How to prove $P\left\{\min(X_1,X_2,\dots,X_n) = X_i\right\} = \frac{\lambda _i}{\lambda_1+\dots+\lambda_n}$ , when $X_i$ is exponentially distributed

Question

I want to prove :

$P\left\{\min(X_1,X_2,\dots,X_n) = X_i\right\} = \frac{\lambda_i}{\lambda_1 + \dots + \lambda_n}$ when $X_i$ is exponentially distributed with parameter $\lambda_i$ I've made some progress by showing below:

$P(X_i > t) = P(X_1 > t)P(x_2>t)\dots P(X_n > t) = e^{-\lambda_1t}e^{-\lambda_2t}\dots e^{-\lambda_nt} = e^{-(\lambda_1 + \lambda_2 +\dots+ \lambda_n)t}$

but I don't know how to find the exact answer. I'm sure the for final answer is should use Laplace transform am i right?

Answer 1

You forgot the assumption $X_1,X_2,\dots,X_n$ are independent.

So for $t>0$ and $\delta t$ small, \begin{align*} &\mathbb{P}(\min(X_1,\dots,X_n)=X_i\text{ and }X_i\in(t-\delta t,t])\\ &=\mathbb{P}(X_i\in(t-\delta t,t])\prod_{j\neq i}[\mathbb{P}(X_j\geq t)+O(\delta t)]\\ &=[\lambda_i e^{-\lambda_i t}\delta t+o(\delta t)]e^{-\sum_{j\neq i}\lambda_j t}\\ &=\lambda_i e^{-(\lambda_1+\dots+\lambda_n)t}\delta t(1+o(1)) \end{align*} Thus summing partition of $(0,\infty)$ into intervals of length $\delta t$, and taking $\delta t\to 0$, $$ \mathbb{P}(\min(X_1,\dots,X_n)=X_i)= \int_0^\infty\lambda_ie^{-(\lambda_1+\dots+\lambda_n)t}\,\mathrm{d}t=\frac{\lambda_i}{\lambda_1+\dots+\lambda_n} $$

Answer 2

Assuming the $\ X_j\ $ are independent, \begin{eqnarray} P\left(\,\min\left(X_1,X_2,\dots,X_n\,\right)=X_i\right)&=&P\left(X_i \le X_j\ \mbox{ for } j\ne i\right)\\ &=& \int_\limits{0}^\infty P\left(t\le X_j\ \mbox{ for } j\ne i\left|X_i=t\right.\right)\lambda_i e^{-\lambda_i t} dt\\ &=& \int_\limits{0}^\infty \prod_\limits{j\ne i}P\left(t\le X_j\right)\lambda_i e^{-\lambda_i t} dt\\ &=& \int_\limits{0}^\infty \lambda_i e^{-\sum_{j=1}^n\lambda_jt} dt\\ &=& \frac{\lambda_i}{\sum_\limits{j=1}^n\lambda_j}\ . \end{eqnarray}