# 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

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?

• Be careful! $\lambda_1\dots\lambda_n$ means the product, not sum. – user10354138 Jun 15 at 13:01
• @user10354138 that was a mistake! – Peyman Tahghighi Jun 15 at 13:02

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 $$\delta t\to 0$$, $$\mathbb{P}(\min(X_1,\dots,X_n)=X_i)= \int_0^\infty\lambda_ie^{-(\lambda_1+\dots+\lambda_n)t}\,\mathrm{d}t=\frac{\lambda_i}{\lambda_1+\dots+\lambda_n}$$
• I cannot understand why $X_i \in [t-\delta t,t]$ – Peyman Tahghighi Jun 15 at 14:09
• $X_i$ has to take a value somewhere in $(0,\infty)$, we just make the additional assumption $X_i$ is in the interval $(t-\delta t,t]$. – user10354138 Jun 15 at 14:10
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}$$