Question

3.3.11 Suppose you roll two fair six-sided dice. Let X be the number showing on the
first die, and let Y be the sum of the numbers showing on the two dice. Compute E(X),
E(Y), E(XY), and Cov(X, Y).

Answer 1

Let X be the number showing on the first die

Y be the sum of the numbers showing on the two dice

The random variable X takes values 1,2,3,4,5 and 6 with probability 1/6

The PMF of X can be written in the table form as follows

$E(X) = \sum XP(X=x) \\ = 1 \times \frac{1}{6} + 2 \times \frac{1}{6} + 3 \times \frac{1}{6} + 4 \times \frac{1}{6} + 5 \times \frac{1}{6} + 6 \times \frac{1}{6} \\ = 0.16 + 0.33 + 0.5 + 0.66 + 0.83 + 1 \\ = 3.48 \\ E(X) = 3.48$

The random variable X takes values 2,3,4,5,6,7,8,9,10,11,12

The PMF of X can be written in the table form as follows

$E(Y) = \sum YP(Y=y) \\ = 2 \times \frac{1}{36} + 3 \times \frac{2}{36} + 4 \times \frac{3}{36} + 5 \times \frac{4}{36} + 6 \times \frac{5}{36} + 7 \times \frac{6}{36} + 8 \times \frac{5}{36} + 9 \times \frac{4}{36} + 10 \times \frac{3}{36} + 11 \times \frac{2}{36} + 12 \times \frac{1}{36} \\ = \frac{1}{36}(2+6+12+20+30+42+40+36+30+22+12) \\ = \frac{252}{36} \\ = 7 \\ E(Y) = 7$

The joint PMF of X and Y as follows

$E(XY) = \sum \sum XYP(X=x, Y=y) \\ = (1 \times 2 \times \frac{1}{36}) + (1 \times 3 \times \frac{1}{36}) + (1 \times 4 \times \frac{1}{36}) +….+ (6 \times 12 \times \frac{1}{36}) \\ = \frac{1}{36}(2+3+4+5+6….+66+72) \\ = \frac{987}{36} \\ = 27.41667 \\ E(XY) = 24.41667 \\ COV(XY)=E(XY)-E(X)E(Y) \\ =\frac{329}{12}-3.48 \times 7 \\ =3.0566 \\ Cov(XY)=3.0566$