Question

9.8 Let X and Y be independent random variables with probability distributions given by

P(X = 0) = P(X = 1) = 1/2 and P(Y = 0) = P(Y = 2) = 1/2 .

a. Compute the distribution of Z = X + Y .

b. Let Y˜ and Z˜ be independent random variables, where Y˜ has the same distribution as Y , and Z˜ the same distribution as Z. Compute the distribution of X˜ = Z˜ − Y

Answer 1

The Probability distribution of X is

P ( X=0) = P (X=1) = 1/2.

and probability distribution of Y is

P(Y=0) = P(Y=1) = 1/2.

Since X and Y are independent.

P(X=x, Y=y) = P(X=x) * P(Y=y)

a) Consider the random variable Z is

Z = X + Y

the random variable Z takes value 0 ,1 ,2

P ( Z =0) = P ( X=0, Y=0) = P(X=0) * P(Y=0) = 1/2 *1/2 = 1/4.

P ( Z =1) = P ( X=0, Y=1) + P ( X=1, Y=0) = P(X=0) * P(Y=1) +P(X=1) * P(Y=0) = 1/2 *1/2 + 1/2 *1/2 = 1/2.

P(Z=2) = P ( X=1, Y=1) = P(X=1) * P(Y=1) = 1/2 *1/2 = 1/4.

Hence Probability distribution of Z = X+Y is

z	0	1	2	Total
P(Z=z)	1/4	1/2	1/4	1

b) Since Y^~ has same distribution as Y

i.e. P(Y^~ = 0) = 1/2 and P(Y^~ =1) = 1/2

and Z ^~ has same distribution as Z

P(Z^~ = 0) = 1/4, P(Z^~ = 1) = 1/2 and P(Z^~ =2) = 1/4.

Y^~ and Z^~ are independent.

consider the random variable X^~ = Z^~ - Y

X ^~ take values -1,0,1,2.

P(X^~ = -1) = P(Z^~ = 0, Y=1) = P(Z^~ =0) * P(Y=1) = 1/4*1/2=1/8

P(X^~ = 0) = P(Z^~ = 0, Y=0) + P(Z^~ = 1, Y=1) = P(Z^~ =0) * P(Y=0) + P(Z^~ =1) * P(Y=1)

= 1/4 * 1/2 + 1/2 * 1/2 = 1/8 + 1/4 = 3/8

P(X^~ = 1) = P(Z^~ = 1, Y=0) + P(Z^~ = 2, Y=1) = P(Z^~ =1) * P(Y=0) + P(Z^~ =2) * P(Y=1)

= 1/2 * 1/2 + 1/4 *1/2 = 1/4 + 1/8 = 3/8

P(X^~ = 2) = P(Z^~ = 2, Y=0) = P(Z^~ =2) * P(Y=2) = 1/4*1/2=1/8

The probability distribution of X^~ is

x~	-1	0	1	2	Total
P(X~ =x~)	1/8	3/8	3/8	1/8	1