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Page 1 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
STA 5325, Homework #3
Due June 15, 2009
Ramin Shamshiri
UFID # 90213353 Note: Problem numbers are according to the 6
th text edition. Selected problem are highlighted.
Assignment for Monday, June 1st , 2009
5.4. Given here is the joint probability function associated with data obtained in a study of automobile accidents
in which a child (under age 5) was in the car and at least one fatality occurred. Specifically, the study focused
on whether or not the child survived and what type of seatbelt (if any) he or she used. Define:
π1 = 0, ππ π‘ππ πππππ π π’ππ£ππ£ππ1, ππ πππ‘
and π1 =
0, ππ no belt used1, ππ adult belt used2, ππ car β seat belt used.
Notice that Y1 is the number of fatalities per child and, since childrenβs car seats usually utilize two belts, Y2 is
the number of seatbelts in use at the time of accident.
y2
y1
0 1
0 0.38 0.17 0.55
1 0.14 0.02 0.16
2 0.24 0.05 0.29
0.76 0.24 1.00
a. Verify that the preceding probability function satisfies theorem 5.1.
b. Find F(1,2). What is the interpretation of this value?
Solution a:
Theorem 5.1 states that if Y1 and Y2 are discrete random variables with joint probability function p(y1,y2),
then
π(π¦1, π¦2) β₯ 0 for all y1,y2
π π¦1, π¦2 = 1π¦1 ,π¦2, where the sum is over all values (y1,y2) that are assigned nonzero probabilities.
It can be seen from the table values that all of the probabilities are at least 0 and sum of them are 1.
Solution b:
According to definition 5.2, for any random variable Y1 and Y2, the joint distribution function F(y1,y2) is
given by: πΉ π¦1, π¦2 = π(π1 β€ π¦1, π2 β€ π¦2).
Therefore,
πΉ 1,2 = π π1 β€ 1, π2 β€ 2 = π 0,0 + π 0,1 + π 0,2 + π 1,0 + π 1,1 + π(1,2)
= 0.38 + 0.14 + 0.24 + 0.17 + 0.02 + 0.05 = π. ππ + π. ππ = π
It means that every child in the experiment either survived or didnβt and use either 0,1 or 2 seatbelts.
Page 2 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
5.8. An environment engineer measures the amount (by weight) of particular pollution in air samples of a
certain volume collected over two smokestacks at a coal-operated power plant. One of the stacks is equipped with a cleaning device. Let Y1 denote the amount of pollutant per sample collected above the stack that has no
cleaning device, and let Y2 denote the amount of pollutant per sample collected above the stack that is equipped
with the cleaning device. Suppose that the relative frequency behavior of Y1 and Y2 can be modeled by
π(π¦1, π¦2) = π, 0 β€ y1 β€ 2, 0 β€ y2 β€ 1, 2y2 β€ y1
0, πππ ππ€ππππ
That is, Y1 and Y2 are uniform distributed over the region inside the triangle bounded by y1=2, y2=0, and
2y2=y1
a. Find the value of k that makes this function a probability density function.
b. Find π(π1 β₯ 3π2). That is, find the probability that the cleaning device reduces the amount of pollutant by one-their or more.
Solution a:
π2
2π¦2
1
0
ππ¦1ππ¦2 = 1
π(2 β 2π¦2)1
0
ππ¦2 = 1
2ππ¦2 β π¦22
0
1= 1
2π β 1 = 1 => π = π
Solution b:
π¦2 =1
3π¦1
π π1 β₯ 3π2 = 1
13π¦1
0
2
0
ππ¦2ππ¦1 = 1
3π¦1
2
0
ππ¦1
= π¦12
6
0
2
=4
6=
π
π
Page 3 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
5.13. The management at a fast-food outlet is interested in the joint behavior of the random variables Y1,
defined as the total time between a customerβs arrival at the store and departure from the service window, and Y2, the time a customer waits in line before reaching the service windows. Because Y1 includes the time a
customer waits in line, we must have π1 β₯ π2. The relative frequency distribution of observed values of Y1 and
Y2 can be modeled by the probability density function
π(π¦1, π¦2) = πβπ¦1 , 0 β€ y2 β€ y1 < β
0, elsewhere
With time measured in minutes.
a. Find π(π1 < 2, π2 > 1). b. Find π(π1 β₯ 2π2). c. Find π(π1 β π2 β₯ 1). (Notice that Y1-Y2 denotes the time spent at the service window)
Solution a:
π π1 < 2, π2 > 1 = πβπ¦1
2
π¦2
2
1
ππ¦1ππ¦2
Or we could say π π1 < 2, π2 > 1 = πβπ¦1π¦1
1
2
1ππ¦2ππ¦1
Both are the same. Solving the integral,
π π1 < 2, π2 > 1 = πβ1 β 2πβ2
Solution b:
π π1 β₯ 2π2 = πβπ¦1
β
2π¦2
β
0
ππ¦1ππ¦2
Or we could say:
π π1 β₯ 2π2 = πβπ¦1
β
0
β
2π¦2
ππ¦2ππ¦1
Both are the same. Solving the integral
π π1 β₯ 2π2 =1
2
Solution c:
π π1 β π2 β₯ 1 = πβπ¦1
β
π¦2+1
β
0
ππ¦1ππ¦2 = πβ1
Page 4 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
5.20. In exercise 5.4, you were given the following joint probability function for
π1 = 0, ππ π‘ππ πππππ π π’ππ£ππ£ππ1, ππ πππ‘
and π1 =
0, ππ no belt used1, ππ adult belt used2, ππ car β seat belt used.
y2
y1
0 1
0 0.38 0.17 0.55
1 0.14 0.02 0.16
2 0.24 0.05 0.29
0.76 0.24 1.00
a. Give the marginal probability function for Y1 and Y2. b. Give the conditional probability function for Y2 given Y1=0.
c. What is the probability that a child survived given that he or she was in a car-seat belt?
Solution a:
π1 π¦1 = π(π¦1, π¦2)
π¦2
P1(Y1=0)=P(0,0)+P(0,1)+P(0,2)=0.38+0.14+0.24=0.76 P1(Y1=1)=P(1,0)+P(1,1)+P(1,2)=0.17+0.02+0.05=0.24
So:
π1(π1) = 0.76, π¦1 = 00.24, π¦1 = 1
π2 π¦2 = π(π¦1, π¦2)
π¦1
π2(π2) = 0.55, π¦0 = 00.16, π¦1 = 10.29, π¦2 = 2
Solution b:
π π2 π1 = 0 =π(π¦1 = 0, π¦2)
π1(π¦1 = 0)=
π 0,0 = 0.38
0.76= 0.5 , π¦2 = 0
π 0,1 = 0.14
0.76= 0.184, π¦2 = 1
π 0,2 = 0.24
0.76= 0.315, π¦2 = 2
Solution c:
π π1 = 0 π2 = 2 =π(0,2)
π2(2)=
0.24
0.29= 0.827
Page 5 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
5.21. In example 5.4 and exercise 5.5 we considered the joint density of Y1, the proportion of the capacity of the
tank that is stocked at the beginning of the week, and Y2, the proportion of the capacity sold during the week, given by
π(π¦1, π¦2) = 3π¦1, 0 β€ y2 β€ y1 < 1
0, elsewhere
a. Find the marginal density function for Y2.
b. For what values of y2 is the conditional density π( π¦1 π¦2) defined?
c. What is the probability that more than half of a tank is sold given that three-fourths of a tank is stocked?
Solution a:
π2 π¦2 = π(π¦1, π¦2)β
ββ
ππ¦1 = 3π¦1
1
π¦2
ππ¦1 = 3π¦12
2 π¦2
1
=3
2β
3π¦22
2
hence:
π2 π¦2 = 3
2β
3π¦22
2, 0 β€ y2 β€ 1
0, elsewhere
Solution b: Denominator should not be zero,
π π¦1 π¦2 =π(π¦1, π¦2)
π2(π¦2) β 0
Therefore, 0 β€ y2 β€ 1
Solution c:
π1 π¦1 = π(π¦1, π¦2)β
ββ
ππ¦2 = 3π¦1
π¦1
0
ππ¦2 = 3π¦1π¦2 0π¦1 = 3π¦1
2
π π2 >1
2 π1 =
3
4 = π 0.5 0.75 =
π(0.75,0.5)
π1(0.75)=
3π¦1
π1(0.75)=
94
3π¦12 π¦1=0.75
=
94
3 34
2 =π
π
Page 6 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
Assignment for Tuesday, June 2nd
, 2009
5.42. In exercise 5.4 you were given the following joint probability function for
π1 = 0, ππ πππππ π π’ππ£ππ£ππ1, ππ πππ‘
and π2 =
0, ππ ππ ππππ‘ π’π ππ1, ππ πππ’ππ‘ ππππ‘ π’π ππ2, ππ πππ β π πππ‘ ππππ‘ π’π ππ
y2
y1
0 1
0 0.38 0.17 0.55
1 0.14 0.02 0.16 2 0.24 0.05 0.29
0.76 0.24 1.00
Are Y1 and Y2 independent? Why or why not?
Solution: We should check P(y1,y2)=P1(y1).P2(y2)
Lets pick examine P(0,1), from table, we can see that P(0,1)=0.14. P1(0)=0.76 and P2(1)=0.16
Since 0.14β 0.76)(0.16), we conclude that Y1 and Y2 are not independent. So they are dependent.
5.43. In example 5.4 and exercise 5.5 we considered the joint density of Y1, the proportion of the capacity of
the tank that is stocked at the beginning of the week and Y2, the proposition of the capacity sold during the week, given by
π(π¦1, π¦2) = 3π¦1, 0 β€ π¦2 β€ π¦1 β€ 1
0, πππ ππ€ππππ
Show that Y1 and Y2 are dependent.
Solution: Quick check:
We examine the joint density function and see if it satisfies:
a. The region in which the joint density is positive must be a rectangle (possibly infinite) b. The function equation must factor into a product of two functions-one depends only on y1 and the other
depends only on y2. (One or both could be constant)
Since the region in which the joint density is positive is not a rectangle (it is triangle), then part (a) fails and we
conclude that Y1 and Y2 are dependent. (However part (b) holds)
Detailed solution:
π2 π¦2 = π(π¦1, π¦2)β
ββ
ππ¦1 = 3π¦1
1
π¦2
ππ¦1 = 3π¦12
2 π¦2
1
=3
2β
3π¦22
2
hence:
π2 π¦2 = 3
2β
3π¦22
2, 0 β€ y2 β€ 1
0, elsewhere
π1 π¦1 = π(π¦1, π¦2)β
ββ
ππ¦2 = 3π¦1
π¦1
0
ππ¦2 = 3π¦1π¦2 0π¦1 = 3π¦1
2
π1 π¦1 = 3π¦1
2, 0 β€ y1 β€ 10, elsewhere
π π¦1, π¦2 = 3π¦1 β π1 π¦1 . π2 π¦2
Hence, Y1 and Y2 are dependent.
Page 7 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
5.44. In exercise 5.6, we derived the fact that
π(π¦1, π¦2) = 4π¦1π¦2, 0 β€ π¦1 β€ 1, 0 β€ π¦2 β€ 1
0, πππ ππ€ππππ
is a valid joint probability density function. Are Y1 and Y2 independent?
Quick check: We examine the joint density function and see if it satisfies:
a. The region in which the joint density is positive must be a rectangle (possibly infinite)
b. The function equation must factor into a product of two functions-one depends only on y1 and the other depends only on y2. (One or both could be constant)
Since the region in which the joint density is positive is a rectangle and since the function equation can be
factored into a product of two functions 4y1 and y2m both criteria hold and we conclude that Y1 and Y2 are
independent. (However part (b) holds)
Detailed solution:
π2 π¦2 = π(π¦1, π¦2)β
ββ
ππ¦1 = 4π¦1π¦2
1
0
ππ¦1 = 2π¦12π¦2
0
1= 2π¦2
hence:
π2 π¦2 = 2π¦2, 0 β€ y2 β€ 1
0, elsewhere
π1 π¦1 = π(π¦1, π¦2)β
ββ
ππ¦2 = 4π¦1π¦2
1
0
ππ¦2 = 2π¦1π¦22
0
1= 2π¦1
π1 π¦1 = 2π¦1, 0 β€ y1 β€ 1
0, elsewhere
π ππ, ππ = πππππ = ππ ππ . ππ ππ
Page 8 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
5.65. In exercise 5.7, we determined that
π(π¦1, π¦2) = 6(1 β π¦2), 0 β€ π¦1 β€ π¦2 β€ 1
0, πππ ππ€ππππ
is a valid joint probability density function.
a. Find E(Y1) and E(Y2).
b. Find V(Y1) and V(Y2).
c. Find E(Y1-3 Y2).
Solution a:
πΈ π1 = π¦1
β
ββ
β
ββ
π(π¦1, π¦2)ππ¦1ππ¦2
π¦1
π¦2
0
1
0
6 β 6π¦2 ππ¦1ππ¦2 =π
π
πΈ π2 = π¦2
β
ββ
β
ββ
π(π¦1, π¦2)ππ¦1ππ¦2
π¦2
π¦2
0
1
0
6 β 6π¦2 ππ¦1ππ¦2 =π
π
Solution b:
π π1 = πΈ π12 β πΈ(π1) 2
πΈ π12 = π¦1
2π¦2
0
1
0
6 β 6π¦2 ππ¦1ππ¦2 =π
ππ
π π1 =1
10β
1
4
2
=3
80
π π2 = πΈ π22 β πΈ(π2) 2
πΈ π22 = π¦2
2π¦2
0
1
0
6 β 6π¦2 ππ¦1ππ¦2 =π
ππ
π π2 =3
10β
1
2
2
=1
20
Solution c:
πΈ π1 β 3π2 = πΈ π1 β 3πΈ π2 =1
4β
3
2= β
π
π
Page 9 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
5.70. In exercise 5.32 we determined that the joint density function for Y1, the weight in tons of a bulk item
stocked by a supplier, and Y2, the weight of the item sold by the supplier, has joint density
π(π¦1, π¦2) =
1
π¦1, 0 β€ π¦2 β€ π¦1 β€ 1
0, πππ ππ€ππππ
In this case, the random variable Y1-Y2 measures the amount of stock remaining at the end of the week, a quantity of great importance to the supplier. Find E(Y1-Y2).
Solution:
πΈ π1 β π2 = πΈ π1 β πΈ π2
πΈ π1 = π¦1
β
ββ
β
ββ
π π¦1, π¦2 ππ¦1ππ¦2
= 11
π¦2
1
0
ππ¦1ππ¦2 =1
2
πΈ π2 = π¦2
β
ββ
β
ββ
π π¦1, π¦2 ππ¦1ππ¦2
= π¦2
π¦1
1
π¦2
1
0
ππ¦1ππ¦2 =1
4
πΈ π1 β π2 =1
2β
1
4=
π
π
Page 10 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
Assignment for Wednesday, June 3rd
, 2009
5.75. In Exercise 5.1 we determined that the joint distribution of Y1, the number of contracts awarded to firm A,
and Y2, the number of contracts awarded to firm B, is given by the entries in the following table
y2
y1
0 1 2
0 1/9 2/9 1/9 1 2/9 2/9 0
2 1/9 0 0
Find Cov(Y1, Y2). Does it surprise you that Cov(Y1, Y2) is negative? Why?
Solution:
Cov(Y1, Y2) = E(Y1Y2)-E(Y1).E(Y2)
πΈ π1π2 = π¦1π¦2π π¦1, π¦2 =
π¦2π¦1
0 0 1
9 + 1 0
2
9 + 2 0
1
9 + 0 1
2
9 + 1 1
2
9
+ 0 2 1
9 =
2
9
E(Y1)= E(Y2)=2(1/3)=2/3 since Y1 and Y2 are both binomial with n=2 and p=1/3
Cov(Y1, Y2) = 2/9-(2/3)(2/3)= -2/9
The covariance is negative; it means that the Y2 decreases as the value of Y1 increases.
5.78. In Exercise 5.7, we determined that
π π¦1, π¦2 = 6(1 β π¦2), 0 β€ π¦1 β€ π¦2 β€ 1
0, πππ ππ€ππππ
is a valid joint probability density function. Find Cov(Y1,Y2). Are Y1 and Y2 independent?
Solution:
From Exercise 5.65, E(Y1)=1/4 and E(Y2)=1/2
πΈ π1π2 6π¦1π¦2 1 β π¦2 ππ¦1ππ¦2
π¦2
0
1
0
= 3 π¦23 β π¦2
4 ππ¦2 =3
4β
3
5=
3
20
1
0
Cov(Y1, Y2) = 3/20-1/8=1/40
Since Cov(Y1, Y2) β 0, Y1 and Y2 are not independent.
Page 11 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
5.87. Assume that Y1,Y2 and Y3 are random variables, with
E(Y1)=2 E(Y2)= -1 E(Y3)= 4
V(Y1)=4 V(Y2)=6 V(Y3)= 8
Cov(Y1,Y2)= 1 Cov(Y1,Y3)= -1 Cov(Y2,Y3)= 0
Find E(3Y1+4Y2-6Y3) and V(3Y1+4Y2-6Y3)
Solution:
If U=a1Y1+a2Y2+β¦+anYn Then
E U = ππ . πΈ[ππ]
π
π=1
E(3Y1+4Y2-6Y3) = E (3Y1)+E(4Y2)+E(-6Y3)
= 3E (Y1)+4E(Y2)-6E(Y3) =3(2)+4(-1)-6(4)=-22
V U = ππ2. π[ππ]
π
π=1
+ 2 πππππΆππ£[ππ , ππ ]
V(3Y1+4Y2-6Y3)= 9(4)+16(6)+36(8)+2.3.4.1+2.3.6(-1)+0=480
Page 12 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
5.92. If Y1 is the total time between a customerβs arrival in the store and departure from the service window and
if Y2 is the time spent in line before reaching the window, the joint density of these variables was given in Exercise 5.13 to be
π π¦1, π¦2 = πβπ¦1, 0 β€ π¦2 β€ π¦1 β€ 1
0, πππ ππ€ππππ
The random variable Y1-Y2 represents the time spent at the service window. Find E(Y1-Y2) and V(Y1-Y2). Is it
highly likely that a randomly selected customer would spend more than 4 minutes at the service window?
Solution:
From exercise 5.29,
π1 π¦1 = π¦1πβπ¦1 is a gamma distribution with πΌ = 2, π½ = 1.
Hence E(Y1)=2(1)=2 and V(Y1)= πΌπ½2 = 2.
π2 π¦2 = πβπ¦1
β
π¦2
ππ¦1 = βπβπ¦1 π¦2β = πβπ¦2
Which has a gamma distribution with πΌ = π½ = 1. Hence E(Y2)=V(Y2)=1.
πΈ π1π2 = π¦1π¦2πβπ¦1ππ¦2ππ¦1
π¦1
0
1
0
= π¦1
3
2πβπ¦1ππ¦1 =
Ξ 4 4
2= 3
β
0
Cov(Y1, Y2) = 3-(1)(2)=1
E(Y1-Y2) = 2-1=1
V(Y1-Y2) = 2+1-2(1)=1
Page 13 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
Assignment for Monday, June 8th
, 2009
5.99. A learning experiment requires a rat to run a maze (a network of pathways) until it locates one of three
possible exists. Exit 1 presents a reward of food, but exit 2 and 3 do not. (If the rat eventually select exit 1
almost every time, learning may have taken place.) Let Yi denote the number of times exit i is chosen in successive runnings. For the following, assume that the rat chooses an exit at random on each run.
a. Find the probability than n=6 runs result in Y1=3, Y2=1 and Y3=2.
b. For general n, find E(Y1) and V(Y1). c. Find Cov(Y2,Y3) for general n.
d. To check for the ratβs preference between exit 2 and 3, we may look at Y2-Y3. Find E(Y2-Y3) and V(Y2-
Y3) for general n.
Solution:
Y1 denote the number of times that exit 1 is chosen. The probability that exit 1, 2 or 3 is chosen is equal and is 1/3.
Solution a:
π π¦1, π¦2, π¦3 = π 3,1,2 =6!
3! .1! .2!
1
3
3
. 1
3
1
. 1
3
2
= 0.823
Solution b:
π ππ = πππ => πΈ π1 = ππ1 =1
3π
π ππ = πππ . ππ => π π1 = π 1
3
2
3 =
2
9π
Solution c:
πΆππ£ ππ , ππ‘ = βπ. ππ ππ‘
πΆππ£ π2, π3 = βπ. 1
3
1
3 =
1
9π
Solution d:
πΈ π2 β π3 = πΈ π2 β πΈ π3 = ππ2 β ππ3 =π
3β
π
3= 0
π π2 β π3 = π π2 + π π3 + 2πΆππ£ π2, π3 =2
9π +
2
9π β 2 β
1
9π =
6
9π
Page 14 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
5.103. The National Fire Incident Reporting Service stated that, among residential fires, 73% are in family
homes, 20% are in apartments, and 7% are in other types of dwellings. If four residential fires are independently reported on a single day, what is the probability that two are in family homes, one is in an apartment, and one is
in another type of dwelling?
Solution:
n=4
π π¦1, π¦2, π¦3 = π 2,1,1 =4!
2! .1! .1! 0.73 2. 0.2 1. 0.07 1 = 0.08953
5. 115. In Exercise 5.35, we considered a quality control plan that calls for randomly selecting three items from
the daily production (assumed large) of a certain machine and observing the number of defectives. The
proportion π of defectives produced by the machine varies from day to day and has a uniform distribution on
the interval (0,1).
a. Find the expected number of defectives observed among the three samples items.
b. Find the variance of the number of defectives among the three sampled.
Solution:
Y=(# of defectives)
P= (Proportion of defectives)
n=3
P is uniform on interval (0,1), π π = 1, 0 β€ π β€ 10, πππ ππ€ππππ
E[P]=1/2
Solution a:
πΈ π = πΈ[πΈ[ π π]], given that n=3, πΈ[ π π]=np=3p
πΈ π =E[3P]=3E[P]=3(1/2)=3/2
Solution b:
π π = π πΈ π π + πΈ[π[ π π]] π[ π π]=np(1-p)=3p(1-p)
π π =V[3p]+E[3P(1-P)]=9V[P]+3[E(P)-E(P2)]
E[P]=1/2 and V[P]=1/12 (since P is uniform (0,1)) and E[P2]=V[P]+(E[P])
2=1/12+(1/2)
2=1/3
Therefore,
π π = 9 1
12 + 3
1
2β
1
3 = π. ππ
Page 15 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
5.118. Assume that Y denotes the number of bacteria per cubic centimeter in a particular liquid and that Y has a
Poisson distribution with parameter π. Further assume that π varies from location to location and has a gamma
distribution with parameters πΌ and π½, where πΌ is a positive integer. If we randomly select a location,
a. What is the expected number of bacteria per cubic centimeter? b. What is the standard deviation of the number of bacteria per cubic centimeter?
Solution a:
π π¦ π =ππ¦πβπ
π¦!
πΈ π π = π π has a gamma distribution with parameters πΌ and π½. The expected number of bacteria per cubic centimeter is
given by πΌπ½.
Solution b:
πΈ π π = π and π π π = π.
πΈ π = πΌπ½
π π = πΌπ½2
π π = πΈ π π π + π πΈ π π = πΈ π + π π = πΌπ½ + πΌπ½2 Therefore:
π = πΌπ½(1 + π½)
Page 16 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
6.7. Suppose that a unit of mineral ore contains a proportion Y1 of metal A and a proportion Y2 of metal B.
Experience has shown that the joint probability density function of Y1 and Y2 is uniform over the region
0 β€ π¦1 β€ 1, 0 β€ π¦2 β€ 1, 0 β€ π¦1 + π¦2 β€ 1. Let U=Y1+Y2, the proportion of either metal A or B per unit.
a. Find the probability density function for U. b. Find E(U) by using the answer to (a).
c. Find E(U) by using only the marginal density of Y1 and Y2.
Solution: Y1=(Proportional of metal A)
Y2=(Proportional of metal B)
π π¦1, π¦2 = 2, 0 β€ π¦1 β€ 1 , 0 β€ π¦2 β€ 1 , 0 β€ π¦1 + π¦2 β€ 1 0 , πππ ππ€ππππ
Solution a:
The range of U is 0 β€ π β€ 1 , taking u such that 0 β€ π’ β€ 1 , we will have:
πΉπ π’ = π π β€ π’ = π π1 + π2 β€ π’ = π π1 β€ π’ β π2 = π[π2 β€ π’ β π1]
= 2ππ¦1ππ¦2
π’βπ2
0
= 2π¦1 0π’βπ¦2ππ¦2 = 2 π’ β π¦2 ππ¦2
π’
0
π’
0
π’
0
= 2(π’π¦ βπ¦2
2
2)
0
π’
= 2 π’2 βπ’2
2 = 2
π’2
2 = ππ
ππ π’ =π
ππ’πΉπ π’ =
ππ, 0 β€ π’ β€ 1 π, πππππππππ
Solution b:
πΉπ = π’. 2π’. ππ’ = 2π’2ππ’1
0
= 2π’3
3
0
1
=π
π
1
0
Page 17 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
6.30. A density function sometimes used by engineers to model lengths of life of electronic components is the
Rayleigh density, given by
π π¦ = 2π¦
π π
βπ¦2
π , π¦ > 0
0 , πππ ππ€ππππ.
a. If Y has the Rayleigh density, find the probability density function for U=Y
2.
b. Use the result of (a) to find E(Y) and V(Y)
Solution a:
Y has the Rayleigh density and U=Y2
Let y2=h(y)=u, then π¦ = πβ1 π’ = π’ and
π
ππ’πβ1 π’ =
1
2π’β
1
2
ππ π’ = ππ πβ1 π’ π
ππ’πβ1 π’ =
2 π’
π π
β π’
2
π 1
2π’β
12 =
2π’12
π
π’β12
2 π
βπ’π
ππΌ π = π
π½π
βππ½ , π’ β₯ 0
π , πππ ππ€ππππ.
This is an exponential density with mean π
Solution b:
πΈ π = πΈ π = π’.π
βπ’π
π
β
0
ππ’ =Ξ
32 π3/2
π π’.
πβπ’π
Ξ 32 π3/2
β
0
ππ’ = Ξ 3
2 π1/2 =
1
2Ξ
1
2 π1/2
Ξ 1
2 = π
β΄ π¬ π = π π½
π
πΈ π2 = πΈ π = π
πΈ π = πΈ π2 β πΈ π 2 = π βππ
4= π½(π β
π
π)
Page 18 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
6.46. Let Y1, Y2, β¦Yn be independent Poisson random variables with means π1, π2, β¦,ππ , respectively.
a. Find the probability function of ππππ=1
b. Find the conditional probability function of Y1 given ππππ=1 = π
c. Find the conditional probability function of Y1+Y2 given ππππ=1 = π
Solution a:
Let π = ππππ=1
Since each Yi is Poisson distributed with mean ππ; we have: ππ1 π‘ = ππ(π π‘β1). By Theorem 6.2:
ππ π‘ = ππ1 π‘ . ππ2
π‘ β¦πππ π‘ = πππ
π‘
π
π=1
= ππ1 π π‘β1 . ππ2 π π‘β1 β¦πππ π π‘β1
= π(ππβπ) ππππ=π
This is the moment generating function of a Poisson with mean ππππ=1 .
Solution b: Let k be an arbitrary value for Y1, then:
π π1 = π ππ = π
π
π=1
=π π1 = π, ππ = ππ
π=1
π[ ππ = πππ=1 ]
=π[π1 = π, ππ = π β ππ
π=2 ]
π[ ππ = πππ=1 ]
=π π1 = π π[ ππ = π β ππ
π=2 ]
π[ ππ = πππ=1 ]
by independence
=
π1πππ1
π!. ππ
ππ=2 πβππ ππ
ππ=2
(π β π)!
ππππ=1
ππ ππ
ππ=1
π!
=π!
π! π β π !.π1
π ππππ=2 πβπ
ππππ=1
π = π
π .
π1π . ππ
ππ=2 πβπ
ππππ=1
π ππ
ππ=1
πβπ
= π
π
π1
ππππ=1
π
ππ
ππ=2
ππππ=1
πβπ
This is a binomial probability function with probability of success π = π1
ππππ=1
π
Page 19 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
Solution c:
Let W= Y1+Y2 and k be an arbitrary value for W. Then:
π π = π ππ = π
π
π=1
=π π = π, ππ = ππ
π=1
π[ ππ = πππ=1 ]
=π π = π π[ ππ = π β ππ
π=3 ]
π[ ππ = πππ=1 ]
by independence
=
π1 + π2 πππ1+π2
π!. ππ
ππ=2 πβππ ππ
ππ=3
(π β π)!
ππππ=1
ππ ππ
ππ=1
π!
=π!
π! π β π !. π1 + π2 π ππ
ππ=3 πβπ
ππππ=1
π
= π
π .
π1 + π2 π . ππππ=2 πβπ
ππππ=1
π ππ
ππ=1
πβπ
= π
π
π1 + π2
ππππ=1
π
ππ
ππ=3
ππππ=1
πβπ
This is a binomial probability function with probability of success π = π1+π2 ππ
ππ=1
Page 20 of 20
Ramin Shamshiri STA 5325, HW #3 Due 06/15/09
6.58. Let Y1 and Y2 be independent and uniformly distributed over the interval (0,1).
a. Find the probability density function of U1=min(Y1,Y2)
b. Find E(U1) and V(U1)
Solution:
Given that Y1 and Y2 are independent and each uniformly distributed over the interval (0,1)
π π¦ = 1, 0 < π¦ < 10, πππ ππ€ππππ
πΉ π¦ = 0, π¦ < 0π¦, 0 β€ π¦ β€ 11, π¦ > 0
Solution a:
U1=min(Y1,Y2)
πΊ1 π’ = π π1 β€ π’ = 1 β π[π β€ π’] = 1 β π[π1 > π’, π2 > π’] = 1 β π π1 > π’ π[π2 > π’] (by independence)
The density function of U1 is:
π1 π’ = 2 1 β πΉ π’ β π π’ = π(π β π), 0 β€ π’ β€ 1
π, πππππππππ
Solution b:
πΈ[π1] = π’. 2 1 β π’ ππ’1
0
= 2 π’ β π’2 ππ’ = 2 π’2
2β
π’3
3
0
1 1
0
= 1 β2
3=
π
π
πΈ[π12] = π’2 . 2 1 β π’ ππ’
1
0
= 2 π’2 β π’3 ππ’ = 2 π’3
3β
π’4
4
0
1 1
0
=2
3β
1
2=
π
π
Thus
π[π1] = πΈ[π12] β πΈ[π1] 2 =
1
6β
1
3
2
=π
ππ