Joint Probability Distributions – Examples Example 1 A certain ...

f. Determine the marginal pmf of Y .Solution p Y (n) = P (Y = n) = p(0; n) + p(1; n) + p(2; n) + p(3; n) + p(4; n)y 0 1 2 3p Y (y) 0.19 0.30 0.28 0.23g. By inspection of the probabilities P (X = 4), P (Y = 0), and P (X = 4; Y = 0), are X and Yindependent random variables? Explain.Solution P (X = 4) = 0:12, P (Y = 0) = 0:19, and P (X = 4; Y = 0) = 0. Hence there exists anordered pair (x; y) such that p(x; y) 6= p X (x)¢p Y (y) and therefore X and Y are not independentrandom variables.Example 2 An instructor has given a short quiz consisting of two parts. For a randomly selectedstudent, let X be the number of points earned on the …rst part and Y be the number of pointsearned on the second part. Suppose the joint pmf of X and Y is given in the following table:X / Y 0 5 10 150 0.02 0.06 0.02 0.105 0.04 0.15 0.20 0.1010 0.01 0.15 0.14 0.01a. If the score recorded in the grade book is the total number of points earned on the two parts,what is the expected recorded score E(X + Y )?Solution The expected recorded score is:E(X + Y ) = X X(x + y) ¢ p(x; y)x y= (0 + 0)(0:02) + (0 + 5)(0:06) + : : : + (10 + 15)(0:01) = 14:10:Note that E(X + Y ) = E(X) + E(Y ).b. If the maximum of the two scores is recorded, what is the expected recorded score?Solution The expected recorded score is:E [max(X; Y )] = X Xmax(x; y) ¢ p(x; y)x y= (0)(0:02) + (5)(0:06) + : : : + (15)(0:01) = 9:60:c. Compute the covariance for X and Y .Solution The covariance for X and Y is computed as follows:E(X) = X x x ¢ p X(x) = 5:55E(Y ) = X y ¢ p yY (y) = 8:55E(XY ) = X Xxy ¢ p(x; y) = (0)(0:02) + : : : + (150)(0:01) = 44:25x yCov(X; Y ) = E(XY ) ¡ E(X)E(Y ) = 44:25 ¡ (5:55)(8:55) = ¡3:2025:2: