Question

Consider a pharmaceutical company that produces flu vaccines that are good with reliability p = 0.9, independently from vaccine to vaccine. Assume that you three vaccines to test whether they are effective or not. Define the random variable X as the number of effective vaccines circuits that you obtain after testing each of the three selected ones. Define the random variable Y as the number of vaccines that you test from the three before you find an ineffective one. If all three vaccines are found to be effective, let Y = 3. Note both X, Y can take the value 0. (a) Find the possible values of X, Y . (b) Compute the joint PMF of X, Y as a matrix. (c) Compute the for the marginal probability mass function (PMF) of X. (d) Compute the marginal PMF PY (y) of the random variable Y . (e) Compute the conditional pmf PY |X(y|x) for all possible values of x, y. (f) Compute E[Y |X = 1] (g) Compute the expected values E[X], E[Y ].

Answer 1

According to the problem,

X = Number of effective vaccines after testing all of the 3 vaccines.

Y = Number of vaccines tested before getting an ineffective one.

(a)

Number of effective vaccines after testing all of the 3 vaccines can be 0(all vaccines ineffective), 1(one effective and two ineffective), 2(two effective and one ineffective) or 3(all effective and no ineffective).

So, X = 0, 1, 2 and 3.

Number of vaccines tested before getting an (the first one) ineffective one can be 0(first one is ineffective vaccine), 1(first one is effective and second one is ineffective), 2(first two are effective and third one is ineffective) or 3(all of three vaccines are effective).

So, Y = 0, 1, 2 and 3

(b)

Denoting effective vaccine as E and ineffective as I we have sample units and corresponding values of X and Y as follows.

Sample unit	X=x	Y=y	P(X=x, Y=y)
EEE	3	3	0.9*0.9*0.9=0.729
IEE	2	0	0.1*0.9*0.9=0.081
EIE	2	1	0.9*0.1*0.9=0.081
EEI	2	2	0.9*0.9*0.1=0.081
IIE	1	0	0.1*0.1*0.9=0.009
IEI	1	0	0.1*0.9*0.1=0.009
EII	1	1	0.9*0.1*0.1=0.009
III	0	0	0.1*0.1*0.1=0.001

The 4*4 joint PMF matrix of X and Y is as follows.

		Y				Marginal total
		0	1	2	3
X	0	0.001	0.000	0.000	0.000	0.001
	1	0.018	0.009	0.000	0.000	0.027
	2	0.081	0.081	0.081	0.000	0.243
	3	0.000	0.000	0.000	0.729	0.729
Marginal total		0.100	0.090	0.081	0.729	1.000

(c)

Marginal PMF for X is as follows.

X=x	P(X=x)
0	0.001
1	0.027
2	0.243
3	0.729
Total	1.000

(d)

Marginal PMF for Y is as follows.

X=y	P(Y=y)
0	0.100
1	0.090
2	0.081
3	0.729
Total	1.000

(e)

Conditional PMF for all possible values of x and y are as follows.

		Y
		0	1	2	3
X	0	0.001/0.001=1.000	0.000/0.001=0.000	0.000/0.001=0.000	0.000/0.001=0.000
	1	0.018/0.027=0.667	0.009/0.027=0.333	0.000/0.027=0.000	0.000/0.027=0.000
	2	0.081/0.243=0.333	0.081/0.243=0.333	0.081/0.243=0.333	0.000/0.243=0.000
	3	0.000/0.729=0.000	0.000/0.729=0.000	0.000/0.729=0.000	0.729/0.729=1.000

(f)

(g)