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STAT/MA 416 Answers
Homework 7
December 7, 2007
Solutions by Mark Daniel Ward
PROBLEMS
Chapter 7 Problems
34a. Let Xi indicate whether or not the ith wife sits next to her husband. In other words,
define
(
1 if the ith wife sits next to her husband
Xi =
0 otherwise
After the wife is seated, her husband has 19 available seats. Two of these seats are next to
his wife. Since Xi only takes on the values 0 or 1, thus E[Xi ] =P
P (Xi = 1) =P
2/19. So the
10
Xi = 10
expected number of wives who sit next to their husbands is E
i=1 E[Xi ] =
i=1
P10
i=1 2/19 = 20/19.
P
P10
P10
34b. We know that Var
i<j Cov(Xi , Xj ). We see that
i=1 Xi =
i=1 Var(Xi ) + 2
E[Xi2 ] = 12 P (Xi = 1) + 02 P (Xi = 0) = 2/19 for each i. Therefore Var(Xi ) = E[Xi2 ] −
2
2 2
34
(E[Xi ])2 = 19
− 19
= 361
for each i. For i 6= j, we notice that Xi Xj equals 1 if the
ith and jth wives are each seated next to their husbands, and Xi Xj = 0 otherwise. So
E[Xi Xj ] = P (Xi Xj = 1). The probability that the ith wife sits next to her husband is
2/19. Given that the ith wife sits next to her husband, then 18 seats in a row remain
empty. The probability that the wife sits on the end of such a row with her husband next
to her is (2)(1/18)(1/17); the probability that the wife is seated strictly within the row,
and her husband is
to her, is (16)(1/18)(2/17). Thus, E[Xi Xj ] = P (Xi Xj = 1) =
next
2
2
1 2
2
2
1 1
2
2
− 19
.
2 18 17 + 16 18 17 = 171 . So Cov(Xi Xj ) = E[Xi Xj ] − E[Xi ]E[Xj ] = 171
= 3249
19
19
P10 34
P10
P
2
34
2
360
Thus Var
i=1 361 + 2
i=1 Xi =
i<j 3249 = (10) 361 + (2)(45) 3249 = 361 .
36. We write Xi to indicate if the ith roll is a “1” or not. Similarly, we write Yi to indicate
if the ith roll is a “2” or not. In other words,
(
(
1 if the ith roll is a “1”
1 if the ith roll is a “2”
Xi =
Yi =
0 otherwise
0 otherwise
Pn
Pn
Thus X = i=1 Xi and Y = i=1 Yi . We also know that
!
n
n
n X
n
X
X
X
Cov(X, Y ) = Cov
Xi ,
Yj =
Cov(Xi , Yj )
i=1
j=1
i=1 j=1
We recall Cov(Xi , Yj ) = E[Xi Yj ] − E[Xi ]E[Yj ].
For i = j, we see that Xi Yj = 0 since the same roll cannot be both “1” and “2”. So, for
i = j, we have E[Xi Yj ] = 0, and of course E[Xi ] = E[Yj ] = 1/6, so Cov(Xi , Yj ) = −1/36.
For i 6= j, we see that Xi Yj = 1 with probability 1/36, so E[Xi Yj ] = 1/36, and thus
1
− 16 61 = 0. (An immediate way to see that Cov(Xi , Yj ) = 0 is to notice
Cov(Xi , Yj ) = 36
that Xi and Yj are independent when i 6= j.)
1
2
Thus
Cov(X, Y ) =
n X
n
X
Cov(Xi , Yj ) =
i=1 j=1
n
X
Cov(Xi , Yi ) +
n X
X
i=1
=
n
X
i=1
Cov(Xi , Yj )
i=1 j6=i
−
1
+
36
n X
X
0 = −n/36
i=1 j6=i
−2x R∞Rx
R∞Rx
dy dx = 1/4, and E[X] = 0 0 (x) 2e x
dy dx =
38. Note E[XY ] = 0 0 (xy) 2e x
R ∞ R x 2e−2x 1/2 and E[Y ] = 0 0 (y)
dy dx = 1/4, so Cov(X, Y ) = E[XY ] − E[X]E[Y ] =
x
1 1
1
1
− 2 4 = 8.
4
−(y+x/y) R∞R∞
40. We compute E[XY ] = 0 0 (xy) e y
dx dy = 2, and also we have E[X] =
−(y+x/y) −(y+x/y) R∞R∞
R
R
∞ ∞
e
e
(x)
dxdy
=
1,
and
E[Y
]
=
(y)
dxdy = 1, so Cov(X, Y ) =
y
y
0
0
0
0
E[XY ] − E[X]E[Y ] = 2 − (1) (1) = 1.
To perform these integrations, it is helpful to note (as in class) that y1 e−x/y is the density
R∞
of an exponential distribution with λ = 1/y, so 0 y1 e−x/y dx = 1 (just because y1 e−x/y
R∞
is a density function) and also 0 (x) y1 e−x/y dx = λ1 = y is the mean of the exponential
distribution.
−2x
42a. Let Xi indicate whether or not the ith pair consists of a man and a woman. In other
words, define
(
1 if the ith pair consists of a man and a woman
Xi =
0 otherwise
Since Xi only takes on the values 0 or 1, thus E[Xi] = P (Xi = 1). There are 20
= 190
2
10 10
ways that the ith pair can be selected. There are 1 1 = 100 ways that the pair consists
of a man and a woman. So E[Xi ] = 100/190
=
number of pairs
10/19.
P10
P10
P10 So the expected
consisting of a man and a woman is E
i=1 10/19 = 100/19.
i=1 Xi = P i=1 E[Xi ] =
P10
P10
2
We know that Var
i<j Cov(Xi , Xj ). We see that E[Xi ] =
i=1 Xi =
i=1 Var(Xi ) + 2
12 P (Xi = 1) + 02 P (Xi = 0) = 10/19 for each i. Therefore Var(Xi ) = E[Xi2 ] − (E[Xi ])2 =
10
10 2
90
−
= 361
for each i. For i 6= j, we notice that Xi Xj equals 1 if the ith and jth pairs
19
19
each consist of a man and a woman, and Xi Xj = 0 otherwise. So E[Xi Xj ] = P (Xi Xj = 1).
The probability that the ith pair consists of a man and a woman is 10/19. Given that the
ith pair consists of a man
other
and a woman, then 9 men and 9 woman remain to form9the
9
18
9 pairs. So there are 2 = 153 remaining ways to select the jth pair; there are 1 1 = 81
ways that
will also consist of a man and a woman. Thus, E[XiXj ] =
the
jth pair
P (Xi Xj =
10
81
90
90
10
10
1) = 19 153 = 323 . So Cov(Xi Xj ) = E[Xi Xj ] − E[Xi ]E[Xj ] = 323 − 19 19 = 10/6137.
16200
P10 90
P10
P
10
90
10
Thus Var
+
(2)(45)
= 6137 .
X
=
+
2
=
(10)
i
i=1
i=1 361
i<j 6137
361
6137
42b. Let Yi indicate whether or not the ith pair consists of a married couple. In other
words, define
(
1 if the ith pair consists of a married couple
Yi =
0 otherwise
3
Since Yi only takes on the values 0 or 1, thus E[Yi ] = P (Yi = 1). There are 20
= 190
2
ways that the ith pair can be selected. There are 10 ways that the pair consists of a married
couple. So E[Y
= 1/19. So P
the expected number of pairs consisting of a married
Pi10] = 10/190
P10
couple is E
E[Yi ] = 10
1/19 = 10/19.
i=1 Yi =
i=1
P10 P10 i=1
P
2
We know that Var
i=1 Yi =
i=1 Var(Yi ) + 2
i<j Cov(Yi , Yj ). We see that E[Yi ] =
12 P (Yi = 1) + 02 P (Yi = 0) = 1/19 for each i. Therefore Var(Yi ) = E[Yi2 ] − (E[Yi ])2 =
1
1 2
18
−
= 361
for each i. For i 6= j, we notice that Yi Yj equals 1 if the ith and jth pairs
19
19
each consist of a married couple, and Yi Yj = 0 otherwise. So E[Yi Yj ] = P (Yi Yj = 1). The
probability that the ith pair consists of a married couple is 1/19. Given that the ith pair
consists of a married couple, then 9 men and 9 woman remain to form the other 9 pairs. So
there are 18
= 153 remaining ways to select the jth pair; there are 9 waysthat the jth
2
1
9
1
pair will also consist of a married couple. Thus, E[Yi Yj ] = P (Yi Yj = 1) = 19
= 323
.
153
P
10
1
1
1
So Cov(Y Y ) = E[Yi Yj ] − E[Yi ]E[Yj ] = 323 − 19 19 = 2/6137. Thus Var
i=1 Yi =
3240
P10 18 i j P
2
18
2
i=1 361 + 2
i<j 6137 = (10) 361 + (2)(45) 6137 = 6137 .
48a. The number of rolls X necessary to obtain a “6” is geometric with probability of
success p = 1/6, and thus E[X] = 1/p = 6.
48b. Given that Y = 1, we know that the first outcome is a “5”. After the first roll, the
number of additional rolls necessary to obtain a “6” is geometric with probability of success
p = 1/6, so the expected number of additional rolls is 6. So the total number of expected
rolls in this conditional case is E[X | Y = 1] = 1 + 6 = 7.
48b. Given that Y = 5, we know that none of the first four rolls is a “5”, and
the
fifth
role is a “5”. So P (X = 1 | Y = 5) = 15 , and P (X = 2 | Y = 5) = 45 15 , and
2 1 4 3 1
P (X = 3 | Y = 5) = 45
,
and
P
(X
=
4
|
Y
=
5)
=
. Also P (X = 5 | Y =
5
5
5
4 1 , and
5) = 0. For the remaining of the possibilities, we have P (X = 6 | Y = 5) = 45
6
1
4 4 5
4 4 5 2 1
P (X = 7 | Y = 5) = 5
, and P (X = 8 | Y = 5) = 5
, and in general,
6
6
6 6 4 4 5 i−6 1
P (X = i | Y = 5) = 5
for i ≥ 6. Thus
6
6
2 3 1
4
1
4
1
4
1
E[X | Y = 5] = (1)
+ (2)
+ (3)
+ (4)
5
5
5
5
5
5
5
4
i−6
X
4
5
1
+
i
5
6
6
i≥6
4
We can pull the 54 outside of the sum above, and then change the index of “i” by 5
afterwards, to obtain
4 X
i−1 821
4
5
1
E[X | Y = 5] =
+
(i + 5)
625
5
6
6
i≥1
We note that, if Z is geometric with probability of success p = 1/6, then the summation
P
5 i−1 1
found in the last line above is just E[Z + 5] = E[Z] + 5 = 6 + 5.
i≥1 (i + 5) 6
6
Therefore, plugging this result in, we obtain
4
821
4
3637
E[X | Y = 5] =
+
(6 + 5) =
625
5
625
4
50. We first compute the marginal density of Y ,
Z ∞ −x/y −y
e
e
fY (y) =
dx = e−y
y
0
for y > 0 and of course fY (y) = 0 for y ≤ 0. So the conditional density of X given Y is
fX|Y (x|y) =
e−x/y e−y /y
e−x/y
=
e−y
y
Therefore, X conditioned on having Y = y is exponential with parameter λ = 1/y. Now we
compute the desired conditional second moment:
−x/y Z ∞
Z ∞
e
2
2
2
(x )
x fX|Y (x|y) dx =
E[X | Y = y] =
dx = 2y 2
y
0
−∞
A different method, without computing the integral on the previous line, is to just recall
that an exponential random variable with parameter λ has mean λ1 and variance λ12 and thus
2
second moment λ12 + λ1 = λ22 . So the second moment of X conditioned on Y = y is just
2
= 2y 2 .
λ2
53. We write X for the number of days until the prisoner reaches freedom, and we write Y
for the door he selects. Since E[X] = E[E[X|Y ]], then
E[X] = E[X | Y = 1]P (Y = 1) + E[X | Y = 2]P (Y = 2) + E[X | Y = 3]P (Y = 3)
= (E[X] + 2)(.5) + (E[X] + 4)(.3) + (1)(.2)
Thus E[X] = 12.
56. Let Xi indicate whether or not the elevator stops on the ith floor. In other words, define
(
1 if the elevator stops on the ith floor
Xi =
0 otherwise
Since Xi only takes on the values 0 or 1, thus E[Xi ] = P (Xi = 1). We write Y for the
number of people who enter the elevator, so Y is Poisson with parameter λ = 10. Thus
∞
X
P (Xi = 1) =
P (Xi = 1 and Y = y)
=
y=0
∞
X
P (Xi = 1 | Y = y)P (Y = y)
y=0
∞ X
y e−10 10y
y!
y=0
y
∞
∞
X
X
1 y N −1
e−10 10y
−10
=
−e
10
y!
y!
N
y=0
y=0
=
1−
N −1
N
= 1 − e−10 e10(N −1)/N
= 1 − e−10/N
5
So E[Xi ] = 1 − e−10/N . So the expected number of stops is is E
PN
−10/N
) = N (1 − e−10/N ).
i=1 (1 − e
hP
N
i=1
i P
Xi = N
i=1 E[Xi ] =
58a. Let X denote the total number of flips. We write q = 1 − p as the probability of tails.
Let Y denote whether a head or tail appears on the first try, by writing
(
H
Y =
T
if “heads” appears on the first flip
if “tails” appears on the first flip
Then E[X] = E[E[X|Y ]] = E[X | Y = H]P (Y = H) + E[X | Y = T ]P (Y = T ). Of course
P (Y = H) = p and P (Y = T ) = q. Given that Y = H, then the number of additional
flips until the first tail appears is geometric with probability of success q, so we expect 1q
additional flips. Thus E[X | Y = H] = 1 + 1q . Similarly, given that Y = T , then the number
of additional flips until the first head appears is geometric with probability of success p, so
we expect p1 additional flips. Thus E[X | Y = T ] = 1 + p1 . So
E[X] =
1
1+
q
1
(p) + 1 +
p
(q) = 1 +
p q
1 − p + p2
+ =
q p
pq
Another possible method was suggested to me by Ms. Xueyao Chen this week. The
expected value of X can be written directly as
E[X] =
∞
X
n=2
np
n−1
q+
∞
X
nq n−1 p
n=2
We know that theP
expected value of a geometric random variable with probability of success
1
n−1
q = 1q . Subtracting the “n = 1” term on both sides yields
q is q , and thus ∞
n=1 np
P∞
n−1
q = 1q − q. Similarly, we know that the expected value of a geometric random
n=2 np
P
n−1
p = p1 . Subtracting the
variable with probability of success p is p1 , and thus ∞
n=1 nq
P∞
“n = 1” term on both sides yields n=2 nq n−1 p = p1 − p. Therefore, putting these results
together, we have
E[X] =
1
1
1 1
1 − p + p2
−q+ −p= + −1=
q
p
q p
pq
58b. The last flip lands heads if and only if the first flip lands tails. So the last flip
lands heads with probability q. (Remember, there are only two kinds of possibilities in this
problem: either we have H, H, H, . . . , H, T or we have T, T, T, . . . , T, H.)
75a. We note that, since X has moment generating function exp(2et − 2), then X is Poisson
with parameter λ = 2. Since Y has moment generating function ( 43 et + 14 )10 , then Y is
Binomial with parameter n = 10 and p = 3/4. So, also using the fact that X and Y are
6
independent, we compute
P (X + Y = 2) = P (X = 0, Y = 2) + P (X = 1, Y = 1) + P (X = 2, Y = 2)
= P (X = 0)P (Y = 2) + P (X = 1)P (Y = 1) + P (X = 2)P (Y = 2)
2 8
1 9
0 10
e−2 20 10
3
1
e−2 21 10
3
1
e−2 22 10
3
1
=
+
+
0!
2
4
4
1!
1
4
4
2!
0
4
4
467
=
e−2
1048576
75b. We see that XY = 0 if and only if X = 0 or Y = 0. Thus P (XY = 0) = P (X =
−2 0
0) + P (Y = 0) − P (X = 0, Y = 0). We have P (X = 0) = e 0!2 = e−2 and P (Y = 0) =
3 0 1 10
10
1
= 1048576
and since X and Y are independent we have P (X = 0, Y = 0) =
0
4
4
1
e−2
e−2
− 1048576
.
P (X = 0)P (Y = 0) = 1048576 . Thus P (XY = 0) = e−2 + 1048576
75c. Since X and Y are independent, then E[XY ] = E[X]E[Y ]. Since X is Poisson with
λ = 2 then E[X] = 2. Since Y is Binomial with n = 10 and p = 3/4, then E[Y ] = (10)(3/4).
So E[XY ] = (2)(10)(3/4) = 15.
Chapter 8 Problems
1. Since X has mean µ = 20, then
P (0 < X < 40) = P (|X − µ| < 20)
We are also given that X has variance σ 2 = 20, so by using Chebyshev’s inequality, we see
that
σ2
20
1
P (|X − µ| ≥ 20) ≤ 2 = 2 =
20
20
20
So we conclude that
19
P (|X − µ| < 20) ≥
20
X1 +···+Xn
3. Let Xi denote the ith student’s score, so
is the average score of n students on
n
the exam. The class average for n students is within 5 of 75 if
X1 + · · · + Xn
≤ 80
n
The probability of this event can be approximated using the Central Limit Theorem; we use
the fact that √
the average of X1 + · · · + Xn is 75n and the variance is 25n, so the standard
deviation is 5 n. We obtain
X1 + · · · + Xn
P 70 ≤
≤ 80 = P (70n ≤ X1 + · · · + Xn ≤ 80n)
n
= P (−5n ≤ X1 + · · · + Xn − 75n ≤ 5n)
−5n
X1 + · · · + Xn − 75n
5n
√ ≤
√
=P
≤ √
5 n
5 n
5 n
√
√
≈ P (− n ≤ Z ≤ n)
√
√
= Φ( n) − Φ(− n)
√
= 2Φ( n) − 1
70 ≤
7
√
We√want the probability above to be at least .9, so we have 2Φ( n) − 1 ≥ .9, or equivalently,
Φ(
√ n) ≥ .95. Examining the table for the normal distribution on page 222, this implies that
n ≥ 1.65, so n ≥ 2.72. Therefore, since n must be an integer, n ≥ 3 students are sufficient.
4a. Since X1 , . . . , X20 are Poissons, each with mean 1, then X = X1 + · · · + X20 is a random
variable with mean 20, so by Markov’s inequality
20
= 4/3
15
Of course, all probabilities are at most 1 (so certainly less than 4/3), so that was not very
helpful at all. We didn’t gain anything by using Markov’s inequality in this case.
4b. Now we use the Central Limit Theorem to approximate P (X ≥ 15). We already noted
that X has mean 20. Also, each Xi is Poisson with λ = 1,√so each Xi has variance 1, so the
variance of X is also 20, and X has standard deviation 20. The first equality below is a
continuity correction, since we note that each Xi (and thus X too) is an integer. Thus
P (X ≥ 15) ≤
P (X1 + · · · + X20 ≥ 15) = P (X1 + · · · + X20 ≥ 15.5)
= P (X1 + · · · + X20 − 20 ≥ 15.5 − 20)
X1 + · · · + X20 − 20
15.5 − 20
√
=P
≥ √
20
20
≈ P (Z ≥ −1.01)
= P (Z ≤ 1.01)
= Φ(1.01)
≈ .8438
6. We let Xi denote the outcome of the ith die. The desired probability is a bit difficult
to calculate directly. It is easier to calculate the opposite probability, namely, 79 rolls are
sufficient to exceed 300. So the desired probability is
1 − P (X1 + · · · + X79 > 300)
We note that Xi has mean E[Xi ] = 1(1/6) + 2(1/6) + · · · + 6(1/6) = 7/2 and second moment
7 2
E[Xi2 ] = 12 (1/6) + 22 (1/6) + · · · + 62 (1/6) = 91/6, and thus Var(Xi ) = 91
−
= 35/12.
6
2
Now we use the Central Limit Theorem. The first equality below is a continuity correction,
since we note that each Xi is an integer. Thus
P (X1 + · · · + X79 > 300) = P (X1 + · · · + X79 ≥ 300.5)
= P (X1 + · · · + X79 − 79(7/2) ≥ 300.5 − 79(72/2))
!
X1 + · · · + X79 − 79(7/2)
300.5 − 79(7/2)
p
√
√
=P
≥ p
35/12 79
35/12 79
≈ P (Z ≥ 1.58)
So the desired probability is
1 − P (Z ≥ 1.58) = P (Z ≤ 1.58) = Φ(1.58) ≈ .9429
7. Let Xi denote the lifetime of the ith lightbulb. Each lightbulb has exponential lifetime
with mean λ1 = 5 and therefore variance λ12 = 25. So X = X1 + · · · + X100 has mean
8
(100)(5) = 500 and variance (100)(25) = 2500 and standard deviation
√
2500 = 50. Thus
P (X1 + · · · + X100 ≥ 525) = P (X1 + · · · + X100 − 500 ≥ 525 − 500)
X1 + · · · + X100 − 500
525 − 500
=P
≥
50
50
≈ P (Z ≥ 1/2)
= 1 − P (Z ≤ 1/2)
≈ 1 − .6915
= .3085
14. Let Xi denote the lifetime of the ith component. Each component has mean lifetime
100; the variance of each component’s lifetime
is 302 . So Xn = X1 + · · · + Xn has mean 100n
√
variance 302 n and standard deviation 30 n. Thus
P (X1 + · · · + Xn ≥ 2000) = P (X1 + · · · + Xn − 100n ≥ 2000 − 100n)
X1 + · · · + Xn − 100n
2000 − 100n
√
√
=P
≥
30 n
30 n
2000 − 100n
√
≈P Z≥
30 n
We want the system to be in continual operation
for over 2000 hours with
probability
2000−100n
2000−100n
at least .95, so we need P Z ≥ 30√n
≥ .95, i.e., P Z ≤ − 30√n
≥ .95, i.e.,
√
Φ − 2000−100n
≥ .95. Examining the table for the normal distribution on page 222, this
30 n
√
2000−100n
implies that − 30√n
≥ 1.65, which is equivalent to −2000 + 100n ≥ (1.65)(30 n),
√
i.e., 100n − (1.65)(30) n − 2000 ≥ 0. Using the
formula, we see that 100n −
√ quadratic
2
√
√
(1.65)(30)± ((1.65)(30)) −4(100)(−2000)
which is approximi(1.65)(30) n − 2000 = 0 when n =
2(100)
ately
−4.23 or 4.73, depending on whether the “−” or “+” is used (respectively). So we have
√
n ≥ 4.73 and thus n ≥ 22.34. Since n is an integer, this implies that n ≥ 23 is sufficient.
15. Let Xi denote the ith person’s yearly claim. Each claim has mean 240 and variance 8002 ,
so X = X1 + · · · + X10,000 has mean (10,000)(240)
= 2,400,000 and variance (10,000)(8002 ) =
√
6,400,000,000 and standard deviation 6,400,000,000 = 80,000. Thus
P (X1 + · · · + X10,000 ≥ 2,700,000) = P (X1 + · · · + X10,000 − 2,400,000 ≥ 2,700,000 − 2,000,000)
2,700,000 − 2,400,000
X1 + · · · + X10,000 − 2,400,000
=P
≥
80,000
80,000
≈ P (Z ≥ 3.75)
= 1 − P (Z ≤ 3.75)
≈1−1
=0
So the probability that the total yearly claim exceeds $2.7 million is approximately 0. In
other words, this event happens very, very rarely!