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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 $[-\infty,y]\times[-\infty,+\infty]$.

$P(X<Y)=\int\int{}f(x,y)=\int_{-\infty}^ydx\int_{-\infty}^{+\infty}f(x,y)dy$

$f(x,y)=f_{X}(x)f_{Y}(y)$

$\int_{-\infty}^ydx\int_{-\infty}^{+\infty}f(x,y)dy=\int_{-\infty}^ydx\int_{-\infty}^{+\infty}f_{X}(x)f_{Y}(y)dy=\int_{-\infty}^y[f_{X}(x)\int_{-\infty}^{+\infty}f_{Y}(y)dy]dx=\int_{-\infty}^y[f_{X}(x)\int_{-\infty}^{+\infty}f_{Y}(y)dy]dx=\int_{-\infty}^yf_{X}(x)dx=F_X(y)$

But we know that

$P(X<Y)=\int_{-\infty}^{+\infty}f_X(x)F_Y(x)dx$

What's wrong with my calculation?

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  • $\begingroup$ 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
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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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  • $\begingroup$ 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

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