1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Be careful! $\lambda_1\dots\lambda_n$ means the product, not sum. $\endgroup$ – user10354138 Jun 15 at 13:01
  • $\begingroup$ @user10354138 that was a mistake! $\endgroup$ – Peyman Tahghighi Jun 15 at 13:02
0
$\begingroup$

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 $\delta t\to 0$, $$ \mathbb{P}(\min(X_1,\dots,X_n)=X_i)= \int_0^\infty\lambda_ie^{-(\lambda_1+\dots+\lambda_n)t}\,\mathrm{d}t=\frac{\lambda_i}{\lambda_1+\dots+\lambda_n} $$

$\endgroup$
  • $\begingroup$ I cannot understand why $X_i \in [t-\delta t,t]$ $\endgroup$ – Peyman Tahghighi Jun 15 at 14:09
  • $\begingroup$ $X_i$ has to take a value somewhere in $(0,\infty)$, we just make the additional assumption $X_i$ is in the interval $(t-\delta t,t]$. $\endgroup$ – user10354138 Jun 15 at 14:10
0
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.