LetX1, X2 be two independent random variables each with p.d.f. f1(x) = ex for x > 0 and f1(x) = 0 for x 0.

LetX1, X2 be two independent random variables each with p.d.f. f1(x) = eˆ’x for x > 0 and f1(x) = 0 for x ‰¤ 0. Let Z = X1 ˆ’ X2 and W = X1/X2.
a. Find the joint p.d.f. of X1 and Z.
b. Prove that the conditional p.d.f. of X1 given Z = 0 is

c. Find the joint p.d.f. of X1 and W.
d. Prove that the conditional p.d.f. of X1 given W = 1 is

e. Notice that {Z = 0} = {W = 1}, but the conditional distribution of X1 given Z = 0 is not the same as the conditional distribution of X1 given W = 1. This discrepancy is known as the Borel paradox. In light of the discussion that begins on page 146 about how conditional p.d.f.€™s are not like conditioning on events of probability 0, show how €œZ very close to 0€ is not the same as €œW very close to 1.€ Draw a set of axes for x1 and x2, and draw the two sets {(x1, x2): |x1 ˆ’ x2|

2e2 for x>0 otherwise 81(x10) = 4for>0. otherwise 4x1e 0 hi(x11)