Domain probability problem

1. Feb 22, 2008

tronter

If $$f(x,y) = e^{-x-2y}$$ find $$P(X<Y)$$.

So is this equaled to $$1 - P(X>Y) = 1 - \int\limits_{0}^{\infty} \int\limits_{0}^{x} e^{-x-2y} \ dy \ dx = \frac{1}{3}?$$

2. Feb 22, 2008

EnumaElish

What is the domain? $\int_0^\infty\int_0^\infty f(x,y)dydx$=1/2, not 1.

3. Feb 22, 2008

tronter

the domain is $$0 < x < \infty$$, $$0 < y < \infty$$. It is equaled to $$1$$, so its a valid pdf.

4. Feb 22, 2008

EnumaElish

On that domain the probability doesn't add up to 1.

5. Feb 22, 2008

tronter

whoops, I meant $$-\infty < x < \infty$$, $$-\infty < y< \infty$$.

6. Feb 22, 2008

EnumaElish

It looks like the lower bound should be -Log[2]/3.

For a "suitable" lower bound that equates the double integral to 1, your formula is correct.

Alternatively, since integration over [0, +infinity) gives 1/2, you might define the cumulative distribution F*(x,y) = F(x,y) + 1/2, where F(x,y) is the double integral of f(s,t) over s from 0 to x and t from 0 to y. In this case f is a valid pdf.

Last edited: Feb 22, 2008
7. Feb 22, 2008

tronter

forgetting about the bounds for the moment, is the general set up correct?

8. Feb 22, 2008

EnumaElish

See my previous post (edited).