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$X$,$Y$ are independent random variables, whose density function is $f(x,y)$.

To get the Probability of $X<Y$, I use the integration of the area $[-\infty,y]\times[-\infty,+\infty]$.




But we know that


What's wrong with my calculation?

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We want $P(X<Y)$, so $x$ can only go up to $y$. Despite too many years integrating, I still always sketch the region. – André Nicolas May 9 '12 at 14:00

It is not true that you should integrate over $(-\infty,y] \times (-\infty,\infty)$. In order $X < Y$ to be true, for every given $y \in (-\infty,\infty)$ variable $x$ varies over $(-\infty,y]$ hence for every given $y$ you should integrate in $x$ over $(-\infty,y]$. Thus $$ P(X < Y) = \int_\mathbb{R} dy \int_{(-\infty,y)} f(x,y) dx. $$ You should also observe that $P(X < Y)$ is a number. This implies that it cannot depend on $x$ or $y$.

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Oh, what a naive mistake i've made! then how to calculate $P(X < Y) = \int_\mathbb{R} dy \int_{(-\infty,y)} f(x,y) dx$. – Charles Bao May 9 '12 at 10:39
You calculate this in the same way as you did. $\int_{(-\infty,y)} f_X(x) dx = F_X(y)$. – xen May 9 '12 at 11:10
then the answer is still depend on y, the mistake is that i change the integrate order. – Charles Bao May 10 '12 at 0:25
It depends on $y$ but there is also the integral over $y$. – xen May 10 '12 at 4:59

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