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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 $?

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  • $\begingroup$ Your assertion about the distribution of $\min\{X_1,X_2\}$ is right if $X_1,X_2$ are independent. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 28 '15 at 1:34
  • $\begingroup$ It's not clear what you mean by $\arg\min$ in this case. $\endgroup$ – Michael Hardy Jan 28 '15 at 1:35
  • $\begingroup$ 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. $\endgroup$ – Andrew Jan 28 '15 at 8:52
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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}

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  • $\begingroup$ 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} $ $\endgroup$ – Andrew Jan 28 '15 at 8:54
  • $\begingroup$ 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 $x$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 28 '15 at 15:51

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