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SOLUTION FOR HOMEWORK 9, ACTS 4306
Welcome to your 9th homework.
The main task of this homework is to recognize an underlying distribution and then solve
the problem as fast as possible using known facts about the distribution. I will use notation
of the text.
1. Solution: First of all, look how quickly I can figure out the correct answer. The range
of X is larger than the range of Y , so variance of X is also larger. Then, the variance depends
on n, so only Answer A fits.
Now a rigorous solution. For a uniform discrete random variable Z with the pmf pZ (x) =
−1
k I(x ∈ {a + 1, a + 2, . . . , a + k}), where a is an integer, we know that
E(Z) = a +
k+1
,
2
Var(Z) =
k2 − 1
.
12
Using this we get
(n + 1)2 − 1
,
12
and then Var(X) − Var(Y ) = (2n + 1)/12.
Var(X) =
Var(Y ) =
n2 − 1
,
12
Answer: A
2. Notation: X is the number of days that the machine breaks down in 10 days.
Find: P (X ≥ 2).
Solution: Clearly X is Binomial(θ = .2, n = 10). As a result,
P (X ≥ 2) = 1 − P (X < 2) = 1 − [P (X = 0) + P (X = 1)]
(use the formula)
=1−[
10!
10!
(.2)0 (.8)10 +
(.2)1 (.8)9 ] = 1 − .3758 = .6242.
0!10!
1!9!
Answer: D
3. Solution: Because X is P oisson(λ) with E(X) = λ, we get λ = ln(2). Then
E(cos(πX)) =
∞
X
pX (k) cos(πk) = e−λ
k=0
(note that cos(πk) = (−1)k , and we can continue)
= e−λ
∞
X
(−λ)k
k!
k=1
1
∞
X
λk
cos(πk)
k=0 k!
(now I use e−λ =
P∞
k=0
(−λ)k
k!
and then λ = ln(2))
= e−λ e−λ = e−2λ = 1/4.
Answer: D
4. Solution: Here X is Geometric(θ = .02) with the pmf pX (k) = θ(1 − θ)k−1 I(k ∈
{1, 2, . . .}).
The problem is to find the smallest x such that P (X ≤ x) ≥ 1/2. Write
P (X ≤ x) =
x
X
pX (k) =
k=0
x
X
θ(1 − θ)k−1 = θ
k=0
x
X
(1 − θ)k−1
k=0
(we got the geometric sum and using 1 + a + a2 + . . . + ax−1 = [1 − ax ]/[1 − a] we continue)
=θ
1 − (1 − θ)x
1 − (1 − θ)
(plug in θ = 1/6)
=
(1/6)[1 − (5/6)x ]
= 1 − (5/6)x .
1 − 5/6
Now we need to find the smallest x such that
(5/6)x ≤ 1/2.
Try the increasing x = 2, 3, 4, . . . and the minimal x which satisfies the inequality is x = 4.
Answer: C
5. Solution: Let X denote the number of accidents resulting in a claim for the first 3
accidents. Then X is Binomial(θ = 1/4, n = 3). We are asked to calculate P (X ≥ 1).
Write
P (X ≥ 1) = 1 − P (X = 0) = 1 − (3/4)3 = 1 − .422 = .578.
Answer: E
6. Solution: Let X be the number of claims occurring in the year. Then X is Binomial(θ =
.2, n = 5). The total expected claim for the year is
E(106 X) = 106 θn = 106 .
The probability in question is
P (106X > E(106 X)) = P (X > 1) = 1 − P (X = 0) − P (X = 1)
= 1 − (.8)5 − (5)(.2)(.8)4 = 1 − .7373 = .2627.
Note that you do not need to know the face amount on each policy to solve the problem.
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Answer: E.
7. Solution: Let N be P oisson(λ = 2). The probability generating function for N is
G(t) = E(tN ) = e−λ
∞
X
tk λk /k! = e−λ
k=0
(remember the Taylor series ex =
P∞
k
k=0 x /k!
∞
X
(λt)k /k!
k=0
and continue)
= e−λ eλt = eλ(t−1) = e2(t−1) .
Answer: E
8. Notation: X and Y are the completed number of years until death for the two
individuals.
Given: pX (k) = pY (k) = (.01)(.99)k I(k ∈ {0, 1, . . .}), X and Y are independent.
Find: P (X = Y ).
Solution: By the Total Probability Theorem
P (X = Y ) =
∞
X
P (X = Y = k)
k=0
(using independence and identical distribution of X and Y we continue)
=
∞
X
[P (X = k)]2 =
k=0
∞
X
(.01)2 [(.99)2 ]k
k=0
(using formula for the sum of geometric series we continue)
= (.01)2
1
= .005.
1 − (.99)2
Answer: C
9. Solution: Here X is Geometric(θ) with θ being unknown. However, it is given that
E(X|θ) = 12.5 and we also know that for Geometric distribution we have E(X|θ) = 1/θ.
As a result, θ = 1/(12.5). The problem is to find
P (X = 6|θ) = θ(1 − θ)6−1 = (1/12.5)[1 − (1/12.5)]5 = .053.
Answer: B
10. Solution: As we know,
Var(X) = E(X 2 ) − [E(X)]2 ,
and the moments can be calculated from derivatives of the moment generating function,
namely
E(X) = M ′ (t)|t=0 , E(X 2 ) = M ′′ (t)|t=0 .
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Let us take the derivatives:
et
M ′ (t) = 9 [(2 + et )/3]8
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which yields E(X) = 3. Next,
M ′′ (t) = 3[(2 + et )/3]8 + 3et (et /3)(8)[(2 + et )/3]7
which yields E(X 2 ) = 11. We conclude that
Var(X) = 11 − 32 = 2.
Answer: A
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