Question

-1 1 9. Suppose the discrete random variables X and Y are jointly distributed according to the following table: 0 0.1 0.1 0.1 3 0 0.2 0.1 4 0.2 0.1 0.1 2x 1 a. Compute the expected values E(X) and E(Y), variances V(X) and V(Y), and covariance Cov(X,Y) of X and Y. [11] b. Let W = X – Y. Compute E(W) and V(W). [4]

Answer 1

The solution to this problem is given below

The joint mass function of joint probability pandom variables X given by two and Y is Х - 1 0 Total P(y) 10 Y 11 0 1 1 3 4 0. 02. 0.1 0.1 0.2 Oil Oil Oil 0.4 0.3 0.3 0.3 0.4 12 1 1 2 > =0.3. 3 4 Total PC) 0.3 The marginal PMF of X is, P(x= -1) = 0.3 P(x=0} = ObeP(x=1) The marginal PMF of Y is, P(E1) P(y=3) = 0.3 »P (Y= 4) = 0.4. a ;!; E(X)= I x PCX=x) = {1x0.3) + (0x0.6) + (10.3) E(23)- 2 x PCX=x) ={ED*X0.3}+66x0.4) +)*0.3 0.3 > 5 6 = 0 - 0.6 , var (x) = E(83) [: E(X) = 0 ] F(x) =0 , var (X) = 0.6 0.6 2019

Now, ECUE I y PCY=y) у (1X0,3) + (3x0.3)+(4x0.4) 2:8 E(73) { } P ( 7 = y) У (2x0.3) + CX0.3) + (4x0.6) 9.4 i i var (Y) = E(Y²) - [E()7² 9.4 2.82 1.56 var 2. E(Y)= 28 (Y) = 1:56. Now, cou(x, y) - ExY) – EJEC) E(XY) [: E(X)= 0 ] E(XY) = x Es xy p(x=x, Y =Y) ху Or, E(XY) = (-1X1X0.1)+ (Oxi X0.1) + (1xIx0.1) +(-183x0) + (0x3 x 0:2) + (183x 0.1) + C-1X4 X0.2) + (0X4X0:1) + (1x 4 * 0.1)

1 i. Os, E(XY) 0 cor (x,y) = 0 b) Now let Then, E (W) = E(X-Y) = E(X)- EU X-Y. - 0 - 2.8 = -2.8 - - 2.8 Now, Var(W E(W) = var (x-7) var (x) + var (Y)-260V(x,y) [rver (ax+by) = alvar (x) +b Vary) = 0.6 +1.56 - 6x0) +zabcov (X,Y)] 2.16 = -28, var (w) var (W) = 2.16.