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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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You might try also asking this at stats.stackexchange.com. (I'm not saying that this is inappropriate for Math Overflow, but the statistician population here is pretty small.) –  Michael Lugo Aug 4 '11 at 22:37
2  
Are you assuming these are independent? Might it help to recall that the $\chi^2_2$ distribution is the same as an exponential distribution with expectation 2, and the $\chi^2_4$ distribution is the distribution of the sum of two independent exponentials, etc. –  Michael Hardy Aug 5 '11 at 4:09
    
Thanks Michael L; I wasn't sure if stats was more applied stats, but I will try it! Thanks Michael H for the clarification (yes, they are independent; have edited); and thanks for the reminder of the connection to exponentials (which are indeed easier for order statistics)--I had given up on that route since there are still sums of them, but it sounds more promising than anything else so far so I'll revisit it. –  David M Kaplan Aug 5 '11 at 4:27
    
I evaluated the probability that the minimum is at least $1$ numerically as $0.5428253780118192732336667897236961030128174$ which didn't show up in the Inverse Symbolic Calculator's list. –  Douglas Zare Aug 5 '11 at 5:21
    
Douglas: I'm curious how you did that evaluation. –  Michael Lugo Aug 5 '11 at 5:37

2 Answers 2

It might be best to look at the hazard function, since the hazard function of min of independent variates is the sum of individual hazard functions.

Hence the hazard function of your distribution is given by

$$ \sum_{k=1}^\infty \frac{x^{2k-1} 2^{1-k} }{ \Gamma_k (\frac{x^2}{2}) } e^{-\frac{x^2}{2}} = \sum_{k=1}^\infty x \frac{ \left(\tfrac{x^2}{2}\right)^{k-1} e^{-\frac{x^2}{2} } }{ \int_{\tfrac{x^2}{2}}^\infty t^{k-1} e^{-t} dt } $$

where $\Gamma_k(x)$ stands for incomplete gamma function, i.e. $\Gamma_k(z) = \int_z^\infty t^{k-1} e^{-t} dt $.

Consider $ 0 < x < 1$. In this case $\Gamma_k(x) \to (k-1)!$ for large $k$. Hence the hazard function will asymptotically equal $x$ (evaluate the sum with $(k-1)!$ used instead of incomplete gamma). This means that for small $x$ your density looks like $x e^{-x^2/2}$.

Notice also that for large $x$, the individual hazard functions of $\chi^2_{2k}$ also behave as $x$.

Hope this helps.

Below is some Mathematica code I used to visualize your distribution. It computes survival function of your distribution, i.e. $\text{Pr}( X>x)$.

SF[x_Real] := Module[{prod = 1, prev, m = 1},
  While[True, prev = prod; 
   prod *= SurvivalFunction[ChiDistribution[2 (m)], x]; 
   If[prod == prev, Break[]]; m++;]; prod]

Then plotting shows agreement with $x e^{-x^2/2}$ for small $x$.

LogPlot[{SF[x], E^(-(x^2/2))}, {x, 0, 2}]


EDIT

Using closed form expression for incomplete gamma with integral parameter the hazard function can be written as

$$ HF(x) = x \sum_{m=1}^\infty \frac{1}{(m-1)!} \left(\frac{x^2}{2}\right)^{m-1} \frac{1}{ \sum_{k=0}^{m-1} \frac{1}{k!} \frac{x^{2 k}}{2^k} } $$

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That is a helpful perspective, thank you! –  David M Kaplan Aug 9 '11 at 14:50

This is an answer proposed on stats.stackexchange.com (CrossValidated). It is not 100% definitive, but is suggestive enough for my taste. (Suggestive that no closed-form CDF of the minimum exists.)

As mentioned briefly in the original post, this was checking a special case of a (complicated) distribution of an estimator given in another paper. They used the existence of the distribution to justify using bootstrap; my advisor suggested I try to approximate the distribution. It seems like their distribution might be off for this special case (where I know what it should be), so I'll check with my advisor after his grant deadline this week; but potentially, I would be trying to take a higher-order expansion of the $m$th order-statistic (divided by m) as $m\to\infty$, in a more complicated setting. I'll post again if so, but for now I'm satisfied with the (negative) result.

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