Hi everyone,

This is my first time here, so please let me know if I can clarify my question in any way (incl. formatting, tags, etc.). (And hopefully I can edit later!) I tried to find references, and tried to solve myself using induction, but failed at both.

I'm trying to simplify a distribution that seems to reduce to an order statistic of a countably infinite set of independent $\chi^2$ random variables with different degrees of freedom; specifically, what is the distribution of the $m$th smallest value among independent $\chi^2_2,\chi^2_4,\chi^2_6,\chi^2_8,\ldots$?

I would be interested in the special case $m=1$: what is the distribution of the minimum of (independent) $\chi^2_2,\chi^2_4,\chi^2_6,\ldots$?

For the case of the minimum, I was able to write the cumulative distribution function (CDF) as an infinite product, but can't simplify it further. I used the fact that the CDF of $\chi^2_{2m}$ is $$F_{2m}(x)=\gamma(m,x/2)/\Gamma(m)=\gamma(m,x/2)/(m-1)!=1-e^{-x/2}\sum_{k=0}^{m-1}x^k/(2^k k!).$$ (With $m=1$, this confirms the second comment below about equivalence with an exponential distribution with expectation 2.) The CDF of the minimum can then be written as $$F_{min}(x) = 1-(1-F_2(x))(1-F_4(x))\ldots = 1-\prod_{m=1}^\infty (1-F_{2m}(x)) $$ $$= 1- \prod_{m=1}^\infty \left(e^{-x/2}\sum_{k=0}^{m-1}\frac{x^k}{2^k k!}\right).$$ The first term in the product is just $e^{-x/2}$, and the "last" term is $e^{-x/2}\sum_{k=0}^\infty x^k/(2^k k!)=1$. But I don't know how (if possible?) to simplify it from there. Or maybe a totally different approach is better.

If anyone is curious, I am trying to simplify Theorem 1 in this paper for the case of regression on a constant ($x_i=1$ for all $i$). (I have $\chi^2$ instead of $\Gamma$ distributions since I have multiplied by $2\kappa$.)

Thanks for any help!

Dave

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