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Let $X_1,\dots,X_n$ be independent exponentially distributed random variables with respective rates $\lambda_1,\dots,\lambda_n$. According to wikipedia, the following property holds $$ Pr[ X_k = \min\{X_1,\dots,X_n\} ] = \frac{\lambda_k}{\lambda_1+\dots+\lambda_n}. $$ Is there a simple proof?

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  • $\begingroup$ Say do it for $X_1$. If you prefer to avoid many variables, note the easy fact that $Y=\min(X_2,\dots,X_n)$ has exponential distribution parameter $\lambda_2+\cdots+\lambda_n$. So now we need the probability that $X_1\lt Y$. This is straightforward $2$ variable integration, I have written out details on MSE at least twice, as have others, but of course cannot find it. $\endgroup$ – André Nicolas Jul 16 '13 at 3:03
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Note that $$ \begin{align*} P(X_k=\min\{X_1,\ldots,X_n\})&=\int_0^{\infty}\left[\prod_{\substack{1\leq j\leq n\\j\neq k}}\int_{x_k}^{\infty}\lambda_je^{-\lambda_jx_j}\,dx_j\right]\lambda_k e^{-\lambda_k x_k}\,dx_k\\ &=\int_0^{\infty}\left[\prod_{\substack{1\leq j\leq n\\j\neq k}}e^{-\lambda_j x_k}\right]\lambda_ke^{-\lambda_kx_k}\,dx_k\\ &=\int_0^{\infty}\lambda_k e^{-(\lambda_1+\cdots+\lambda_n)x_k}\,dx_k\\ &=\frac{\lambda_k}{\lambda_1+\cdots+\lambda_n}. \end{align*} $$ To get to the first step, we just noticed that the event $\{X_k=\min\{X_1,\ldots,X_n\}\}$ is precisely the event that $X_j\geq X_k$ for all $j$; so, given the value of $X_k$ is $x_k$, it is the event that $X_j\geq x_k$ for all $j\neq k$. We then use independence of the variables to say that their joint density function is simply the product of the density functions, and break up over multiplication.

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