STAT 333 Assignment 2 SOLUTIONS
Type I Problems
1.
Suppose X has probability density function f(x) = 2x
−
3
for x
≥
1. (This is called a powerlaw
or Pareto distribution.) Suppose Y is independent of X and has a gamma distribution with
parameters
α
= 4 and
λ
= 2. Obtain P(X > Y ).
First obtain the cdf of X, F
X
(x) =
for x
≥
1
[]
2
1
2
1
3
1
2
−
−
−
−
=
=
∫
x
u
du
u
x
x
∫
8
y
∫
∫
∫
∞
−
−
−
−
∞
−
+
=
=
>
1
2
3
2
1
0
2
3
0
2
3
0
2
3
4
0
3
)
(
3
8
)
1
(
3
)
(
!
2
)
(
)
(
)

(
dy
e
y
y
dy
e
y
e
X
e
y
X
dy
y
f
y
Y
Y
X
P
y
y
y
Y
∞
=
Then, to find P(X > Y), condition on Y
P(X > Y)
∫
=
>
=
3

dy
y
Y
y
P
substitution
∞
>
=
8
dy
y
y
P
independence
since if y < 1, X is guaranteed to be bigger
You can then evaluate the integrals either by integrating by parts, using Maple/other software, or
with some help from the Gamma function technique. Obviously I can’t ask you something this
hard on a test – the point of this question was to use the conditioning to set up the integral and
recognize that you need different things if y is below or above 1.
2. # 37
Let T represent the typist
a) We know that XT = A ~ Poi(2.6), XT = B ~ Poi(3), and XT = C ~ Poi(3.4)
So E[X] = E[E[XT]]
= E[XT = A]P(T = A) + E[XT = B]P(T = B) + E[XT = C]P(T = C)
= 2.6 * 1/3 + 3 * 1/3 + 3.4 * 1/3
= 3
b) For Poisson random variables, the variance and mean are both equal to the parameter
λ
. So the
second moment of a Poisson is
λ
+
λ
2
E[X
2
T = A] = 2.6 + 2.6
2
etc
So E[X
2
] = E[E[X
2
T]]
= E[X
2
T = A]P(T = A) + E[X
2
T = B]P(T = B) + E[X
2
T = C]P(T = C)
= (2.6 + 2.6
2
) * 1/3 + (3 + 3
2
) * 1/3 + (3.4 + 3.4
2
) * 1/3
= 12.1067
And thus Var(X) = 12.1067  3
2
= 3.1067
3. # 44
Let N represent the number of customers in the store on a given day. N ~ Poi(10)
Let X
i
represent the amount spent by the i
th
customer of the day. X
i
’s ~ X ~ U(0, 100)
Let S represent the total amount of money on the day.
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View Full DocumentWe must assume that the number of customers and the amounts that any one customers spends
are all independent random variables. (This may or may not actually be a good assumption…)
Then S = X
1
+ X
2
+ … + X
N
is a compound random variable since it satisfies the definition.
We know E[N] = 10 and Var(N) = 10 since N is Poisson
We know (or can derive) E[X] = 50 and Var(X) = 100
2
/12 since X is Uniform
So using the properties for compound random variables,
E[S] = E[N]E[X] = 10 * 50 = 500
Var(S) = E[N]Var(X) + (E[X])
2
Var(N) = 10 * 100
2
/12 + 50
2
* 10 = 33,333.33
4. # 62
Let p
A
, p
B
, and p
C
be the probabilities that A, B, and C win overall, respectively.
We must have one winner, so p
A
+ p
B
+ p
C
= 1.
Also, A and B are equivalent (we could just rename them) so p
A
= p
B
.
We need a third equation in order to solve.
Look at the probability A wins overall, conditional on the outcome of the first game.
p
A
= P(A wins overallwins first)*P(wins first) + P(A wins overallloses first)*P(loses first)
Let’s look at P(A wins overallloses first)
½ the time, B will win against C, and the probability of A winning overall is 0.
The other ½ the time, A will be in the exact same starting position as C – about to play the
winner of the current game.
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 Spring '08
 Chisholm
 Probability, Probability theory, probability density function, renewal, Tλ, fλ

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