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

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

$P(X

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

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

But we know that

$P(X

What's wrong with my calculation?

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    We want P(X, so $x$ can only go up to $y$. Despite too many years integrating, I still *always* sketch the region.2012-05-09

1 Answers 1

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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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    It depends on $y$ but there is also the integral over $y$.2012-05-10