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If $X_1,...,X_N$ are independent and identically distributed exponential random variables, what can be said about the distribution of $\text{min}(X_1,...,X_N)$ when $N$ is random and modelled as a Poisson random variable?

Michael M
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Ämin Persialinen
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    http://stats.stackexchange.com/questions/220/how-is-the-minimum-of-a-set-of-random-variables-distributed?rq=1 – Alex R. Mar 22 '16 at 23:55
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    Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. – gung - Reinstate Monica Mar 23 '16 at 00:22
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    I'm not sure if this is really a duplicate of the linked thread, given the Poisson distribution of n. – gung - Reinstate Monica Mar 23 '16 at 00:26
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    No this question is not for course and it is not duplicated as In this case the $n$ is also random which is different from the other question that $n$ is fixed. – Ämin Persialinen Mar 23 '16 at 14:55
  • There might be an incorrect duplicate tag. The size of the sample $n$ is also random here, and modeled with a Poisson. This makes the problem different from the link provided as duplicate. – Greenparker Mar 23 '16 at 15:11
  • @ÄminPersialinen: Please check if my edits are correct. I also replaced the small $n$ by a $N$ to stress the fact that it is a random variable. – Michael M Mar 23 '16 at 15:16
  • The poisson distribution puts positive mass on zero. How is the min of an empty set of random variables defined? – Adrian Mar 23 '16 at 15:50
  • I don't know how this happened -- I didn't VTC as duplicate, I VTC as self-study. – Sycorax Mar 23 '16 at 18:04
  • @AlexR. In the provided link, n is fixed and if you look at my question I considered the case that N is random as well. Thanks for the comment – Ämin Persialinen Mar 23 '16 at 18:55
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    First you'll need to define the minimum when $N=0$ (or use a shifted poisson distribution starting at $N=1$).

    The random $N$ doesn't add much: Given $N=n$, you get $P(\min(X_1,\cdots,X_N)\leq x | N=n)=1-[1-F(x)]^n$. Notice that the minimum of exponential random variables is exponentially distributed with parameter $\lambda= \lambda_1+\cdots\lambda_n$.

    Now just sum over:

    $$P(\min(X_1,\cdots,X_N)\leq x) = \sum_{n\geq 0} [1-(1-F(x))^n ] P(N=n)$$

     – Alex R. Mar 23 '16 at 19:23
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    If the Poisson distribution allows n=0, then there are no samples in that draw. Programmatically I would consider it as either an NA-padded mixture, or a draw that results in no samples added to the pool. I like Alex's idea about shifting the Poisson distribution. – EngrStudent Mar 24 '16 at 00:06

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