Question

1. Suppose you have two random variables, X and Y with joint distribution given by the following tables So, for example, the probability that Y o,x - 0 is 4, and the probability that Y (a) Find the marginal distributions (pmfs) of X and Y, denoted f(x),J(Y). (b) Find the conditional distribution (pmf) of Y give X, denoted f(YX). (c) Find the expected values of X and Y, EX), E(Y). (d) Find the variances of X and Y, Var(X),Var(Y). (e) Find the covariance of X and Y, Cov(X, Y). (f) Find the correlation of X and Y, p(X, Y). (8) Are X and Y independent? Why or why not? 0+2-6

Answer 1

The joint distribution function of two random variables X and Y is given as follows: Y = 0 X=0X=1 0.4 0.2 0 .1 0.3 Y=1 (a) The marginal distributions of X is computed as: P(X=0)=P(X=0.Y = 0)+P( X = 0.Y =1) = 0.4+0.1 = 0.5 P( X =1)= P( X = 1 Y =0)+P(X = 1, Y =1) =0.2+0.3 =0.5 Hence, the marginal distribution function of X is as given in the table below: P(X = x) 0.5 0.5 The marginal distributions of Y is computed as: P(Y = 0)=P(X = 0.Y = 0)+P( X = 1.Y = 0) =0.4+0.2 = 0.6 P(Y =1)=P(X = 0,4 =1)+P( X = 1, Y = 1) = 0.1+0.3 =0.4 Hence, the marginal distribution function of Y is as given in the table below:

The conditional distribution of Y given X is computed as shown below: P(Y =0 X =0) P(Y=0|X =0)=- P(X=0) -0.4 505 = 0.8 P(Y =1 X = 0) P(Y =1|X =0) = P(X=0) 0.1 0.5 = 0.2 P(Y =0|X =1) = P(Y =0|X =1) P(X =1) 0.2 0.5 = 0.4 P(Y =1 X =1) = P(Y =1 X =1) P(X=1) 03 505 (c) The expected values of X and Y are computed as shown below: E(X)= xP(X = x) =(0x P(X = 0))+(1xP(X = 1)) =(0x0.5)+(1+0.5) E(X)=0.51 E(Y)= y.P(Y = y) = (0x P(Y =0))+(1xP(Y = 1)) =(0x0.6)+(1x0.4) E (Y)=0.4

The variance of X and Y are computed as shown below: V(X)=E(X) - E(X) = 2xP(X = x)-0.52 =(0x P(X = 0))+(1+xP(X = 1)) -0.25 =0.5-0.25 V(X)=0.25 V(Y)=E(Y?) - E(Y) = y? P(Y = y)-0.42 = (02xP(Y =0))+(1+xP(T = 1))-0.16 = 0.4-0.16 V(I)=0.24 The covariance of X and Y is calculated as follows: Cov(X,Y)= E(XY)- E(X)E(Y) = EXP(x = x, Y = y) - E(X)E(I) =[(0x0x0.4)+(0x1x0.1)+(1x0x0.2)+(1x1x0.3)]-(0.5x0.4) = 0.3-0.2 = 0.1 Thus, Cov(X,Y)=0.1 The correlation between X and Y is calculated by the following formula: Cov(X,Y) Cor(X,Y)= 0.1 10.250.24 = 0.408 Thus, F(X,Y)=0.408

To determine the independence of X and Y verify the following expression: P(X = x, Y = y)=P(X = x)P(I = y) For different values of X and Y: P(X = 0, y = 0) = 0.4 P(X=0)P(Y = 0) =0.5x0.6 =0.30 P(X=0,=0)=P(X=0)P(Y=0) P(X=0,4 =1)=0.1 P(X = 0) P(Y = 1) =0.5x0.4 = 0.20 P(X=0, Y = 1) =0.1 P(X =0) P(Y =1) P(X=1, Y = 0) =0.2 P( X =1)P(Y = 0) =0.5x0.6 =0.30 P(X = 1, Y =0)=P( X =1) P(Y =0) P( X = 1, Y =1)=0.3 P(X=1) P(Y =1)=0.5x0.4 0.20 P(X = 1, Y = 1)=P(X=1) P(Y = 1) The independence can also be verified using expectations as shown below: EXY)= 0.3 E(X)E(Y)=0.5x0.4 So, EXY +EX E(Y From the above calculations, it is clear that X and Y are not independent.