# Finding distribution $W$

Suppose $X_1,\ldots,X_m\sim E(\mu,\sigma_1)$, $Y_1,\ldots,X_n\sim E(\mu,\sigma_2)$ are two independent random sample. If $W=\min(X_{(1)},Y_{(1)})$, how can find distribution $W$?

Note: $X_{(1)}$ is smallest order statistics in sample $X_1,\ldots,X_m$ and $Y_{(1)}$ is smallest order statistics in sample $Y_1,\ldots,Y_n$ and if ($X\sim E(\mu,\sigma)$ then $f(x)=\frac{1}{\sigma}\displaystyle e^\frac{-(x-\mu)}{\sigma},\ x>\mu$).

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So many questions, so little time to accept the answers so generously provided by participants on this stackexchange. – Dilip Sarwate Mar 25 '12 at 15:07
@hadisanji : I posted an answer to your question about finding a UMVUE. Was that of use to you? If so, could you up-vote or "accept" the answer? As it is, I have no way of knowing whether you ever saw it at all. – Michael Hardy Mar 25 '12 at 15:26

$W=\min\{X_{(1)},Y_{(1)}\}$ is the smallest of all $m+n$ observations.
\begin{align} \Pr(W > w) & = \Pr(X_1>w\ \&\ \cdots\ \&\ X_m>w\ \&\ Y_1>w\ \&\ \cdots\ \&\ Y_n>w) \\ \\ & = \left(\exp\left(-\frac{w-\mu}{\sigma_1}\right)\right)^m\left(\exp\left(-\frac{w-\mu}{\sigma_2}\right)\right)^n \\ \\ & = \exp\left(-m\frac{w-\mu}{\sigma_1} - n\frac{w-\mu}{\sigma_2}\right) =\exp\left(-\frac{\sigma_2 m+\sigma_1 n}{\sigma_1\sigma_2}(x-\mu)\right)\qquad \text{for }w>\mu. \end{align} So $1$ minus that is the CDF.