Expected value of minimum of $n$ Gamma functions?

Question

Suppose I have a set of $n$ Gamma functions, all with different parameters. I then draw one sample from each function. How can I find the expected value of the minimum of these samples?

Answer 1

Given independent nonnegative random variables $X_1, \ldots, X_n$ with continuous distributions, let $M = \min(X_1, \ldots, X_n)$. $$ \mathbb P(M > t) = \mathbb P(\text{all}\ X_i > t) = \prod_{i=1}^n \mathbb P(X_i > t)$$ and then $$ \mathbb E[M] = \int_0^\infty \mathbb P(M > t)\; dt $$ I wouldn't expect a simple general formula in the case of Gamma distributions with different scale and shape parameters. Let's say $X_i$ has shape parameter $k_i$ which is a positive integer and scale parameter $\theta_i$. Then $$\mathbb P(X_i > t) = Q_{k_i}(t/\theta_i) e^{-t/\theta_i}$$ where $Q_{k_i}$ is a polynomial of degree $k_i-1$; $\mathbb P(M > t)$ is a polynomial of degree $\sum_{i} (k_i - 1)$ times $\exp(-rt)$ where $r = \sum_i 1/\theta_i$, and the integration is elementary but complicated.

Thus for the case $n=4$ with all $k_i = 2$, I get the rather horrendous

$${\frac {2 \theta_{{1}}\theta_{{2}}\theta_{{3}}\theta_{{4}} \left( { \theta_{{1}}}^{4}{\theta_{{2}}}^{4}{\theta_{{3}}}^{4}+5\,{\theta_{{1}} }^{4}{\theta_{{2}}}^{4}{\theta_{{3}}}^{3}\theta_{{4}}+8\,{\theta_{{1}} }^{4}{\theta_{{2}}}^{4}{\theta_{{3}}}^{2}{\theta_{{4}}}^{2}+5\,{\theta _{{1}}}^{4}{\theta_{{2}}}^{4}\theta_{{3}}{\theta_{{4}}}^{3}+{\theta_{{ 1}}}^{4}{\theta_{{2}}}^{4}{\theta_{{4}}}^{4}+5\,{\theta_{{1}}}^{4}{ \theta_{{2}}}^{3}{\theta_{{3}}}^{4}\theta_{{4}}+20\,{\theta_{{1}}}^{4} {\theta_{{2}}}^{3}{\theta_{{3}}}^{3}{\theta_{{4}}}^{2}+20\,{\theta_{{1 }}}^{4}{\theta_{{2}}}^{3}{\theta_{{3}}}^{2}{\theta_{{4}}}^{3}+5\,{ \theta_{{1}}}^{4}{\theta_{{2}}}^{3}\theta_{{3}}{\theta_{{4}}}^{4}+8\,{ \theta_{{1}}}^{4}{\theta_{{2}}}^{2}{\theta_{{3}}}^{4}{\theta_{{4}}}^{2 }+20\,{\theta_{{1}}}^{4}{\theta_{{2}}}^{2}{\theta_{{3}}}^{3}{\theta_{{ 4}}}^{3}+8\,{\theta_{{1}}}^{4}{\theta_{{2}}}^{2}{\theta_{{3}}}^{2}{ \theta_{{4}}}^{4}+5\,{\theta_{{1}}}^{4}\theta_{{2}}{\theta_{{3}}}^{4}{ \theta_{{4}}}^{3}+5\,{\theta_{{1}}}^{4}\theta_{{2}}{\theta_{{3}}}^{3}{ \theta_{{4}}}^{4}+{\theta_{{1}}}^{4}{\theta_{{3}}}^{4}{\theta_{{4}}}^{ 4}+5\,{\theta_{{1}}}^{3}{\theta_{{2}}}^{4}{\theta_{{3}}}^{4}\theta_{{4 }}+20\,{\theta_{{1}}}^{3}{\theta_{{2}}}^{4}{\theta_{{3}}}^{3}{\theta_{ {4}}}^{2}+20\,{\theta_{{1}}}^{3}{\theta_{{2}}}^{4}{\theta_{{3}}}^{2}{ \theta_{{4}}}^{3}+5\,{\theta_{{1}}}^{3}{\theta_{{2}}}^{4}\theta_{{3}}{ \theta_{{4}}}^{4}+20\,{\theta_{{1}}}^{3}{\theta_{{2}}}^{3}{\theta_{{3} }}^{4}{\theta_{{4}}}^{2}+60\,{\theta_{{1}}}^{3}{\theta_{{2}}}^{3}{ \theta_{{3}}}^{3}{\theta_{{4}}}^{3}+20\,{\theta_{{1}}}^{3}{\theta_{{2} }}^{3}{\theta_{{3}}}^{2}{\theta_{{4}}}^{4}+20\,{\theta_{{1}}}^{3}{ \theta_{{2}}}^{2}{\theta_{{3}}}^{4}{\theta_{{4}}}^{3}+20\,{\theta_{{1} }}^{3}{\theta_{{2}}}^{2}{\theta_{{3}}}^{3}{\theta_{{4}}}^{4}+5\,{ \theta_{{1}}}^{3}\theta_{{2}}{\theta_{{3}}}^{4}{\theta_{{4}}}^{4}+8\,{ \theta_{{1}}}^{2}{\theta_{{2}}}^{4}{\theta_{{3}}}^{4}{\theta_{{4}}}^{2 }+20\,{\theta_{{1}}}^{2}{\theta_{{2}}}^{4}{\theta_{{3}}}^{3}{\theta_{{ 4}}}^{3}+8\,{\theta_{{1}}}^{2}{\theta_{{2}}}^{4}{\theta_{{3}}}^{2}{ \theta_{{4}}}^{4}+20\,{\theta_{{1}}}^{2}{\theta_{{2}}}^{3}{\theta_{{3} }}^{4}{\theta_{{4}}}^{3}+20\,{\theta_{{1}}}^{2}{\theta_{{2}}}^{3}{ \theta_{{3}}}^{3}{\theta_{{4}}}^{4}+8\,{\theta_{{1}}}^{2}{\theta_{{2}} }^{2}{\theta_{{3}}}^{4}{\theta_{{4}}}^{4}+5\,\theta_{{1}}{\theta_{{2}} }^{4}{\theta_{{3}}}^{4}{\theta_{{4}}}^{3}+5\,\theta_{{1}}{\theta_{{2}} }^{4}{\theta_{{3}}}^{3}{\theta_{{4}}}^{4}+5\,\theta_{{1}}{\theta_{{2}} }^{3}{\theta_{{3}}}^{4}{\theta_{{4}}}^{4}+{\theta_{{2}}}^{4}{\theta_{{ 3}}}^{4}{\theta_{{4}}}^{4} \right) }{ \left( \theta_{{1}}\theta_{{2}} \theta_{{3}}+\theta_{{1}}\theta_{{2}}\theta_{{4}}+\theta_{{1}}\theta_{ {3}}\theta_{{4}}+\theta_{{2}}\theta_{{3}}\theta_{{4}} \right) ^{5}}} $$

Answer 2

It is quite a technical challenge even assuming that our independent random variables $X_1,\ldots,X_n$ have the same distribution $\Gamma(k,\theta)$. In that case have: $$\begin{eqnarray*} \mathbb{E}\left[\min(X_1,\ldots X_n)\right] &=& \int_{0}^{+\infty}\mathbb{P}\left[\min(X_1,\ldots X_n)\geq t\right]\,dt\\&=& \int_{0}^{+\infty}\left(\mathbb{P}[X_1\geq t]\right)^n\,dt\\&=&\int_{0}^{+\infty}\left(\frac{\Gamma\left(k,\frac{t}{\theta}\right)}{\Gamma(k)}\right)^n\,dt\\&=&\frac{\theta}{\Gamma(k)^n}\int_{0}^{+\infty}\Gamma(k,u)^n\,du\end{eqnarray*} $$ and the last integral can be computed by repeated integration by parts.

If $n$ is large, the Fisher-Tippett-Gnedenko theorem gives that $\min(X_1,\ldots,X_n)$ can be approximated by a Gumbel distribution.

The expected value of $\min(X_1,\ldots,X_n)$ should be close to: $$ \frac{1}{n(n+1)}\sum_{k=1}^{n}\mathbb{E}[X_k].$$