$X$,$Y$ 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?

  • $\begingroup$ We want $P(X<Y)$, so $x$ can only go up to $y$. Despite too many years integrating, I still always sketch the region. $\endgroup$ – 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 $y$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.