Answer to Question #151504 in Statistics and Probability for Yosef

Question #151504

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).

Expert's answer

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) \\\\\n\n= 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} \\\\\n\n= 0.16 + 0.33 + 0.5 + 0.66 + 0.83 + 1 \\\\\n\n= 3.48 \\\\\n\nE(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) \\\\\n\n= 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} \\\\\n\n= \\frac{1}{36}(2+6+12+20+30+42+40+36+30+22+12) \\\\\n\n= \\frac{252}{36} \\\\\n\n= 7 \\\\\n\nE(Y) = 7"

The joint PMF of X and Y as follows

"E(XY) = \\sum \\sum XYP(X=x, Y=y) \\\\\n\n= (1 \\times 2 \\times \\frac{1}{36}) + (1 \\times 3 \\times \\frac{1}{36}) + (1 \\times 4 \\times \\frac{1}{36}) +\u2026.+ (6 \\times 12 \\times \\frac{1}{36}) \\\\\n\n= \\frac{1}{36}(2+3+4+5+6\u2026.+66+72) \\\\\n\n= \\frac{987}{36} \\\\\n\n= 27.41667 \\\\\n\nE(XY) = 24.41667 \\\\\n\nCOV(XY)=E(XY)-E(X)E(Y) \\\\\n\n=\\frac{329}{12}-3.48 \\times 7 \\\\\n\n=3.0566 \\\\\n\nCov(XY)=3.0566"