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} $?
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} $?
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.