sta_5325_hw_3_ramin_shamshiri

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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:

𝑌1 = 0, 𝑖𝑓 𝑡𝑕𝑒 𝑐𝑕𝑖𝑙𝑑 𝑠𝑢𝑟𝑣𝑖𝑣𝑒𝑑1, 𝑖𝑓 𝑛𝑜𝑡

and 𝑌1 =

0, 𝑖𝑓 no belt used1, 𝑖𝑓 adult belt used2, 𝑖𝑓 car − seat belt used.

Notice that Y1 is the number of fatalities per child and, since children’s 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

𝑃(𝑦1, 𝑦2) ≥ 0 for all y1,y2

𝑝 𝑦1, 𝑦2 = 1𝑦1 ,𝑦2, 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: 𝐹 𝑦1, 𝑦2 = 𝑃(𝑌1 ≤ 𝑦1, 𝑌2 ≤ 𝑦2).

Therefore,

𝐹 1,2 = 𝑃 𝑌1 ≤ 1, 𝑌2 ≤ 2 = 𝑃 0,0 + 𝑃 0,1 + 𝑃 0,2 + 𝑃 1,0 + 𝑃 1,1 + 𝑃(1,2)

= 0.38 + 0.14 + 0.24 + 0.17 + 0.02 + 0.05 = 𝟎. 𝟕𝟔 + 𝟎. 𝟐𝟒 = 𝟏

It means that every child in the experiment either survived or didn’t and use either 0,1 or 2 seatbelts.

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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

𝑓(𝑦1, 𝑦2) = 𝑘, 0 ≤ y1 ≤ 2, 0 ≤ y2 ≤ 1, 2y2 ≤ y1

0, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒

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 𝑃(𝑌1 ≥ 3𝑌2). That is, find the probability that the cleaning device reduces the amount of pollutant by one-their or more.

Solution a:

𝑘2

2𝑦2

1

0

𝑑𝑦1𝑑𝑦2 = 1

𝑘(2 − 2𝑦2)1

0

𝑑𝑦2 = 1

2𝑘𝑦2 − 𝑦22

0

1= 1

2𝑘 − 1 = 1 => 𝒌 = 𝟏

Solution b:

𝑦2 =1

3𝑦1

𝑃 𝑌1 ≥ 3𝑌2 = 1

13𝑦1

0

2

0

𝑑𝑦2𝑑𝑦1 = 1

3𝑦1

2

0

𝑑𝑦1

= 𝑦12

6

0

2

=4

6=

𝟐

𝟑

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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 customer’s 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 𝑌1 ≥ 𝑌2. The relative frequency distribution of observed values of Y1 and

Y2 can be modeled by the probability density function

𝑓(𝑦1, 𝑦2) = 𝑒−𝑦1 , 0 ≤ y2 ≤ y1 < ∞

0, elsewhere

With time measured in minutes.

a. Find 𝑃(𝑌1 < 2, 𝑌2 > 1). b. Find 𝑃(𝑌1 ≥ 2𝑌2). c. Find 𝑃(𝑌1 − 𝑌2 ≥ 1). (Notice that Y1-Y2 denotes the time spent at the service window)

Solution a:

𝑃 𝑌1 < 2, 𝑌2 > 1 = 𝑒−𝑦1

2

𝑦2

2

1

𝑑𝑦1𝑑𝑦2

Or we could say 𝑃 𝑌1 < 2, 𝑌2 > 1 = 𝑒−𝑦1𝑦1

1

2

1𝑑𝑦2𝑑𝑦1

Both are the same. Solving the integral,

𝑃 𝑌1 < 2, 𝑌2 > 1 = 𝑒−1 − 2𝑒−2

Solution b:

𝑃 𝑌1 ≥ 2𝑌2 = 𝑒−𝑦1

∞

2𝑦2

∞

0

𝑑𝑦1𝑑𝑦2

Or we could say:

𝑃 𝑌1 ≥ 2𝑌2 = 𝑒−𝑦1

∞

0

∞

2𝑦2

𝑑𝑦2𝑑𝑦1

Both are the same. Solving the integral

𝑃 𝑌1 ≥ 2𝑌2 =1

2

Solution c:

𝑃 𝑌1 − 𝑌2 ≥ 1 = 𝑒−𝑦1

∞

𝑦2+1

∞

0

𝑑𝑦1𝑑𝑦2 = 𝑒−1

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5.20. In exercise 5.4, you were given the following joint probability function for

𝑌1 = 0, 𝑖𝑓 𝑡𝑕𝑒 𝑐𝑕𝑖𝑙𝑑 𝑠𝑢𝑟𝑣𝑖𝑣𝑒𝑑1, 𝑖𝑓 𝑛𝑜𝑡

and 𝑌1 =

0, 𝑖𝑓 no belt used1, 𝑖𝑓 adult belt used2, 𝑖𝑓 car − 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:

𝑃1 𝑦1 = 𝑝(𝑦1, 𝑦2)

𝑦2

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:

𝑃1(𝑌1) = 0.76, 𝑦1 = 00.24, 𝑦1 = 1

𝑃2 𝑦2 = 𝑝(𝑦1, 𝑦2)

𝑦1

𝑃2(𝑌2) = 0.55, 𝑦0 = 00.16, 𝑦1 = 10.29, 𝑦2 = 2

Solution b:

𝑃 𝑌2 𝑌1 = 0 =𝑃(𝑦1 = 0, 𝑦2)

𝑃1(𝑦1 = 0)=

𝑃 0,0 = 0.38

0.76= 0.5 , 𝑦2 = 0

𝑃 0,1 = 0.14

0.76= 0.184, 𝑦2 = 1

𝑃 0,2 = 0.24

0.76= 0.315, 𝑦2 = 2

Solution c:

𝑃 𝑌1 = 0 𝑌2 = 2 =𝑃(0,2)

𝑃2(2)=

0.24

0.29= 0.827

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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

𝑓(𝑦1, 𝑦2) = 3𝑦1, 0 ≤ y2 ≤ y1 < 1

0, elsewhere

a. Find the marginal density function for Y2.

b. For what values of y2 is the conditional density 𝑓( 𝑦1 𝑦2) 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:

𝑓2 𝑦2 = 𝑓(𝑦1, 𝑦2)∞

−∞

𝑑𝑦1 = 3𝑦1

1

𝑦2

𝑑𝑦1 = 3𝑦12

2 𝑦2

1

=3

2−

3𝑦22

2

hence:

𝑓2 𝑦2 = 3

2−

3𝑦22

2, 0 ≤ y2 ≤ 1

0, elsewhere

Solution b: Denominator should not be zero,

𝑓 𝑦1 𝑦2 =𝑓(𝑦1, 𝑦2)

𝑓2(𝑦2) ≠ 0

Therefore, 0 ≤ y2 ≤ 1

Solution c:

𝑓1 𝑦1 = 𝑓(𝑦1, 𝑦2)∞

−∞

𝑑𝑦2 = 3𝑦1

𝑦1

0

𝑑𝑦2 = 3𝑦1𝑦2 0𝑦1 = 3𝑦1

2

𝑃 𝑌2 >1

2 𝑌1 =

3

4 = 𝑓 0.5 0.75 =

𝑓(0.75,0.5)

𝑓1(0.75)=

3𝑦1

𝑓1(0.75)=

94

3𝑦12 𝑦1=0.75

=

94

3 34

2 =𝟒

𝟑

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Assignment for Tuesday, June 2nd

, 2009

5.42. In exercise 5.4 you were given the following joint probability function for

𝑌1 = 0, 𝑖𝑓 𝑐𝑕𝑖𝑙𝑑 𝑠𝑢𝑟𝑣𝑖𝑣𝑒𝑑1, 𝑖𝑓 𝑛𝑜𝑡

and 𝑌2 =

0, 𝑖𝑓 𝑛𝑜 𝑏𝑒𝑙𝑡 𝑢𝑠𝑒𝑑1, 𝑖𝑓 𝑎𝑑𝑢𝑙𝑡 𝑏𝑒𝑙𝑡 𝑢𝑠𝑒𝑑2, 𝑖𝑓 𝑐𝑎𝑟 − 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

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 0.14≠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

𝑓(𝑦1, 𝑦2) = 3𝑦1, 0 ≤ 𝑦2 ≤ 𝑦1 ≤ 1


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:

𝑓2 𝑦2 = 𝑓(𝑦1, 𝑦2)∞

−∞

𝑑𝑦1 = 3𝑦1

1

𝑦2

𝑑𝑦1 = 3𝑦12

2 𝑦2

1

=3

2−

3𝑦22

2

hence:

𝑓2 𝑦2 = 3

2−

3𝑦22

2, 0 ≤ y2 ≤ 1

0, elsewhere

𝑓1 𝑦1 = 𝑓(𝑦1, 𝑦2)∞

−∞

𝑑𝑦2 = 3𝑦1

𝑦1

0

𝑑𝑦2 = 3𝑦1𝑦2 0𝑦1 = 3𝑦1

2

𝑓1 𝑦1 = 3𝑦1

2, 0 ≤ y1 ≤ 10, elsewhere

𝑓 𝑦1, 𝑦2 = 3𝑦1 ≠ 𝑓1 𝑦1 . 𝑓2 𝑦2

Hence, Y1 and Y2 are dependent.

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5.44. In exercise 5.6, we derived the fact that

𝑓(𝑦1, 𝑦2) = 4𝑦1𝑦2, 0 ≤ 𝑦1 ≤ 1, 0 ≤ 𝑦2 ≤ 1


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:

𝑓2 𝑦2 = 𝑓(𝑦1, 𝑦2)∞

−∞

𝑑𝑦1 = 4𝑦1𝑦2

1

0

𝑑𝑦1 = 2𝑦12𝑦2

0

1= 2𝑦2

hence:

𝑓2 𝑦2 = 2𝑦2, 0 ≤ y2 ≤ 1

0, elsewhere

𝑓1 𝑦1 = 𝑓(𝑦1, 𝑦2)∞

−∞

𝑑𝑦2 = 4𝑦1𝑦2

1

0

𝑑𝑦2 = 2𝑦1𝑦22

0

1= 2𝑦1

𝑓1 𝑦1 = 2𝑦1, 0 ≤ y1 ≤ 1

0, elsewhere

𝒇 𝒚𝟏, 𝒚𝟐 = 𝟒𝒚𝟏𝒚𝟐 = 𝒇𝟏 𝒚𝟏 . 𝒇𝟐 𝒚𝟐

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5.65. In exercise 5.7, we determined that

𝑓(𝑦1, 𝑦2) = 6(1 − 𝑦2), 0 ≤ 𝑦1 ≤ 𝑦2 ≤ 1


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:

𝐸 𝑌1 = 𝑦1

∞

−∞

∞

−∞

𝑓(𝑦1, 𝑦2)𝑑𝑦1𝑑𝑦2

𝑦1

𝑦2

0

1

0

6 − 6𝑦2 𝑑𝑦1𝑑𝑦2 =𝟏

𝟒

𝐸 𝑌2 = 𝑦2

∞

−∞

∞

−∞

𝑓(𝑦1, 𝑦2)𝑑𝑦1𝑑𝑦2

𝑦2

𝑦2

0

1

0

6 − 6𝑦2 𝑑𝑦1𝑑𝑦2 =𝟏

𝟐

Solution b:

𝑉 𝑌1 = 𝐸 𝑌12 − 𝐸(𝑌1) 2

𝐸 𝑌12 = 𝑦1

2𝑦2

0

1

0

6 − 6𝑦2 𝑑𝑦1𝑑𝑦2 =𝟏

𝟏𝟎

𝑉 𝑌1 =1

10−

1

4

2

=3

80

𝑉 𝑌2 = 𝐸 𝑌22 − 𝐸(𝑌2) 2

𝐸 𝑌22 = 𝑦2

2𝑦2

0

1

0

6 − 6𝑦2 𝑑𝑦1𝑑𝑦2 =𝟑

𝟏𝟎

𝑉 𝑌2 =3

10−

1

2

2

=1

20

Solution c:

𝐸 𝑌1 − 3𝑌2 = 𝐸 𝑌1 − 3𝐸 𝑌2 =1

4−

3

2= −

𝟓

𝟔

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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

𝑓(𝑦1, 𝑦2) =

1

𝑦1, 0 ≤ 𝑦2 ≤ 𝑦1 ≤ 1


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:

𝐸 𝑌1 − 𝑌2 = 𝐸 𝑌1 − 𝐸 𝑌2

𝐸 𝑌1 = 𝑦1

∞

−∞

∞

−∞

𝑓 𝑦1, 𝑦2 𝑑𝑦1𝑑𝑦2

= 11

𝑦2

1

0

𝑑𝑦1𝑑𝑦2 =1

2

𝐸 𝑌2 = 𝑦2

∞

−∞

∞

−∞

𝑓 𝑦1, 𝑦2 𝑑𝑦1𝑑𝑦2

= 𝑦2

𝑦1

1

𝑦2

1

0

𝑑𝑦1𝑑𝑦2 =1

4

𝐸 𝑌1 − 𝑌2 =1

2−

1

4=

𝟏

𝟒

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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)

𝐸 𝑌1𝑌2 = 𝑦1𝑦2𝑝 𝑦1, 𝑦2 =

𝑦2𝑦1

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

𝑓 𝑦1, 𝑦2 = 6(1 − 𝑦2), 0 ≤ 𝑦1 ≤ 𝑦2 ≤ 1


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

𝐸 𝑌1𝑌2 6𝑦1𝑦2 1 − 𝑦2 𝑑𝑦1𝑑𝑦2

𝑦2

0

1

0

= 3 𝑦23 − 𝑦2

4 𝑑𝑦2 =3

4−

3

5=

3

20

1

0

Cov(Y1, Y2) = 3/20-1/8=1/40

Since Cov(Y1, Y2) ≠ 0, Y1 and Y2 are not independent.

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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+…+anYn Then

E U = 𝑎𝑖 . 𝐸[𝑌𝑖]

𝑛

𝑖=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 = 𝑎𝑖2. 𝑉[𝑌𝑖]

𝑛

𝑖=1

+ 2 𝑎𝑖𝑎𝑗𝐶𝑜𝑣[𝑌𝑖 , 𝑌𝑗 ]

V(3Y1+4Y2-6Y3)= 9(4)+16(6)+36(8)+2.3.4.1+2.3.6(-1)+0=480

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5.92. If Y1 is the total time between a customer’s 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

𝑓 𝑦1, 𝑦2 = 𝑒−𝑦1, 0 ≤ 𝑦2 ≤ 𝑦1 ≤ 1


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,

𝑓1 𝑦1 = 𝑦1𝑒−𝑦1 is a gamma distribution with 𝛼 = 2, 𝛽 = 1.

Hence E(Y1)=2(1)=2 and V(Y1)= 𝛼𝛽2 = 2.

𝑓2 𝑦2 = 𝑒−𝑦1

∞

𝑦2

𝑑𝑦1 = −𝑒−𝑦1 𝑦2∞ = 𝑒−𝑦2

Which has a gamma distribution with 𝛼 = 𝛽 = 1. Hence E(Y2)=V(Y2)=1.

𝐸 𝑌1𝑌2 = 𝑦1𝑦2𝑒−𝑦1𝑑𝑦2𝑑𝑦1

𝑦1

0

1

0

= 𝑦1

3

2𝑒−𝑦1𝑑𝑦1 =

Γ 4 4

2= 3

∞

0

Cov(Y1, Y2) = 3-(1)(2)=1

E(Y1-Y2) = 2-1=1

V(Y1-Y2) = 2+1-2(1)=1

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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 rat’s 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:

𝑃 𝑦1, 𝑦2, 𝑦3 = 𝑃 3,1,2 =6!

3! .1! .2!

1

3

3

. 1

3

1

. 1

3

2

= 0.823

Solution b:

𝑃 𝑌𝑖 = 𝑛𝑃𝑖 => 𝐸 𝑌1 = 𝑛𝑃1 =1

3𝑛

𝑉 𝑌𝑖 = 𝑛𝑝𝑖 . 𝑞𝑖 => 𝑉 𝑃1 = 𝑛 1

3

2

3 =

2

9𝑛

Solution c:

𝐶𝑜𝑣 𝑌𝑠 , 𝑌𝑡 = −𝑛. 𝑝𝑠𝑝𝑡

𝐶𝑜𝑣 𝑌2, 𝑌3 = −𝑛. 1

3

1

3 =

1

9𝑛

Solution d:

𝐸 𝑌2 − 𝑌3 = 𝐸 𝑌2 − 𝐸 𝑌3 = 𝑛𝑃2 − 𝑛𝑃3 =𝑛

3−

𝑛

3= 0

𝑉 𝑌2 − 𝑌3 = 𝑉 𝑌2 + 𝑉 𝑌3 + 2𝐶𝑜𝑣 𝑌2, 𝑌3 =2

9𝑛 +

2

9𝑛 − 2 −

1

9𝑛 =

6

9𝑛

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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

𝑃 𝑦1, 𝑦2, 𝑦3 = 𝑃 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 𝜌 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), 𝑓 𝑝 = 1, 0 ≤ 𝑝 ≤ 10, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒

E[P]=1/2

Solution a:

𝐸 𝑌 = 𝐸[𝐸[ 𝑌 𝑃]], given that n=3, 𝐸[ 𝑌 𝑃]=np=3p

𝐸 𝑌 =E[3P]=3E[P]=3(1/2)=3/2

Solution b:

𝑉 𝑌 = 𝑉 𝐸 𝑌 𝑃 + 𝐸[𝑉[ 𝑌 𝑃]] 𝑉[ 𝑌 𝑃]=np(1-p)=3p(1-p)

𝑉 𝑌 =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,

𝑉 𝑌 = 9 1

12 + 3

1

2−

1

3 = 𝟏. 𝟐𝟓

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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 𝜆. Further assume that 𝜆 varies from location to location and has a gamma

distribution with parameters 𝛼 and 𝛽, where 𝛼 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:

𝑝 𝑦 𝜆 =𝜆𝑦𝑒−𝜆

𝑦!

𝐸 𝑌 𝜆 = 𝜆 𝜆 has a gamma distribution with parameters 𝛼 and 𝛽. The expected number of bacteria per cubic centimeter is

given by 𝛼𝛽.

Solution b:

𝐸 𝑌 𝜆 = 𝜆 and 𝑉 𝑌 𝜆 = 𝜆.

𝐸 𝜆 = 𝛼𝛽

𝑉 𝜆 = 𝛼𝛽2

𝑉 𝑌 = 𝐸 𝑉 𝑌 𝜆 + 𝑉 𝐸 𝑌 𝜆 = 𝐸 𝜆 + 𝑉 𝜆 = 𝛼𝛽 + 𝛼𝛽2 Therefore:

𝜍 = 𝛼𝛽(1 + 𝛽)

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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 ≤ 𝑦1 ≤ 1, 0 ≤ 𝑦2 ≤ 1, 0 ≤ 𝑦1 + 𝑦2 ≤ 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)

𝑓 𝑦1, 𝑦2 = 2, 0 ≤ 𝑦1 ≤ 1 , 0 ≤ 𝑦2 ≤ 1 , 0 ≤ 𝑦1 + 𝑦2 ≤ 1 0 , 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒

Solution a:

The range of U is 0 ≤ 𝑈 ≤ 1 , taking u such that 0 ≤ 𝑢 ≤ 1 , we will have:

𝐹𝑈 𝑢 = 𝑃 𝑈 ≤ 𝑢 = 𝑃 𝑌1 + 𝑌2 ≤ 𝑢 = 𝑃 𝑌1 ≤ 𝑢 − 𝑌2 = 𝑃[𝑌2 ≤ 𝑢 − 𝑌1]

= 2𝑑𝑦1𝑑𝑦2

𝑢−𝑌2

0

= 2𝑦1 0𝑢−𝑦2𝑑𝑦2 = 2 𝑢 − 𝑦2 𝑑𝑦2

𝑢

0

𝑢

0

𝑢

0

= 2(𝑢𝑦 −𝑦2

2

2)

0

𝑢

= 2 𝑢2 −𝑢2

2 = 2

𝑢2

2 = 𝒖𝟐

𝑓𝑈 𝑢 =𝑑

𝑑𝑢𝐹𝑈 𝑢 =

𝟐𝒖, 0 ≤ 𝑢 ≤ 1 𝟎, 𝒆𝒍𝒔𝒆𝒘𝒉𝒆𝒓𝒆

Solution b:

𝐹𝑈 = 𝑢. 2𝑢. 𝑑𝑢 = 2𝑢2𝑑𝑢1

0

= 2𝑢3

3

0

1

=𝟐

𝟑

1

0

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6.30. A density function sometimes used by engineers to model lengths of life of electronic components is the

Rayleigh density, given by

𝑓 𝑦 = 2𝑦

𝜃 𝑒

−𝑦2

𝜃 , 𝑦 > 0

0 , 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒.

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 𝑦 = 𝑕−1 𝑢 = 𝑢 and

𝑑

𝑑𝑢𝑕−1 𝑢 =

1

2𝑢−

1

2

𝑓𝑈 𝑢 = 𝑓𝑌 𝑕−1 𝑢 𝑑

𝑑𝑢𝑕−1 𝑢 =

2 𝑢

𝜃 𝑒

− 𝑢

2

𝜃 1

2𝑢−

12 =

2𝑢12

𝜃

𝑢−12

2 𝑒

−𝑢𝜃

𝒇𝑼 𝒖 = 𝟏

𝜽𝒆

−𝒖𝜽 , 𝑢 ≥ 0

𝟎 , 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒.

This is an exponential density with mean 𝜃

Solution b:

𝐸 𝑌 = 𝐸 𝑈 = 𝑢.𝑒

−𝑢𝜃

𝜃

∞

0

𝑑𝑢 =Γ

32 𝜃3/2

𝜃 𝑢.

𝑒−𝑢𝜃

Γ 32 𝜃3/2

∞

0

𝑑𝑢 = Γ 3

2 𝜃1/2 =

1

2Γ

1

2 𝜃1/2

Γ 1

2 = 𝜋

∴ 𝑬 𝒀 = 𝝅𝜽

𝟐

𝐸 𝑌2 = 𝐸 𝑈 = 𝜃

𝐸 𝑌 = 𝐸 𝑌2 − 𝐸 𝑌 2 = 𝜃 −𝜋𝜃

4= 𝜽(𝟏 −

𝝅

𝟒)

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6.46. Let Y1, Y2, …Yn be independent Poisson random variables with means 𝜆1, 𝜆2, …,𝜆𝑛 , respectively.

a. Find the probability function of 𝑌𝑖𝑛𝑖=1

b. Find the conditional probability function of Y1 given 𝑌𝑖𝑛𝑖=1 = 𝑚

c. Find the conditional probability function of Y1+Y2 given 𝑌𝑖𝑛𝑖=1 = 𝑚

Solution a:

Let 𝑈 = 𝑌𝑖𝑛𝑖=1

Since each Yi is Poisson distributed with mean 𝜆𝑖; we have: 𝑚𝑌1 𝑡 = 𝑒𝜆(𝑒 𝑡−1). By Theorem 6.2:

𝑚𝑉 𝑡 = 𝑚𝑌1 𝑡 . 𝑚𝑌2

𝑡 …𝑚𝑌𝑛 𝑡 = 𝑚𝑌𝑖

𝑡

𝑛

𝑖=1

= 𝑒𝜆1 𝑒 𝑡−1 . 𝑒𝜆2 𝑒 𝑡−1 …𝑒𝜆𝑛 𝑒 𝑡−1

= 𝒆(𝒆𝒕−𝟏) 𝝀𝒊𝒏𝒊=𝟏

This is the moment generating function of a Poisson with mean 𝜆𝑖𝑛𝑖=1 .

Solution b: Let k be an arbitrary value for Y1, then:

𝑃 𝑌1 = 𝑘 𝑌𝑖 = 𝑚

𝑛

𝑖=1

=𝑃 𝑌1 = 𝑘, 𝑌𝑖 = 𝑚𝑛

𝑖=1

𝑃[ 𝑌𝑖 = 𝑚𝑛𝑖=1 ]

=𝑃[𝑌1 = 𝑘, 𝑌𝑖 = 𝑚 − 𝑘𝑛

𝑖=2 ]


=𝑃 𝑌1 = 𝑘 𝑃[ 𝑌𝑖 = 𝑚 − 𝑘𝑛

𝑖=2 ]


by independence

=

𝜆1𝑘𝑒𝜆1

𝑘!. 𝜆𝑖

𝑛𝑖=2 𝑚−𝑘𝑒 𝜆𝑖

𝑛𝑖=2

(𝑚 − 𝑘)!

𝜆𝑖𝑛𝑖=1

𝑚𝑒 𝜆𝑖

𝑛𝑖=1

𝑚!

=𝑚!

𝑘! 𝑚 − 𝑘 !.𝜆1

𝑘 𝜆𝑖𝑛𝑖=2 𝑚−𝑘

𝜆𝑖𝑛𝑖=1

𝑚 = 𝑚

𝑘 .

𝜆1𝑘 . 𝜆𝑖

𝑛𝑖=2 𝑚−𝑘

𝜆𝑖𝑛𝑖=1

𝑘 𝜆𝑖

𝑛𝑖=1

𝑚−𝑘

= 𝑚

𝑘

𝜆1

𝜆𝑖𝑛𝑖=1

𝑘

𝜆𝑖

𝑛𝑖=2

𝜆𝑖𝑛𝑖=1

𝑚−𝑘

This is a binomial probability function with probability of success 𝑃 = 𝜆1

𝜆𝑖𝑛𝑖=1

𝑘

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Solution c:

Let W= Y1+Y2 and k be an arbitrary value for W. Then:

𝑃 𝑊 = 𝑘 𝑌𝑖 = 𝑚

𝑛

𝑖=1

=𝑃 𝑊 = 𝑘, 𝑌𝑖 = 𝑚𝑛

𝑖=1


=𝑃 𝑊 = 𝑘 𝑃[ 𝑌𝑖 = 𝑚 − 𝑘𝑛

𝑖=3 ]


by independence

=

𝜆1 + 𝜆2 𝑘𝑒𝜆1+𝜆2

𝑘!. 𝜆𝑖

𝑛𝑖=2 𝑚−𝑘𝑒 𝜆𝑖

𝑛𝑖=3

(𝑚 − 𝑘)!

𝜆𝑖𝑛𝑖=1

𝑚𝑒 𝜆𝑖

𝑛𝑖=1

𝑚!

=𝑚!

𝑘! 𝑚 − 𝑘 !. 𝜆1 + 𝜆2 𝑘 𝜆𝑖

𝑛𝑖=3 𝑚−𝑘

𝜆𝑖𝑛𝑖=1

𝑚

= 𝑚

𝑘 .

𝜆1 + 𝜆2 𝑘 . 𝜆𝑖𝑛𝑖=2 𝑚−𝑘

𝜆𝑖𝑛𝑖=1

𝑘 𝜆𝑖

𝑛𝑖=1

𝑚−𝑘

= 𝑚

𝑘

𝜆1 + 𝜆2

𝜆𝑖𝑛𝑖=1

𝑘

𝜆𝑖

𝑛𝑖=3

𝜆𝑖𝑛𝑖=1

𝑚−𝑘

This is a binomial probability function with probability of success 𝑃 = 𝜆1+𝜆2 𝜆𝑖

𝑛𝑖=1

of 20


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)

𝑓 𝑦 = 1, 0 < 𝑦 < 10, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒

𝐹 𝑦 = 0, 𝑦 < 0𝑦, 0 ≤ 𝑦 ≤ 11, 𝑦 > 0

Solution a:

U1=min(Y1,Y2)

𝐺1 𝑢 = 𝑃 𝑈1 ≤ 𝑢 = 1 − 𝑃[𝑈 ≤ 𝑢] = 1 − 𝑃[𝑌1 > 𝑢, 𝑌2 > 𝑢] = 1 − 𝑃 𝑌1 > 𝑢 𝑃[𝑌2 > 𝑢] (by independence)

The density function of U1 is:

𝑔1 𝑢 = 2 1 − 𝐹 𝑢 − 𝑓 𝑢 = 𝟐(𝟏 − 𝒖), 0 ≤ 𝑢 ≤ 1

𝟎, 𝒆𝒍𝒔𝒆𝒘𝒉𝒆𝒓𝒆

Solution b:

𝐸[𝑈1] = 𝑢. 2 1 − 𝑢 𝑑𝑢1

0

= 2 𝑢 − 𝑢2 𝑑𝑢 = 2 𝑢2

2−

𝑢3

3

0

1 1

0

= 1 −2

3=

𝟏

𝟑

𝐸[𝑈12] = 𝑢2 . 2 1 − 𝑢 𝑑𝑢

1

0

= 2 𝑢2 − 𝑢3 𝑑𝑢 = 2 𝑢3

3−

𝑢4

4

0

1 1

0

=2

3−

1

2=

𝟏

𝟔

Thus

𝑉[𝑈1] = 𝐸[𝑈12] − 𝐸[𝑈1] 2 =

1

6−

1

3

2

=𝟏

𝟏𝟖

sta_5325_hw_3_ramin_shamshiri

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