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I am looking for the the mean of the maximum of N independent but not identical exponential random variables. I found the CDF and the pdf but I couldn't compute the integral to find the mean of the maximum.

The CDF : $ \mathbf{F_{X_{max}}}(x)= \prod_{k=1}^{K} (1-exp(-\lambda_k x)) $ and The pdf is : $ \mathbf{f_{X_{max}}}(x)= \sum_{k=1}^{K}\lambda_k exp(-\lambda_k x) \prod_{q=1,q\neq k}^{K} (1-exp(-\lambda_q x)) $ so the mean that I am trying to find is : $ \mathbb{E} [X_{max}] = \int_{0}^{\infty} x \mathbf{f_{X_{max}}}(x) dx $

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  • $\begingroup$ Could you write down the integral you have to evaluate? $\endgroup$ – StubbornAtom May 23 '18 at 10:47
  • $\begingroup$ Do you have some restrictions on the parameters of the exponential? $\endgroup$ – kjetil b halvorsen May 23 '18 at 11:15
  • $\begingroup$ I added the expression of the integral. $\endgroup$ – irched May 23 '18 at 11:25
  • $\begingroup$ I don't understand what you mean by restrictions on the exponentials parameters ? $\endgroup$ – irched May 23 '18 at 11:27
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Using this formula for expectation of positive random variables in terms of the survival function and expanding the product in your formula for the cdf, \begin{align} EX_\text{max} &=\int_0^\infty P(X_\text{max}>x)dx \\&=\int_0^\infty1-\prod_{i=1}^n(1-e^{-\lambda_i x})dx \\&=\sum_{S\subseteq\{1,2,\dots,n\}}(-1)^{|S|} \int_0^\infty e^{-x\sum_{j\in S}\lambda_j}dx \\&=\sum_{S\subseteq\{1,2,\dots,n\}}(-1)^{|S|-1} \frac1{\sum_{j\in S}\lambda_j}, \end{align} where the outer sum is over all non-empty subsets $S$ of $\{1,2,\dots,n\}$ and $|S|$ denotes the number of elements of $S$.

So for $n=1$, this simplifies to the usual formula $$ E X_\text{max} = \frac1{\lambda_1}, $$ for $n=2$, $$ E X_\text{max} = \frac1{\lambda_1} + \frac1{\lambda_2} - \frac1{\lambda_1+\lambda_2}, $$ for $n=3$, $$ E X_\text{max} = \frac1{\lambda_1} + \frac1{\lambda_2} + \frac1{\lambda_3} - \frac1{\lambda_1+\lambda_2} - \frac1{\lambda_1+\lambda_3} - \frac1{\lambda_2+\lambda_3} + \frac1{\lambda_1+\lambda_2+\lambda_3}, $$ and so on.

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