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I can't seem to grasp the following:

Let $X_1 \sim \exp(\lambda_1), X_2 \sim \exp(\lambda_2)$ and independent.

Then $$ \mathbb{E}\left[X_1 | X_1 < X_2\right] = \frac{1}{\lambda_1 + \lambda_2} $$

Why? How do I get this result?

Also, is this somehow related to $ \mathbb{E}\left[\min(X_1,X_2)\right] = \frac{1}{\lambda_1 + \lambda_2} $? If so, why are they the same?

I would prefer answers that solve it using identities rather than pdf/CDF.

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  • $\begingroup$ @HenningMakholm: Yes, that's a typo. I'll correct it now. $\endgroup$ – Jean-Paul Feb 20 '15 at 22:42
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    $\begingroup$ The two results are clearly related in that one follows from the other: $E(X_1\mid X_1<X_2)$ is the same as $E(\min(X_1,X_2)\mid X_1<X_2)$ and then if the expectation of the minimum is the same no matter whether $X_1<X_2$ or $X_2<X_1$, then this common value must also be the unconditional expectation of the minimum (because $X_1=X_2$ almost never). $\endgroup$ – Henning Makholm Feb 20 '15 at 22:48
  • $\begingroup$ I get the idea but I would love to prove it based on writing out the conditional expectation opposed to disregarding it based on the premise that the condition disappears because it's redundant (same in both cases). $\endgroup$ – Jean-Paul Feb 20 '15 at 22:52
  • $\begingroup$ x @Jean: Well, $\min(X_1,X_2)$ is defined by cases according to whether $X_1<X_2$ or not -- the first thing to do when trying to compute its expectation would be to split into those two cases anyway. It's just serendipity that those conditional expectations turn out to be symmetric in $\lambda_1$ and $\lambda_2$ (if indeed they are; I have no independent knowledge that your formula is even right). $\endgroup$ – Henning Makholm Feb 20 '15 at 23:00
  • $\begingroup$ Note that the claim about $E(X_1\mid X_1<X_2)$ does not follow from that about $E(\min(X_1,X_2))$ -- only the other way around. $\endgroup$ – Henning Makholm Feb 20 '15 at 23:03
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Here is a way to do it using pdf's. I'll use $X,Y$ instead of $X_1,X_2$. $$ E[X|X<Y] =\frac{\int_0^\infty\int_x^\infty x\cdot\lambda_1e^{-\lambda_1x}\cdot \lambda_2e^{-\lambda_2 y}\,dy\,dx}{P(X<Y)} $$ We first compute the intergral. \begin{align} \int_0^\infty\int_x^\infty x\cdot\lambda_1e^{-\lambda_1x}\cdot \lambda_2e^{-\lambda_2 y}\,dy\,dx &=\int_0^\infty x\lambda_1e^{-\lambda_1x}(-e^{-\lambda_2y})\big|^\infty_x\,dx\\ &=\int_0^\infty x\lambda_1e^{-(\lambda_1+\lambda_2)x}\,dx\\ &=-x\frac{\lambda_1}{\lambda_1+\lambda_2}e^{-(\lambda_1+\lambda_2)x}\bigg|^\infty_0+\frac{1}{\lambda_1+\lambda_2}\int_0^\infty \lambda_1e^{-(\lambda_1+\lambda_2)x}\,dx\\ &=\frac{\lambda_1}{(\lambda_1+\lambda_2)^2} \end{align} Now we compute $P(X<Y)$. \begin{align} P(X<Y) &=\int_0^\infty \int_x^\infty \lambda_1e^{-\lambda_1x}\cdot \lambda_2e^{-\lambda_2 y}\,dy\,dx\\ &=\int_0^\infty \lambda_1e^{-(\lambda_1+\lambda_2)x}\,dx=\frac{\lambda_1}{\lambda_1+\lambda_2} \end{align} Dividing the last two gives that $E[X|X<Y]=\frac1{\lambda_1+\lambda_2}$.

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