
= 0,
the correlation between X1 + X2 and X3 + X4 is 0.
Problem 48
A fair die is successively rolled. Let X and Y denote, respectively, the number of rolls
necessary to obtain a 6 and a 5. Find
(a) E[X].
E[X] =
∞
X
xP {X = x} =
x=1
since
P∞
x=1
xz
x−1
∞
X
x(5/6)x−1 (1/6) = (1/6) ·
x=1
=
1
(1−z)2
for |z| < 1.
3
1
= 6.
(1 − 5/6)2
(b) E[X|Y = 1].
E[X|Y = 1] =
∞
X
xP ({X = x}|{Y = 1})
x=1
=
=
=
∞
X
x(5/6)x−2 (1/6)
x=2
∞
X
6
5
1
5
x=2
∞
X
x(5/6)x−1 (1/6)
x(5/6)x−1
x=2
∞
1 1X
=− +
x(5/6)x−1
5 5 x=1
1 1
1
=− + ·
5 5 (1/6)2
∞
X
since
xz x−1 =
x=1
1
for |z| < 1
(1 − z)2
= 7.
(c) E[X|Y = 5].
We have
E[X|Y = 5] =
∞
X
xP ({X = x}|{Y = 5})
x=1
∞
X
P {X = x, Y = 5}
x
=
P {Y = 5}
x=1
∞
X
1
=
xP {X = x, Y = 5}.
P {Y = 5} x=1
We know P {Y = 5} = (5/6)4 (1/6).
If x ≤ 4, then
P {X = x, Y = 5} = (4/6)x−1 (1/6)(5/6)4−x (1/6) = 4x−1 54−x 6−5 .
If x = 5, then P {X = x, Y = 5} = 0. If x ≥ 6, then
P {X = x, Y = 5} = (4/6)4 (1/6)(5/6)x−6 (1/6) = 44 5x−6 6−x .
4
Thus
∞
X
1
xP {X = x, Y = 5}
E[X|Y = 5] =
P {Y = 5} x=1
1
=
(5/6)4 (1/6)
=
4
X
x−1 4−x −5
x·4
5
6
+
x=1
∞
X
!
4 x−6 −x
x·4 5
6
x=6
3637
.
625
Problem 59
There are n + 1 participants in a game. Each person independently is a winner with
probability p. The winners share a total prize of 1 unit. Let A denote a specified one of
the players, and let X denote the amount that is received by A.
(a) Compute the expected total prize shared by the players.
Let Y be the total prize shared by the players. If there is no winner, then Y = 0.
If there is at least one winner, then Y = 1. The probability that there is no winner
is (1 − p)n+1 . Therefore the the expected total prize shared by the players is
E[Y ] = P {Y = 1} = 1 − (1 − p)n+1 .
(b) Argue that
E[X] =
1 − (1 − p)n+1
.
n+1
Label the players
Pn+1 1 through n + 1 and let Yi denote the prize of the i-th player.
Note Y = i=1 Yi is the total prize shared by the players. By part (a), E[Y ] =
1 − (1 − p)n+1 . Since the players win independently and with the same probability,
E[Yi ] is independent of i. In particular, E[Yi ] = E[X] is the expected prize of A.
Therefore
n+1
X
n+1
1 − (1 − p)
= E[Y ] =
E[Yi ] = (n + 1)E[X],
i=1
so
E[X] =
1 − (1 − p)n+1
.
n+1
(c) Compute E[X] by conditioning on whether A is a winner, and conclude that
E[(1 + B)−1 ] =
1 − (1 − p)n+1
(n + 1)p
when B is a binomial random variable with parameters n and p.
Let I = 1 if A wins and I = 0 otherwise. Then E[X] = E[E[X|I]]. If I = 0, then
E[X|I] = 0 since A can only get a prize if A wins. If I = 1, then E[X|I] = (1+B)−1 ,
5
where B denotes the number of winners excluding I. Note that B is a binomial
random variable with parameters n and p. Therefore
E[X] = E[E[X|I]]
= E[X|I = 0] · P {I = 0} + E[X|I = 1] · P {I = 1}
= 0 + E[(1 + B)−1 ] · p
= E[(1 + B)−1 ] · p.
So using part (b), we get
E[(1 + B)−1 ] =
E[X]
1 − (1 − p)n+1
=
.
p
(n + 1)p
Chapter 7 Theoretical Exercises (pages 380-384)
Problem 33
A coin that lands on heads with probability p is continually flipped. Compute the expected
number of flips that are made until a string of r heads in a row is obtained.
Let X denote the number of flips that are made until a string of r heads in a row is
obtained. Let Y denote the number of flips until the first occurrence of tails. Then
E[X] = E[E[X|Y ]] =
∞
X
E[X|Y = i]P {Y = i}.
i=1
Note P {Y = i} = pi−1 (1 − p). If i ≤ r, then on the i + 1 flip we start over trying to get
r consecutive heads and thus E[X|Y = i] = i + E[X]. If i > r, then the first r flips are
heads and thus E[X|Y = i] = r. Hence
E[X] =
∞
X
E[X|Y = i]P {Y = i}
i=1
= (1 − p)
r
X
pi−1 (i + E[X]) + (1 − p)
i=1
∞
X
pi−1 r
i=r+1
r
r
X
X
= (1 − p)
ipi−1 + (1 − p)
pi−1 E[X] + rpr
i=1
i=1
r
X
= (1 − p)
ipi−1 + E[X](1 − pr ) + rpr
i=1
6
We can calculate
r
X
ip
i−1
=
i=1
∞
X
ip
i=1
i−1
−
∞
X
ipi−1
i=r+1
∞
X
=
∞
X
1
i−1
−
rp
−
(i − r)pi−1
2
(1 − p)
i=r+1
i=r+1
=
X
rpr
1
−
−
ipr+i−1
(1 − p)2 1 − p i=1
∞
1
pr
rpr
−
−
(1 − p)2 1 − p (1 − p)2
1 − (1 − p)rpr − pr
=
.
(1 − p)2
=
So
r
X
ipi−1 + E[X](1 − pr ) + rpr
E[X] = (1 − p)
i=1
1 − (1 − p)rpr − pr + E[X](1 − pr ) + rpr
= (1 − p)
(1 − p)2
1 − pr
=
+ E[X](1 − pr ).
1−p
Solving for E[X], we get
E[X] =
1 − pr
.
pr (1 − p)
Problem 34
For another approach to Theoretical Exercise 33, let Tr denote the number of flips required
to obtain a run of r consecutive heads.
(a) Determine E[Tr |Tr−1 ].
Let j be an integer with j ≥ r − 1. Let I = 1 if the j + 1 flip is heads and I = 0 if
it is tails. Then
E[Tr |Tr−1 = j] = E[Tr |Tr−1 = j, I = 0]P {I = 0} + E[Tr |Tr−1 = j, I = 1]P {I = 1}.
If I = 0, that is the next flip is tails, we start over trying to get r consecutive
heads. Thus E[Tr |Tr−1 = j, I = 0] = j + 1 + E[Tr ]. If I = 1, then we’ve succeeded
at getting r consecutive heads, and so E[Tr |Tr−1 = j, I = 1] = j + 1. Hence
E[Tr |Tr−1 = j] = E[Tr |Tr−1 = j, I = 0]P {I = 0} + E[Tr |Tr−1 = j, I = 1]P {I = 1}
= (j + 1 + E[Tr ])(1 − p) + (j + 1)p
= j + 1 + (1 − p)E[Tr ].
That is, E[Tr |Tr−1 ] is the function of Tr−1 whose value at Tr−1 = j is
j + 1 + (1 − p)E[Tr ].
7
(b) Determine E[Tr ] in terms of E[Tr−1 ].
We have
E[Tr ] = E[E[Tr |Tr−1 ]]
X
=
E[Tr |Tr−1 = j]P {Tr−1 = j}
j
=
X
j + 1 + (1 − p)E[Tr ] P {Tr−1 = j}
j
=
X
j · P {Tr−1 = j} + 1 + (1 − p)E[Tr ]
j
= E[Tr−1 ] + 1 + (1 − p)E[Tr ]
Hence
E[Tr ] =
E[Tr−1 ] 1
+ .
p
p
(c) What is E[T1 ]?
E[T1 ] =
∞
X
iP {T1 = i} =
i=1
since
P∞
n=1
nxn−1 =
1
(1−x)2
∞
X
i(1 − p)i−1 p =
i=1
for |x| < 1.
(d) What is E[Tr ]?
E[Tr−1 ] 1
+
p
p
E[Tr−2 ]
1
1
=
+ 2+
2
p
p
p
= ...
E[Tr ] =
r−1
E[T1 ] X 1
= r−1 +
p
pi
i=1
1
p−1 − p−r
+
pr
1 − p−1
pr − 1
= r
.
p (p − 1)
=
8
1
p
=
2
p
p