AMS 311 Homeworks 9,10,11

AMS 311 HW#9 Chapter 7

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] = (1/2)2(7/2) + 1/2(1/2)(7/2), where 7/2 is expect value of a die, = (5/4)(7/2)

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) = (b) P(W < 0) = P(N > 2) = 1/4
(c) E[W] = 2 E[N] = 0, where E[N] 1/(1/2) = 2, a geometric distrib with p = 1/2

5. Take positive quadrant E{X+Y] = E[X] + E[Y] = 3/4 + 3/4 = 3/2

6. Just multiply expected value of one toss by 10 = 10(7/2) = 35.

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 = sum {Xi}, Y = sum {Yi}, and W = sum {Wi}, a) E[X] = sum{E[Xi] = 10(3/10)(3/10) = .9
b) E[Y] = sum{E[Yi] = 10(7/10)(7/10) = 4.9 c) E[W] = sum{E[Wi] = 10(21)(3/10)(7/10) = 4.2

11. Let Xi equal 1 if a changeover occurs on the ith 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] = sum{E[Xi] = (n-1)2p(1-p).

13. Just like Example 2h. Answer = 1000(1/1000) = 1
26. Use Lemma 2.1 (p. 191) E[Y] = Int {0.oo; P(Y > y) dy
a) So E(max) = Int{0,1; P(max > y)dy = Int{0,1; 1 - P(max y)dy =
Int{0,1; 1 y
n
dy = 1 1/(n+1) = n/(n+1)
b) E(min) = Int{0,1; P(min > y)dy = Int{0,1; (1-y)
n
= 1/(n+1).

30. Recalling that E(Z
2
) =
2
+
2
, then
E[(X-Y)
2
] = E(X
2
XY + Y
2
) = E(X
2
) E(X)E(Y) (by independence) + E(Y
2
)
= (
2
+
2
) - + (
2
+
2
) = 2
2


33. (a) E[X2 + 4X + 4] = E[X
2
] + 4E[X] + 4 = Var(X) + (E[X])
2
+ 4E[X] + 4 = 14
(b) Var(4 + 3X) = Var(3X) = 9Var(X) = 45

37 Wi, i = 1, 2, denote the ith outcome.
Cov(X, Y) = Cov(W1 + W2 , W1 W2) = Cov(W1, W1) Cov(W2, W2) (other terms cancel)
= Var(W1) Var(W2) = 0

45. Cor(X
1
+X
2
, X
2
+X
3
) = Cov(X
1
+X
2
, X
2
+X
3
)/{sqrt(Var(X
1
+X
2
) sqrt(Var(X
2
+X
3
)}
Since variances sum, the denominator becomes sqrt(1+1)sqrt(1+1) = 2.
For the numerator, since means are 0, Cov(X
1
+X
2
, X
2
+X
3
) = E(X
1
+X
2
)E(X
2
+X
3
) =
E(X
1
X
2
) + E(X
1
X
3
) + E(X
2
X
2
) + E(X
2
X
3
) . By independence all these expectations except E(X
2

X
2
) of products of individual expectations (which are all 0). So Cor(X
1
+X
2
, X
2
+X
3
) = E(X
2
2
)/2 =
.

AMS311 Homework 10 - Chapter 7

53. Just like 5c: E[X] = E[XI = 1](.5) + E[XI = 2](.3) + E[XI = 3](.2)
= (2 + E[X])(.5) + (4 + E[X])(.3) + .2. Therefore, E[X] = 12.

56. Let S be the number of floors where the elevator stops. Let I
i
= 1 if elevator stops at i-th stop,
=0 otherwise, and so S = sum{i=1,N; I
i
}. Let X be the number that enter on the ground floor. We
compute E[S} by conditioning on X:
E[S|X=k] = sum{i=1,N; E[I
i
|X=k] = N[1 (1 1/N)
k
]
where (1-1/N)
k
is the probability that no one gets off at the i-th floor.
Then E[S] = sum{k=1,oo; E[S|X=k}Pr(X=k)
= sum{k=1,oo; N[1 (1 1/N)
k
]e
-10
10
k
/k!}
= N - N e
-10
sum{k=1,oo; ((1-1/N)10)
k
/k!}
= N - N e
-10
e
10(1-1/N)
= N Ne
-10/N
= N(1-e
-10/N
)

57. Let X
i
be number of workers injured in the th accident during a week. Let N be number of
accidents during the week. This is like Example 5d.
We want E[sum{i=1,N: I
i
} = E[ E [sum{i=1,n; X
i
| N=n}]] = E[n 2.5] = 12.5.

58. Let X denote the number of flips required. Condition on outcome of the first flip
E[X] = E[X|1
st
heads]p + E[X|1
st
tails](1-p)
= [1 + 1/(1-p)]p + [1 + 1/p](1-p),
where 1/(1-p) and 1/p are expected values of geometric distribution
for number of flips until the first occurrence of the other outcome.
= 1 + p/(1-p) + (1-p)/p.

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[XG]P(G) + E[XB]P(B) = 3(.4) + 5(.6) = 4.2
If Y is Poisson with mean , then E[Y
2
] = +
2
= 12 (good year) or 30 (bad year) 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
] = (1/3){E[X
2
|Y=1] + E[X
2
|Y=2 + E[X
2
|Y=3] }
= (1/3){ 9 + E[(5+X)
2
] + E[(7+X)
2
] }
= (1/3){83 + 24E[X] + 2E[X
2
]}
= (1/3){443 + 2E[X
2
]}, since E[X} = 15.
Multiplying both sides by 3 and simplifying, we have E[X
2
] = 443.
Then Var[X] = 443 (15)
2
= 218.

Theory
48. M
Y
(t) =E[e
tY
] = E[e
t(aX+b)
] = e
tb
E[e
taX
] = e
tb
M
X
(at).

50. (t) = M(t)/M(t), and (t) = {M(t)M(t) M(t)
2
/M(t)
2

Then at t=0, (0) = {1E[X
2
] (E[X])
2
}/ 1 = Var(X)

AMS 311 Homework 11 Chapter 8

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) P(|sum{1,n; X
i
}/n 75| > 5) 25/25n. So need n = 10.

3. Let Z be a standard normal random variable. Then,
P(|sum{1,n; X
i
}/n 75| > 5) is approx. P(|Z| > 1xsqrt(n) ) .1. From Z table., we need sqrt(n)
1.63 --> n 3
NOTE: n =3 is really too small for Central Limit Thm.

4. (a) P{sum{1,20; X
i
} > 15) 20/15
(b) P{sum{1,20; X
i
} > 15) = P(sum{1,20; X
i
} > 15.5)
is approx. P{Z > (15.5 20)/sqrt(20)} = P{Z > -1.006} is approx. .843

5. Letting X
i
denote the i th roundoff error, it follows that E[{sum{1,50;X
i
}] = 0 and
Var[{sum{1,50;X
i
}] = 50Var[X
1
] = 50(1/12), where the last equality uses the fact that .5 + X is
uniform (0,1) and so Var[X] = Var[.5 + X} = 1/12.
Hence P(|{sum{1,50;X
i
}| > 3) is approx. P{| Z| > 3(12/50)
1/2
} by Central Limit Th.
= 2P{|Z| > 1.47} = .142

6. If X
i
is the outcome of the ith roll then E[X
i
] = 7/2 Var(X
i
) = 35/12 and so
P({sum{1,79;X
i
} 300) = P({sum{1,79;X
i
} 300.5)
is approx.. P( Z (300.5 79(7/2))/(79x35/12)
1/2
) = P( Z 1.58) = .943

7. P({sum{1,100;X
i
} > 525} is approx. P{Z > (525-500)/sqrt(100x25) } =
P{Z > .5 } = .308, given an exponential with mean 5 has variance 25.

14. Suppose n components are in stock. The probability they will last for at least 2000 hours is p =
P(({sum{1,n;X
i
} 2000} is approx. P(Z(2000 100n)/(30sqrt(n))
Since .95 = P( Z -1.64}, it follows that p .95 if
(2000 100n)/(30sqrt(n) -1.64 ==> (2000 100n)/sqrt(n) =49.2. By trial-and-error, we get
n 23.

15. P({sum{1,10000;X
i
} > 2,700,000) is approx.
P{Z (2,700,000 2,400,000)/(800x100) } = P{Z 3.75} = about 0.