# 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, $X_1 < X_2$?

• Your assertion about the distribution of $\min\{X_1,X_2\}$ is right if $X_1,X_2$ are independent. ${}\qquad{}$ – Michael Hardy Jan 28 '15 at 1:34
• It's not clear what you mean by $\arg\min$ in this case. – Michael Hardy Jan 28 '15 at 1:35
• Yes, of course you're right, they should be independent. By $\arg\min$ I meant which one of the two variables $X_1, X_2$ will be the minimal of this two. – Andrew Jan 28 '15 at 8:52

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 $\Pr(X_1<X_2)=\lambda_1/(\lambda_1+\lambda_2)$.
• Thank you very much! I've corrected your answer a bit, as $\Pr(\min\{X_1,X_2\}>x)$ $=1 - e^{-(\lambda_1+\lambda_2)x}$, not just $e^{-(\lambda_1+\lambda_2)x}$ – Andrew Jan 28 '15 at 8:54
• You're mistaken. $1-e^{-(\lambda_1+\lambda_2)x}$ is the probability that the minimum is LESS than $x$, not that it is greater than $x$. ${}\qquad{}$ – Michael Hardy Jan 28 '15 at 15:51