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How can I find this probability $P(X ? knowing that X and Y are independent random variables.

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    In general, just knowing they are independent isn't useful, since $X$ might be a random number from $-1$ to $0$ and $Y$ might be a random number from $1$ to $2$, or visa versa, giving you probabilities anywhere between $1$ and $0$. If they are identical and independent continuous random variables, then the probability will be $1/2$. (Continuity implies $P(X=Y)=0$ and identical implies $P(XY)$.)2012-12-17
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    What if both of them are exponentially distributed random variables?2014-02-18

2 Answers 2

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Assuming both variables are real-valued and $Y$ is absolutely continuous with density $f_Y$ and $X$ has cumulative distribution function $F_X$ then it is possible to do the following

$$ \Pr \left[ X < Y \right] = \int \Pr \left[ X < y \right] f_Y \left( y \right) \mathrm{d} y = \int F_X \left( y \right) f_Y \left( y \right) \mathrm{d} y $$

Otherwise, as @ThomasAndrews said in a comment, it is case-by-case.

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I think we can control everything by the following general solution.

Consider $Z:=X-Y$. Then, by putting condition on the value of X, we get

$$\begin{align} P(X

You may also put a condition on the value of $Y$ to get a similar result. So, the solution of this problem depends on what you want.