Question

3. You are testing two integrated circuits. In each test you either accept (designate by A) or reject (designate by R), with probabilities of 0.9 and 0.1 respectively. Define x-# of acceptable ICs # of tests before you get the first reject. a. Find the joint PMF of X and Y. Also find the Marginal PMFs of each R.V. Are X and Y independent? b. What is the Probability that X is equal to Y? c. Find the Joint Expectation of X and Y.

Answer 1

(a)

Here X can take values 0, 1, and 2.

When X=0, it means number of acceptable ICs is zero. That is at first test we reject the ICs. So when X=0 then Y must be zero and both ICs rejected. That is

P(X=0, Y=0) = 0.1 *0.1 = 0.01

When X=1, it means number of acceptable ICs is one. That is one IC is rejected. IC can be rejected either at first test or on second test. When IC is rejected at first test then Y is zero and we have

P(X=1, Y=0) = 0.1 *0.9 = 0.09

When IC is rejected at second test then Y is one and we have

P(X=1, Y=1) = 0.9 *0.1 = 0.09

When X=2, it means number of acceptable ICs is two. That is no IC is rejected. So

P(X=2, Y=2) = 0.9 *0.9 = 0.81

Following table shows the joint PMF and marginal pdfs of X and Y:

			X
		0	1	2	P(Y=y)
	0	0.01	0.09	0	0.1
Y	1	0	0.09	0	0.09
	2	0	0	0.81	0.81
	P(X=x)	0.01	0.18	0.81	1

Since for each X and Y following is the not true

$P(X=x, Y=y)\neq P(X=x)P(Y=y)$

So X and Y are not independent.

b)

P(X=Y) = P(X=0, Y=0)+P(X=1, Y=1)+P(X=2, Y=2) = 0.01 + 0.09 +0.81 = 0.91

c)

Following table shows the joint E(XY):

X	Y	P(X=x, Y=y)	xyP(X=x, Y=y)
0	0	0.01	0
0	1	0	0
0	2	0	0
1	0	0.09	0
1	1	0.09	0.09
1	2	0	0
2	0	0	0
2	1	0	0
2	2	0.81	3.24
Total			3.33

So,

E(XY) = 〉·.ryP(X = х. У = y) = 3.33