# Order statistics (e.g., minimum) of infinite collection of chi-square variates?

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.

Another potentially helpful reminder: $\chi^2_2$ is the same as an exponential distribution with expectation 2, and $\chi^2_4$ is the sum of two such exponentials, etc.

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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Does this answer your question? –  mpiktas Aug 5 '11 at 10:52
@mpiktas: thanks for the suggestion. It's similar, except instead of exponentials with different rate parameters, I have chi-squares with different degrees of freedom (and an infinite number of them, not finite). And while $\chi^2_2$ is an exponential, $\chi^2_4,\chi^2_6,\ldots$ are not; they are sums of exponentials, but sums of exponentials are not exponential themselves. (And ideally I'm hoping for a general order statistic, though the min would be a great start.) –  David M Kaplan Aug 5 '11 at 15:51
I doubt there's a closed form for this. It does have a curious characterization, however: when $X_k$ are iid Poisson($\lambda/2$) variates, $k=1,2,\ldots$, then $1-F_{min}(\lambda)$ is the chance that all $X_k \le k$. –  whuber Aug 5 '11 at 17:11
@whuber: It's perhaps not quite so curious when thought of in terms of a Poisson process, which is the formulation I had been playing with. Let $T_1, T_2, \ldots$ be i.i.d. $\mathrm{Exp}(1/2)$ random variables, with corresponding Poisson process $N(t) := \sup\{n: \sum_{i=1}^n T_i \leq t\}$ of rate $1/2$. Let $U_1 = T_1$, $U_2 = T_2 + T_3$, $U_3 = T_4 + T_5 + T_6$, etc. Then, the $U_i\sim\chi_{2i}^2$ are independent and by the stationary independent-increments property of a Poisson process, we have that $\mathbb{P}(U_i \geq t) = \mathbb{P}( N(t) \leq i)$. –  cardinal Aug 6 '11 at 12:38
@Cardinal Of course: that's a good way to see it. The curiosity isn't in the relationship between Poissons and Gammas; it lies in the description of the event itself! –  whuber Aug 6 '11 at 13:36
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The zeros of the infinite product will be the union of the zeros of the terms. Computing out to the 20th term shows the general pattern:

This plot of the zeros in the complex plane distinguishes the contributions of the individual terms in the product by means of different symbols: at each step, the apparent curves are extended further and a new curve is started even further left.

The complexity of this picture demonstrates there exists no closed-form solution in terms of well-known functions of higher analysis (such as gammas, thetas, hypergeometric functions, etc, as well as the elementary functions, as surveyed in a classic text like Whittaker & Watson).

Thus, the problem might be more fruitfully posed a little differently: what do you need to know about the distributions of the order statistics? Estimates of their characteristic functions? Low order moments? Approximations to quantiles? Something else?

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Why zeroes of the product are of importance? I feel I am missing something trivial. –  mpiktas Aug 5 '11 at 20:03
@mp The zeros show something about the complexity of the function. Rational functions have a finite number of zeros. Elementary functions typically have a line of zeros, such as at $2 i \pi n$, $n$ integral, for $\exp()$; typical "transcendental" functions have slightly more complex patterns of zeros, such as at all the non-positive integers (Gamma function) or on a polygonal lattice of points (theta functions and elliptic functions generally). The complicated pattern exhibited here suggests it will be difficult or impossible to express the CDF in terms of these familiar functions. –  whuber Aug 6 '11 at 21:28
@whuber (1/2), thanks! I did not know about the different classes of functions having those different patterns of zeroes in the complex plane; that sounds very useful, and your graph seems to answer my question (as posed). –  David M Kaplan Aug 7 '11 at 4:39
@whuber (2/2), 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 w/ my advisor after his grant deadline; but potentially, I'd be trying to take a higher-order expansion of the $m$th order-stat (divided by $m$) as $m\to\infty$, in a more complicated setting. Will post again if so; thanks again! –  David M Kaplan Aug 7 '11 at 4:49