1) Consider the joint pmf of Exercise 1 on p. 332 of the text (an exercise cover

Question

1) Consider the joint pmf of Exercise 1 on p. 332 of the text (an exercise covered in class during the 11th lecture). Give the conditional pmf of Y given that X = 0, pY |X(y|0). (Note: Your answer should be a relatively simple function of y. It shouldn’t involve x since it’s for the case of X = 0, and any xs you may have had at one point should have each been replaced by a 0. As a check of your work, I’ll give you that the mean of the conditional distribution, E(Y |X = 0), is equal to 9/5.)

1. Let the joint probability mass function of random variables X and Y be given by (x2 +y?) if x = 1,2, y=0,1,2 25x 0 p(x, y) = elsewhere. Are X and Y independent? Why or why not?

Explanation / Answer

Here to know that p(x,y) is independent or not. We will calculate that the correlation coefficient is zero or not.

Here,

Corr(X,Y) = Cov(X,Y)/ X Y

Here the joint distribution table is : I have calculated this by putting values of X and Y in joint distribution

Here px (x)= 0.32; X = 1

= 0.68 ; X =2

E(X) = 1 * 0.32 + 2 * 0.68 = 1.68

E(X2) = 1 * 0.32 + 4 * 0.68 = 3.04

Var(X) = 3.04 - 1.682 = 0.2176

similarly,

py(y) = 0.2 ; Y = 0

= 0.28 ; Y = 1

= 0.52 ; Y =2

E(Y) = 0.2 *0 + 0.28 * 1 + 0.52 * 2 = 1.32

E(Y2) = 0.2 * 0 + 0.28 * 1 + 0.52 * 4 = 2.36

Var(Y) = 2.36 - 1.322 = 0.6176

Here Cov (X,Y) = E(XY) - E(X)E(Y)

E(XY) = 0.04 * 0 * 1 + 0.16 * 0 * 2 + 0.08 * 1 * 1 + 0.2 * 1 * 2 + 0.2 * 2 * 1 + 0.32 * 2 * 2 = 2.16

so Cov(X,Y) = 2.16 - 1.68 * 1.32 = -0.0576

so yess here the value of covariance is not equal to zero so we shall say that X and Y are not independent in nature.

X/Y 0 1 2 Total 1 0.04 0.08 0.2 0.32 2 0.16 0.2 0.32 0.68 Total 0.2 0.28 0.52 1

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