Find and sample minimum of two exponential distribtions I have two (or more) independent exponential variables $ X_1 \sim \exp(\lambda_1) $ and $ X_2 \sim \exp(\lambda_2) $. I want to get both the value of $ \min(X_1, X_2) $ and $ \arg\min(X_1, X_2) $. Can I just draw them independently? Is such approach correct? 
I.e. it is well-known that $ \min(X_1, X_2) \sim \exp(\lambda_1 + \lambda_2) $ and $ P(X_1 < X_2) = \frac{\lambda_1}{\lambda_1 + \lambda_2} $, but what about the distribution of $ \min(X_1, X_2) $ if we know that, for example, $ X_1 < X_2 $?
 A: I find I am slightly surprised by the answer I get.  The event that $\min\{X_1,X_2\}>x$ and the event that $X_1<X_2$ are actually independent of each other.  Consequently the conditional distribution of the minimum given that $X_1<X_2$ is the same as the unconditional distribution of the minimum.  I'd have expected that if $X_1,X_2$ had the same distribution, but as it is I expected the answer to depend on the ratio of the two $\lambda$s.  But it doesn't.
One of the results that you say you already know can be stated by saying $\Pr(\min\{X_1,X_1\}>x)$ $=e^{-(\lambda_1+\lambda_2)x}$, for all $x>0$. The other one says $\Pr(X_1<X_2)=\lambda_1/(\lambda_1+\lambda_2)$.
So now consider
\begin{align}
& \Pr(x<\min\{X_1,X_2\}\ \&\ X_1<X_2)=\Pr(x<\min=X_1<X_2) \\[8pt]
= {} & \int_x^\infty\cdots\,dx_1 = \int_x^\infty\left(\int_{x_1}^\infty\cdots\,dx_2\right)\,dx_1
= \int_x^\infty\left(\int_{x_1}^\infty \lambda_1 e^{-\lambda_1 x_1} \lambda_2 e^{-\lambda_2 x_2} \,dx_2\right)\,dx_1 \\[8pt]
= {} & \int_x^\infty\left( \lambda_1 e^{-\lambda_1 x_1} \int_{x_1}^\infty e^{-\lambda_2 x_2} (\lambda_2\,dx_2) \right) dx_1 = \int_x^\infty\left( \lambda_1 e^{-\lambda_1 x_1} 
e^{-\lambda_2 x_1} \right) dx_1 \\[8pt]
= {} & \lambda_1 \int_x^\infty e^{-(\lambda_1+\lambda_2)x_1}\,dx_1 =\lambda_1 \frac{e^{-(\lambda_1+\lambda_2)x}}{\lambda_1+\lambda_2} \\[8pt]
= {} &  \frac{\lambda_1}{\lambda_1+\lambda_2} \cdot e^{-(\lambda_1+\lambda_2)x} \\[8pt]
= {} & \Pr(X_1<X_2)\cdot \Pr(\min>x).
\end{align}
