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Assume that $X$ is $Exp(\lambda)$ distributed and $Y$ is $Exp(\mu)$, and they are independent. I want to know how I can calculate $P(X<Y)$. I don't understand why

$$ P(X < Y)=\int_{-\infty}^\infty P(X<Y|X=x)\cdot f_X(x)dx, $$

where $f_X$ is the density of $X$.

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  • 2
    $\begingroup$ This is the law of total probability. Essentially, what you are doing is breaking things down by cases of $X$: For all possible values of $X$, what is the joint probability that $X$ holds that value and that $Y < X$? The only wrinkle is that for continuous distributions of $X$, we can't just do ordinary addition, like we would for discrete distributions; we have to integrate. But otherwise, it's the same basic idea. $\endgroup$ – Brian Tung Mar 12 at 17:26
  • $\begingroup$ Are you familiar with conditional expectation? $\endgroup$ – rubikscube09 Mar 12 at 17:46
  • $\begingroup$ I know the law of total probability and it helps to understand, but i don't think it's enough to proof it. And yes i learned something conditional expectation. $\endgroup$ – Daniel Banov Mar 12 at 17:48
  • $\begingroup$ You can prove that $$ P(X < Y)=\int_{0}^\infty \int_{0}^y f_{X,Y}(x,y) dx dy $$ and use that X,Y are independent and so $f_{X,Y}(x,y)=f_X(x)f_Y(y)$, but the easiest way is the law of total probability. $\endgroup$ – papasmurfete Mar 12 at 20:53
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Assuming $X$ and $Y$ are independent with pdfs $f_X(x)$ and $f_Y(y)$ we have \begin{align} P\{X<Y\}&=\int_{y=-\infty}^{\infty}\left(\int_{x=-\infty}^{y}f_X(x)dx\right)f_Y(y)dy\\ &=\int_{y=-\infty}^{\infty}F_X(y)f_Y(y)dy \end{align} where $F_X(x)$ is the CDF of $X$.

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