Question

A bag contains four balls labelled 0,1,2 and 3. Two balls are chosen at random, one after the other and without replacement. Let X be the label on the first one and Y be the label on the second ball.

1. Find joint probability mass function of X and Y?

2. Calculate E(x), E(Y) , E(XY), and Cov (XY)?

Please explain answer.

Answer 1

1) The joint PMF for X and Y here is obtained as:

(Note that each of the outcome is equally likely and total outcomes = 4*3 = 12

P(X = 0, Y = 1) = 1/12
P(X = 0, Y = 2) = 1/12
P(X = 0, Y = 3) = 1/12
P(X = 1, Y = 0) = 1/12
P(X = 1, Y = 2) = 1/12
P(X = 1, Y = 3) = 1/12
P(X = 2, Y = 1) = 1/12
P(X = 2, Y = 0) = 1/12
P(X = 2, Y = 3) = 1/12
P(X = 3, Y = 1) = 1/12
P(X = 3, Y = 2) = 1/12
P(X = 3, Y = 0) = 1/12

This is the required PMF for (X, Y)

2) From the above distribution, we get the marginal distributions as:

P(X = 0) = P(X = 1) = P(X = 2) = P(X = 3) = 1/4
P(Y = 0) = P(Y = 1) = P(Y = 2) = P(Y = 3) = 1/4

Therefore, E(X) = E(Y) = (0 + 1 + 2 + 3)/4 = 1.5

Therefore E(X) = E(Y) = 1.5

The value for XY for each combination of X and Y is computed first as:

P(XY = 0) = 6/12 = 1/2
P(XY = 2) = 2/12 = 1/6
P(XY = 3) = 2/12 = 1/6
P(XY = 6) = 2/12 = 1/6

Therefore E(XY) = 0*(1/2) + (1/6)*(2 + 3 + 6) = 11/6

Therefore E(XY) = 11/6

Cov(X, Y) = E(XY) - E(X)E(Y) = (11/6) - 1.5² = - 0.4167

Therefore Cov(X, Y) = -0.4167