Question

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.

2. (b) Find the marginal probability of Y1 and Y2.

3. (c) Are Y1 and Y2 independent? Why? (d) Find E(Y1 − Y2).

4. (e) Find Cov(Y1, Y2)

Answer 1

(b)

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.

Assignment	(y1, y2)
AA	(2,0)
AB	(1,1)
AC	(1,0)
BA	(1,1)
BB	(0,2)
BC	(0,1)
CA	(1,0)
CB	(0,1)
CC	(0,0)

The joint probability of Y1 and Y2 is,

Y2 \Y1	0	1	2
0	1/9	2/9	1/9
1	2/9	2/9	0
2	1/9	0	0

(b)

The marginal probability of Y1 is,

P(Y1 = 0) = 1/9 + 2/9 + 1/9 = 4/9

P(Y1 = 1) = 2/9 + 2/9 = 4/9

P(Y1 = 2) = 1/9

The marginal probability of Y2 is,

P(Y2 = 0) = 1/9 + 2/9 + 1/9 = 4/9

P(Y2 = 1) = 2/9 + 2/9 = 4/9

P(Y2 = 2) = 1/9

(c)

P(Y1 = 0, Y2 = 0) = 1/9

P(Y1 = 0) P(Y2 = 0) = (4/9) * (4/9) = 16/25

Since P(Y1 = 0, Y2 = 0) P(Y1 = 0) P(Y2 = 0) , Y1 and Y2 are not independent.

(d)

E(Y1) = 0 * 4/9 + 1 * 4/9 + 2 * 1/9 = 2/3

E(Y2) = 0 * 4/9 + 1 * 4/9 + 2 * 1/9 = 2/3

E(Y1 − Y2) = E(Y1) - E(Y2) = 2/3 - 2/3 = 0

(e)

E(Y1 * Y2) = 0 * 0 * 1/9 + 1 * 0 * 2/9 + 2 * 0 * 1/9 + 0 * 1 * 2/9 + 1 * 1 * 2/9 + 2 * 1 * 0 + 0 * 2 * 1/9 + 1 * 2 * 0 + 2 * 2 * 0

= 2/9

Cov(Y1, Y2) = E(Y1 * Y2) - E(Y1) E(Y2) = 2/9 - (2/3) * (2/3) = -2/9