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) = ½
(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 – yn 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(Z2) = μ2+ σ2, then
E[(X-Y)2] = E(X2 – XY + Y2) = E(X2)– E(X)E(Y) (by independence) + E(Y2)
= (μ2+ σ2) - μμ + (μ2+ σ2) = 2σ2
33. (a) E[X2 + 4X + 4] = E[X2] + 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(X1+X2, X2+X3) = Cov(X1+X2, X2+X3)/{sqrt(Var(X1+X2) sqrt(Var(X2+X3)}
Since variances sum, the denominator becomes sqrt(1+1)sqrt(1+1) = 2.
For the numerator, since means are 0, Cov(X1+X2, X2+X3) = E(X1+X2)E(X2+X3) =
E(X1 X2) + E(X1 X3) + E(X2 X2) + E(X2X3) . By independence all these expectations except E(X2
X2) of products of individual expectations (which are all 0). So Cor(X1+X2, X2+X3) = E(X22)/2 =
½.
AMS311 Homework 10 - Chapter 7
53. Just like 5c: 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.
56. Let S be the number of floors where the elevator stops. Let Ii = 1 if elevator stops at i-th stop,
=0 otherwise, and so S = sum{i=1,N; Ii}. 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[Ii|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 10k/k!}
= N - N e-10sum{k=1,oo; ((1-1/N)10)k/k!}
= N - N e-10e10(1-1/N) = N – Ne-10/N= N(1-e-10/N)
57. Let Xi 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: Ii } = E[ E [sum{i=1,n; Xi| 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|1st heads]p + E[X|1st 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[XG]P(G) + E[XB]P(B) = 3(.4) + 5(.6) = 4.2
If Y is Poisson with mean λ, then E[Y2] = λ + λ2 = 12 (good year) or 30 (bad year) Therefore,
E[X2] = E[X2G]P(G) + E[X2B]P(B) = 12(.4) + 30(.6) = 22.8
Consequently, Var(X) = 22.8 − (4.2)2 = 5.16.
66. E[X2] = (1/3){E[X2|Y=1] + E[X2|Y=2 + E[X2|Y=3] }
= (1/3){ 9 + E[(5+X)2] + E[(7+X)2] }
= (1/3){83 + 24E[X] + 2E[X2]}
= (1/3){443 + 2E[X2]}, since E[X} = 15.
Multiplying both sides by 3 and simplifying, we have E[X2] = 443.
Then Var[X] = 443 – (15)2 = 218.
Theory
48. MY(t) =E[etY] = E[et(aX+b)] = etbE[etaX] = etbMX(at).
50. Ψ’(t) = M’(t)/M(t), and Ψ’’(t) = {M(t)M’’(t) – M’(t)2/M(t)2
Then at t=0, Ψ’’(0) = {1E[X2] – (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; Xi}/n – 75| > 5) ≤ 25/25n. So need n = 10.
3. Let Z be a standard normal random variable. Then,
P(|sum{1,n; Xi}/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; Xi} > 15) ≤ 20/15
(b) P{sum{1,20; Xi} > 15) = P(sum{1,20; Xi} > 15.5)
is approx. P{Z > (15.5 – 20)/sqrt(20)} = P{Z > -1.006} is approx. .843
5. Letting Xi denote the i th roundoff error, it follows that E[{sum{1,50;Xi}] = 0 and
Var[{sum{1,50;Xi}] = 50Var[X1] = 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;Xi}| > 3) is approx. P{| Z| > 3(12/50)1/2} by Central Limit Th.
= 2P{|Z| > 1.47} = .142
6. If Xi is the outcome of the ith roll then E[Xi] = 7/2 Var(Xi) = 35/12 and so
P({sum{1,79;Xi} ≤ 300) = P({sum{1,79;Xi} ≤ 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;Xi} > 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;Xi} ≥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;Xi} > 2,700,000) is approx.
P{Z ≥ (2,700,000 – 2,400,000)/(800x100) } = P{Z ≥ 3.75} = about 0.