Exponential random variables, given one smaller 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.
 A: 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 $E[X|X<Y]=\frac1{\lambda_1+\lambda_2}$.
