1. ## probability

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

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

2. never mind about the limits; is the general process correct?

3. I get $\displaystyle \int_{-\infty}^{\infty} \int_{-\infty}^x e^{-x-2y}~dy~dx$

4. I took the complement.

5. What were the domains of $\displaystyle X$ and $\displaystyle Y$ given in the problem?

6. $\displaystyle -\infty < x < \infty$, $\displaystyle -\infty < y < \infty$.

7. While the integral over the whole positive domain is $\displaystyle \frac{1}{2}$, the function is NOT even. For this to be a density function, $\displaystyle x,y \in\left[{\frac{{ - \ln (2)}}{3},\infty } \right)$ since $\displaystyle \int\limits_{\frac{{ - \ln (2)}}{3}}^\infty {\int\limits_{\frac{{ - \ln (2)}}{3}}^\infty {e^{ - x - 2y} } } ~dy~dx = 1$.

Then again, there are infinitely many possible limits for $\displaystyle X$ and $\displaystyle Y$ that will work. I am not 100% confident in my Prob/Stats knowledge though, so don't quote me on this.

8. yeah I knew the limits were wrong, I was just asking whether my method of finding the probability was correct (e.g. $\displaystyle P(X <Y) = 1 - P(X>Y)$)?

9. Oh. You are correct.