I want to show that if $\{E_i:i\in I\}$ are independent exponential random variables with parameters $\lambda_i$, then $\inf_{i\in I}{E_i}$ is also exponentially distributed.
But I am not really sure what is meant by $\inf_{i\in I}{E_i}$, because wouldn't all exponential random variables have infimum $0$?
Can someone also give me a hint for the question and finding such distribution?
Best Answer
If $\sum_{i\in I}\lambda_i<\infty$ (necessarily implying that $I$ is finite or countably infinite), then for any $t>0$ we have \begin{align} \mathbb P(E>t) &= \mathbb P\left(\bigcap_{i\in I} E_i>t\right)\\ &= \bigcap_{i\in I}\mathbb P(E_i>t)\\ &= \prod_{i\in I}\mathbb P(E_i>t)\\ &= \prod_{i\in I} e^{-\lambda_i t}, \end{align} so that $E:=\inf_{i\in I}E_i\sim\mathsf{Expo}(\prod_{i\in I}\lambda_i)$. If the sum of the $\lambda_i$ does not converge, then $E=0$ with probability $1$.