Re: Conditional probability

Hey MagisterMan.

After looking at this problem, the only reason these are considered independent is because it should meet the criteria (i.e. P(A and B) = P(A)P(B)). Normally given the situation, Y2 would not be independent of Y1 since Y2 is a function of Y1 and N). This is a very special case in that a uniform distribution is completely random.

If you want to do the joint distribution, then show that all probabilities are the same (i.e. given some Y1 = y1, then all probabilities of Y2 = y2 are the same no matter what value of y1) and one way to do this is by using probability generating functions.

As a suggestion, one proof you might want to do is show that if Y1 is uniformly distributed then Y1 and Y2 will be considered independent. Then you can prove that Y2 is uniform and thus your proof is done.