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Suppose $X_i$ are distributed exponentially with rate $\lambda_i,i=1,2.$ Calculate $\Pr(X_1<X_2)$.

I tried to change it into $\int_{0}^{\infty}Pr(X_2>x|X_1=x)dx$, but I have no idea how to do the integration.

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Your integral formulation is incorrect. One cannot add conditional probabilities conditioned on different events: $P(B\mid A) + P(D\mid C)$ makes no sense even if $B$ and $D$ are mutually exclusive events. Similarly, the integral does not give the right answer. What will give the right answer (in this case) is $$\int_{0}^\infty P\{X_2 > x\mid X_1 = x\}f_{X_1}(x)\,\mathrm dx$$ while more generally you will need to use a lower limit of $-\infty$ on the integral – Dilip Sarwate Dec 18 '12 at 5:14
up vote 0 down vote accepted

It will be convenient to call $X_1$ by the name $X$, and $X_2$ by the name $Y$.

To compute the probability, we must make certain assumptions. You are probably expected to assume that $X$ and $Y$ are independent. Then you can easily find the joint density $f(x,y)$ of $X$ and $Y$: it is the product of the individual densities.

Draw the line $y=x$. We want the probability of lying above that line. This is $$\int_{x=0}^\infty \left(\int_{y=x}^\infty f(x,y)dy \right)\,dx.$$

In this case, you can probably immediately write down the inner integral.

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