Let $X_i, i = 1, 2, 3,$ be independent exponential random variables with rates $\lambda_i, i = 1,2,3.$

How does one derive the following:

$$\mathbb P \{\min(X_1, X_2, X_3) = X_1\} = \frac{\lambda_1}{\lambda_1 +\lambda_2 + \lambda_3}?$$

I see this used all of the time, and I'm familiar with the fact that

$$\mathbb P \{X_1 < X_2\} = \frac{\lambda_1}{\lambda_2 +\lambda_2},$$

which I assume is property one uses to get from the latter to the former; but, I've been working with the definitions, and whatnot, with no luck. Any help here would be appreciated. Thanks!

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    $\begingroup$ For every positive $x$, $$P(X_2>x,X_3>x)=e^{-(\lambda_2+\lambda_3)x},$$ hence, by independence, $$P(X_2>X_1,X_3>X_1)=E(e^{-(\lambda_2+\lambda_3)X_1})=\ldots$$ $\endgroup$ – Did Feb 27 '16 at 20:03
  • $\begingroup$ Well, $E(e^{-(\lambda_2+\lambda_3)X_1}) = M_{X_1}(-\lambda_2 - \lambda_3)$, i.e., the moment-generating function, if that's what you're getting at... $\endgroup$ – thisisourconcerndude Feb 27 '16 at 20:09
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    $\begingroup$ ...And this evaluates to? $\endgroup$ – Did Feb 27 '16 at 20:11
  • $\begingroup$ Ah, I see... it's just $\int_{0}^{\infty}e^{-(\lambda_2+\lambda_3)x_1}\lambda_1e^{-\lambda_1x_1}dx_1 = \lambda_1\int_{0}^{\infty}e^{-(\lambda_1 + \lambda_2+\lambda_3)x_1}dx_1=\frac{-\lambda_1}{\lambda_1 + \lambda_2 + \lambda_3}(0 - 1) = \frac{\lambda_1}{\lambda_1 + \lambda_2 + \lambda_3}$. Thanks! $\endgroup$ – thisisourconcerndude Feb 27 '16 at 20:25

Observe that $P(\min(X_1,X_2,X_3)=X_1) = P( \min (X_2,X_3)> X_1)$. Since $\min (X_2,X_3)$ is exponential with parameter $\lambda_2+\lambda_3$, and is also independent of $X_1$, the result follows from the stated formula for the minimum of two independent exponential random variables.

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For $t>0$ we have $$\{X_1\wedge X_2 > t\}=\{X_1>t\}\cap\{X_2>t\}, $$ so by independence, \begin{align} \mathbb P(X_1\wedge X_2 > t) &= \mathbb P(X_1 > t)\mathbb P(X_2>t)\\ &= e^{-\lambda_1 t}e^{-\lambda_2t}\\ &= e^{-(\lambda_1+\lambda_2)t}, \end{align} and hence $X_1\wedge X_2\sim\operatorname{Exp}(\lambda_1+\lambda_2)$. Now, given $n\geqslant1$, $$\bigwedge_{i=1}^{n+1} X_i = \left(\bigwedge_{i=1}^n X_i\right)\wedge X_{n+1}, $$ so by induction, $X_{n+1}\sim\operatorname{Exp}\left(\sum_{i=1}^{n+1} \lambda_i\right)$. The result follows from the fact that if $X\sim\operatorname{Exp}(\lambda)$, $Y\sim\operatorname{Exp}(\mu)$, then \begin{align} \mathbb P(X<Y) &= \int_0^\infty \mathbb P(X<Y\mid X=t)f_X(t)\ \mathsf dt\\ &= \int_0^\infty\mathbb P(Y>t)f_X(t)\ \mathsf dt\\ &= \int_0^\infty e^{-\mu t} \lambda e^{-\lambda t}\ \mathsf dt\\ &= \frac\lambda{\lambda+\mu}\int_0^\infty (\lambda+\mu)e^{-(\lambda+\mu) t} \mathsf dt\\ &= \frac\lambda{\lambda+\mu}. \end{align}

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