Bivariate and Multivariate Probability Distributions

[Pages:6]Stat 366 Lab 3 Solutions (September 26, 2006)

page 1

TA: Yury Petrachenko, CAB 484, yuryp@ualberta.ca,

Bivariate and Multivariate Probability Distributions

5.1 Contracts for two construction jobs are randomly assigned to one or more of three firms, A, B, and C. Let Y1 denote the number of contracts assigned to firm A, and Y2 the number of contracts assigned to firm B. Recall that each firm can receive 0, 1, or 2 contracts.

(a) Find the joint probability function for Y1 and Y2. (b) Find F (1, 0).

Solution. (a) Let's list all nine possible assignments of construction jobs to the three firms: AA, AB, AC, BA, BB, BC, CA, CB, CC. (The first symbol signifies the selection for the first job, the second -- for the second one). These assignments are equally likely. Hence, the probability of each of them is 1/9. Now, let's determine the values of Y1 and Y2 in each case:

Assignment AA AB AC BA BB BC CA CB CC

(y1, y2) (2, 0) (1, 1) (1, 0) (1, 1) (0, 2) (0, 1) (1, 0) (0, 1) (0, 0)

p(y1, y2) 1/9 1/9 1/9 1/9 1/9 1/9 1/9 1/9 1/9

To find the joint probability function for Y1 and Y2, we need to rearrange these data into the following table:

y1

0

1

2

0 1/9 2/9 1/9

y2 1 2/9 2/9

0

2 1/9

0

0

(b) Compute

F (1, 0)

=

P (Y1

1,

Y2

0)

=

p(0, 0)

+ p(1, 0)

=

2 9

+

1 9

=

1 3

.

Stat 366 Lab 3 Solutions (September 26, 2006)

5.6 Let Y1 and Y2 have the joint probability density function given by

f (y1,

y2)

=

k

y1y2, 0,

0 y1 1, elsewhere.

0 y2 1,

(a) Find the value of k that makes this a probability density function. (b) Find the joint distribution function for Y1 and Y2. (c) Find P (Y1 1/2, Y2 3/4).

page 2

Solution. (a) We must have

f (y1, y2) dy1dy2 = 1.

- -

Let's compute:

11

1

1=k

y1y2 dy1dy2 = k

00

0

y2

y12 2

1 y1=0

dy2

=

k 2

1 0

y2

dy2

=

k 4

.

So, k = 4 makes f a probability density function.

(b) By definition, if 0 y1 1 and 0 y2 1:

F (y1, y2) =

y1 0

y2

y1

4 t1t2 dy2dy1 = 4 t1 dy1

0

0

y2 0

t2 dy2

=

4

y12 2

y22 2

=

y12y22.

So, overall

1, y12y22,

F (y1, y2) =

y12, y22, 0,

y1 1, y2 1, 0 y1 1, 0 y2 1, 0 y1 1, y2 1, 0 y2 1, y1 1, y1 0, y2 0.

(c) We have P (Y1 1/2, Y2 3/4) = F (1/2, 3/4) =

12 2

3 4

2

=

9 64

.

Stat 366 Lab 3 Solutions (September 26, 2006)

page 3

5.12 Suppose that the random variables Y1 and Y2 have joint probability density function f (y1, y2)

given by

f

(y1,

y2)

=

6y12y2, 0,

0 y1 y2, elsewhere.

y1 + y2 2,

(a) Verify that this is a valid joint density function. (b) What is the probability that Y1 + Y2 is less than 1?

Solution. Let's integrate the density:

1 2-y1

1

1

6y12y22 dy2dy1 =

3y12 (2 - y1)2 - y12 dy1 =

3y12(4 - 4y1) dy1

0 y1

0

0

= 12

1

(y12 - y13) dy1 = 4y13 - 3y14

0

1

= 4 - 3 = 1.

0

Hence, f (y1, y2) 0 is a valid density function.

Now, to find the probability P (Y1 + Y2 < 1) we need to evaluate the following integral:

1/2 1-y1

1/2

6y12y22 dy2dy1 =

3y12 (1 - y1)2 - y12 dy1

0

y1

0

=

1/2 0

(3y12

-

6y13)

dy1

=

y13

-

3 2

y14

1/2 0

=

1 8

-

3 32

=

1 32

.

5.14 Let Y1 and Y2 denote the proportions of time (out of one working day) during which employees I and II, respectively, perform their assigned tasks. The joint relative frequency behavior of

Y1 and Y2 is modeled by the density function

f

(y1,

y2)

=

y1

+

y2, 0,

0 y1 1, elsewhere.

0 y2 1,

(a) Find P (Y1 < 1/2, Y2 > 1/4). (b) Find P (Y1 + Y2 1).

Stat 366 Lab 3 Solutions (September 26, 2006)

page 4

Solution. This is almost a pure integration technique problem. For the first probability consider the integral

1 1/2

1 1/2

1 1/2

(y1 + y2) dy1dy2 =

y1 dy1dy2 +

y2 dy1dy2

1/4 0

1/4 0

1/4 0

=

3 4

1/2 0

y1 dy1

+

1 2

1 1/4

y2 dy2

=

3 4

?

1 2

?

1 2

2

+

1 2

?

1 2

?

1-

12 4

=

21 64

.

Now, the second probability:

1 1-y2

1

(y1 + y2) dy1dy2 =

00

0

1

=

0

y12 2

+

y1y2

1-y2 y1=0

1

dy2 =

0

(1

- y2)2 2

+

(1

-

y2)y2

dy2

1 2

-

1 2

y22

dy2

=

1 2

?

y2

-

y23 3

1 0

=

1 3

.

Marginal and Conditional Probability Distributions

5.17 Continuing Exercise 5.1. (a) Find the marginal probability distribution of Y1. (b) According to results in Chapter 4, Y1 has a binomial distribution with n = 2 and p = 1/3. Is there any conflict between this result and the answer you provided in (a)?

Solution. (a) Refer to the table in the solution to Exercise 5.1 (a) above. To find the marginal probability distribution of Y1, we just need to add up numbers in respective columns of that table. We have

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

=

p(0, 0)

+

p(0, 1)

+

p(0,

2)

=

4 9

,

P (Y1

=

1)

=

4 9

,

P (Y2

=

2)

=

1 9

.

(b) If Y1 is binomial with such parameters, then it must be p(y1) =

2 y1

1 y1 3

2 3

2-y1 .

We can

easily check that this equality holds for y1 = 0, 1, 2. Hence, no conflict.

5.22 Continuing Exercise 5.6.

Stat 366 Lab 3 Solutions (September 26, 2006)

(a) Find the marginal density functions for Y1 and Y2. (b) Find P (Y1 1/2 | Y2 3/4). (c) Find the conditional density function of Y1 given Y2 = y2. (d) Find the conditional density function of Y2 given Y1 = y1. (e) Find P (Y1 3/4 | Y2 = 1/2).

page 5

Solution. (a) Let's integrate for all y2:

1

f1(y1) = 4y1y2 dy2 = 2y1, 0 y1 1.

0

Because the joint density is symmetric in y1 and y2, the other marginal density is

f2(y2) = 2y2, 0 y2 1.

(b) Let's use the definition of the conditional probability:

P (Y1

1/2 | Y2

3/4)

=

P (Y1

1/2 , P (Y2

Y2 3/4)

3/4)

=

1/2 0

1 3/4

4y1y2 dy2dy1

1 3/4

2y2 dy2

=

1/2 0

2y1 dy1

?

1 3/4

2y2 dy2

1 3/4

2y2 dy2

=

1/2

2y1 dy1 = y12

0

1/2

=

0

1 4

.

(c)&(d) Again, by definition

f1(y1|Y2

=

y2)

=

f1(y1|y2)

=

f (y1, y2) f (y2)

=

4y1y2 2y2

=

2y1.

f2(y2|Y1

=

y1)

=

f2(y2|y1)

=

f (y1, y2) f (y1)

=

4y1y2 2y1

=

2y2.

(e) Since we already know the conditional density f1(y1|y2), we can use it now:

P

Y1

3 4

| Y2

=

1 2

=

3/4

f1

0

y1

Y2

=

1 2

dy1 =

3/4 0

2y1 dy1

=

9 16

.

5.28 Continuing Exercise 5.12.

Stat 366 Lab 3 Solutions (September 26, 2006)

page 6

(a) Show that the marginal density of Y1 is a beta density with = 3 and = 2. (b) Derive the marginal density of Y2. (c) Derive the conditional density of Y2 given Y1 = y1. (d) Find P (Y2 < 1.1 | Y1 = .60).

Solution. (a) Compute

2-y1

f1(y1) =

6y12y2 dy2 = 12y12(1 - y1), 0 y1 1.

y1

This is indeed a Beta(2,3) random variable: = 2, = 3 in y-1(1 - y)-1.

(b) We should consider two separate cases. First, 0 y2 1:

y2

6y12y2 dy1 = 2y24.

0

Second, 1 y2 2: Combine the two:

2-y2

6y12y2 dy1 = 2y2(2 - y2)3.

0

2y24,

f2(y2)

=

2y2(2

-

y2)3, 0,

0 y2 1, 1 y2 2, elsewhere.

(c) We have

f2(y2|y1)

=

f (y1, y2) f1(y1)

=

2

y2 - 2y1

,

0,

y1 y2 2 - y1, elsewhere.

(d) Let's use the result in (c): we know that y1 = .60. So,

f2(y2|y1

=

.60)

=

y2 .8

,

.6 y2 1.4.

To compute the probability we should integrate with this density:

P (Y2 < 1.1 | Y1 = .60) =

1.1 .6

y2 .8

dy2

=

(1.12 - 1.6

.62)

=

17 32

.

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download