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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 $2\kappa$.)

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  • $\begingroup$ Does this answer your question? $\endgroup$
    – mpiktas
    Commented Aug 5, 2011 at 10:52
  • $\begingroup$ @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.) $\endgroup$
    – David M Kaplan
    Commented Aug 5, 2011 at 15:51
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    $\begingroup$ 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 $X_k \le k$. $\endgroup$
    – whuber ♦
    Commented Aug 5, 2011 at 17:11
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    $\begingroup$ @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 $\mathbb{P}(U_i \geq t) = \mathbb{P}( N(t) \leq i)$. $\endgroup$
    – cardinal
    Commented Aug 6, 2011 at 12:38
  • $\begingroup$ @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! $\endgroup$
    – whuber ♦
    Commented Aug 6, 2011 at 13:36

2 Answers 2

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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:

plot of complex zeros

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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  • $\begingroup$ Why zeroes of the product are of importance? I feel I am missing something trivial. $\endgroup$
    – mpiktas
    Commented Aug 5, 2011 at 20:03
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    $\begingroup$ @mp The zeros and poles show something about the complexity of the function. Rational functions have a finite number of them. Elementary functions typically have a line of zeros, such as at $2i\pi n$, $n$ integral, for $\exp()$; typical "transcendental" functions have slightly more complex patterns of zeros, such as at all the non-positive integers (reciprocal of the Gamma function) or on a lattice of points (theta functions and elliptic functions). The complicated pattern exhibited here suggests it will be difficult or impossible to express the CDF in terms of these familiar functions. $\endgroup$
    – whuber ♦
    Commented Aug 6, 2011 at 21:28
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    $\begingroup$ @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). $\endgroup$
    – David M Kaplan
    Commented Aug 7, 2011 at 4:39
  • $\begingroup$ @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! $\endgroup$
    – David M Kaplan
    Commented Aug 7, 2011 at 4:49
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what is the distribution of the minimum of (independent) $\chi^2_2,\chi^2_4,\chi^2_6,\ldots$?

Apologies for arriving some 6 years late. Even though the OP has likely now moved onto other problems, the question remains fresh, and I thought I might suggest a different approach.

We are given $(X_1, X_2, X_3, \dots)$ where $X_i \sim \text{Chisquared}(v_i)$ where $v_i= 2i$ with pdf's $f_i(x_i)$:

enter image description here

Here is a plot of the corresponding pdf's $f_i(x_i)$, as the sample size increases, for $i = 1 \text{ to } 8$:

enter image description here

We are interested in the distribution of $\text{min}(X_1, X_2, X_3, \dots)$.

Each time we add an extra term, the pdf of the marginal last term added shifts out further and further to the right, so that the effect of adding more and more terms becomes not only less and less relevant, but after a just a few terms, becomes almost negligible -- on the sample minimum. This means, in effect, that only a very small number of terms are likely to actually matter ... and adding additional terms (or the presence of an infinite number of terms) is largely irrelevant for the sample minimum problem.

Test

To test this, I have calculated the pdf of $\text{min}(X_1, X_2, X_3, \dots)$ to 1 term, 2 terms, 3 terms, 4 terms, 5 terms, 6 terms, 7 terms, 8 terms, to 9 terms, and to 10 terms. To do this, I have used the OrderStatNonIdentical function from mathStatica, instructing it here to calculate the pdf of the sample minimum (the $1^{\text{st}}$ order statistic) in a sample of size $j$, and where parameter $i$ (instead of being fixed) is $v_i$:

enter image description here enter image description here

It gets a bit complicated as the number of terms increase ... but I have shown the output for 1 term (1st row), 2 terms (second row), 3 terms (3rd row) and 4 terms above.

The following diagram compares the pdf of the sample minimum with 1 term (blue), 2 terms (orange), 3 terms, and 10 terms (red). Note how similar the results are with just 3 terms vs 10 terms: enter image description here

The following diagram compares 5 terms (blue) and 10 terms (orange) -- the plots are so similar, they obliterate each other, and one cannot even see the difference:

enter image description here

In other words, increasing the number of terms from 5 to 10 has almost no discernible visual impact on the distribution of the sample minimum.

Half-Logistic Approximation

Finally, an excellent simple approximation of the pdf of the sample min is the half-Logistic distribution with pdf:

$$g(x) = \frac{2 e^{-x}}{\left(e^{-x}+1\right)^2} \quad \text{ for } x>0$$

The following diagram compares the exact solution with 10 terms (which is indistinguishable from 5 terms or 20 terms) and the half-Logistic approximation (dashed):

enter image description here

Increasing to 20 terms makes no discernible difference.

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