The following table presents the joint probability mass function pmf of variable

Question

The following table presents the joint probability mass function pmf of variables X and Y. 0.14 0.06 0.21 0.09 0.35 0.15 2 (a) Compute the probability that P(X Y2). (b) Compute the expected value of the function G(X,Y) -3. (c) Compute the marginal probability distributions of X and Y (d) Compute the variances of X and Y (e) Compute the covariance and correlation of X and Y (f) Are X and Y statistically independent? (g) Compute the conditional distribution of Y given X - 2 (h) Compute ETYIX -2 (i) Suppose M(X,Y)-2X-Y+10: i. Compute the expected value of M ii. Compute the variance of M. Page 1 a

Explanation / Answer

(a) P( X + Y<= 2) = P(X = 1, Y =0) + Pr(X = 1, Y =1) + Pr(X = 2,Y= 0)

= 0.14 + 0.06 + 0.09 = 0.29

(b) Expected value of function G(X,Y) = Y2 /X - 3

Here the domian of G for each value of X and Y is  

Now E(G) = -3 * 0.14 -2 * 0.06 + 1 * 0.21 - 3 * 0.09 - 2.5 * 0.35 - 1 * 0.15 = -1.625

(c) Marginal density

f(X) = (0.14 + 0.06 + 0.21) = 0.41 ; X =1

= 0.09 + 0.35 + 0.15 = 0.59 ; X = 2

f(Y) = (0.14 + 0.09) = 0.23 ; Y = 0

= (0.06 + 0.35) = 0.41 ; Y = 1

= (0.21 + 0.35) = 0.36 ; Y =2

(d) Here E(X) = 0.41 * 1 + 0.59 * 2 = 1.59

E(X2) = 0.41 * 1 + 0.59 * 22 = 2.77

Var(X) = E(X2) - E(X) 2 = 2.77 - 1.592 = 0.2419

E(Y) = 0.23 * 0 + 0.41 * 1 + 0.36 * 2= 1.13

E(Y2) = 0.23 * 02 + 0.41 * 12 + 0.36 * 22= 1.85

Var(Y) = E(Y2) - E(Y) 2 = 1.85 - 1.132 = 0.5713

X/Y 0 1 2 1 -3 -2 1 2 -3 -2.5 -1

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