# How to prove P{min(X_1,X_2, ldots ,X_n)=X_i}=(lambda_i)/(lambda_1+cdots+lambda_n) , when X_i is exponentially distributed

How to prove $P\left\{min\left({X}_{1},{X}_{2},\dots ,{X}_{n}\right)={X}_{i}\right\}=\frac{{\lambda }_{i}}{{\lambda }_{1}+\cdots +{\lambda }_{n}}$ , when ${X}_{i}$ is exponentially distributed
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Francis Blanchard
cuuhorre76
You forgot the assumption ${X}_{1},{X}_{2},\dots ,{X}_{n}$ are independent.
So for t>0 and $\delta t$ small,
$\left(0,\mathrm{\infty }\right)$
Thus summing partition of $\left(0,\mathrm{\infty }\right)$ into intervals of length $\delta t$, and taking $\delta t\to 0$,
$\mathbb{P}\left(min\left({X}_{1},\dots ,{X}_{n}\right)={X}_{i}\right)={\int }_{0}^{\mathrm{\infty }}{\lambda }_{i}{e}^{-\left({\lambda }_{1}+\cdots +{\lambda }_{n}\right)t}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t=\frac{{\lambda }_{i}}{{\lambda }_{1}+\cdots +{\lambda }_{n}}$