Assume $X_1,X_2,\ldots,X_n$ are identical independent random variables, all distributed with $\text{Poisson}(\lambda)$ (same $\lambda$). I am interested to find out the distribution of their minimum: $X_\min=\min_i\{X_i\}$. Especially I am interested in the mean and variance of that random variable.
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$\begingroup$ These probably have no expression in terms of usual functions. $\endgroup$– DidCommented Mar 15, 2014 at 17:57
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$\begingroup$ Even just the mean? And: I'm ok with "non-usual" functions as well. Any sort of closed form, or even a good approximation might be helpful here. $\endgroup$– BachCommented Mar 15, 2014 at 17:59
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2 Answers
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$$E(X_{\mathrm{min}})=\sum_{k\geqslant1}P(X_{\mathrm{min}}\geqslant k)=\sum_{k\geqslant1}P(X_1\geqslant k)^n=\mathrm e^{-n\lambda}\sum_{k\geqslant1}\left(\sum_{i=k}^\infty\frac{\lambda^i}{i!}\right)^n$$
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The cumulative distribution function is a really good place to start!
You know that $$ Pr(X_{min} \leq x) = 1 - Pr(X_{min} > x)$$ where the latter occurs if and only if $X_i > x$ for each $i$.
Now since $X_{min}$ is non-negative you can compute its expectation using the wonderful formula $$E(X_{min}) = \sum_{x}Pr(X_{min} > x)$$
