# Thread: Find P(X<Y) for two independent exponential random variables

1. ## Find P(X<Y) for two independent exponential random variables

Suppose that X and Y are independent exponential random variables with means m and n, respectively. Demonstrate that P (X < Y) = n /(m + n) using two different methods:
(a) Set up and evaluate a double integral whose value is the desired probability.
(b) Use a conditioning argument: P (X < Y) = E [P (X < Y | Y )] .
To do this, note that in P(X < Y |Y), the Y should be considered a constant (since we’re conditioning on it), which means the conditional probability may be found directly using the c.d.f. of X. The conditional probability will then be a function of Y , so its expectation may be found using the usual techniques.

Since X and Y are independent, I multiplied their pdf's to get
f(x,y) = (1/(mn))e^((-x/m) - (y/n))
When I set up the double integral for (a), I did it wrong and ended up proving that the area under the whole thing is 1 (which it obviously should be). How should I set up the double integral here?

And for (b), the cdf of X should be F(x) = 1 - e^(-x/m), and I know you would just plug y in for x to find P(X<Y) if Y was just a constant, but what do you do for P(X<Y | Y)?

2. show your double integral, the bounds are half of the first quadrant.
If you integrate over the entire first quadrant you get one.

Use $\int_0^{\infty}\int_0^y f(x,y)dxdy$

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