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franz32
Sep1-04, 11:18 AM
I just need a guide to this problem... found in one of the books in the library...
Given the joint pdf f(x,y) = 2e^[-(x+y)] where 0 < x < y, y > 0

find P(Y < 1 / x < 1). Note that "/" means given that.

I got the formula when P(a < Y < b / X = x) is given, i.e., in terms of the integral from a to b of f(y/x)dy. But how about this? Is there a formula to transform this? =) The answer is in the book... so I want to know how to begin with...

ehild
Sep3-04, 01:39 AM
I just need a guide to this problem... found in one of the books in the library...
Given the joint pdf f(x,y) = 2e^[-(x+y)] where 0 < x < y, y > 0

find P(Y < 1 / x < 1). Note that "/" means given that.

I got the formula when P(a < Y < b / X = x) is given, i.e., in terms of the integral from a to b of f(y/x)dy. But how about this? Is there a formula to transform this? =) The answer is in the book... so I want to know how to begin with...

I think the probability density function was given as

f(x,y) = 2e^[-(x+y)] where 0 < x < y, y > 0
and f(x,y)=0 if x>y.

Let be the event A: y<1 and the event B: x<1.
By the definition of conditional probability P(A/B) = P(AB)/P(B). (AB means A AND B). You have to integrate f(x,y) to get the probabilities. Find the integration domains in the picture: for P(AB), it is 0<x<y, 0<y<1. For P(B), it is x<y<infinity, 0<x<1.

ehild