As a follow-up to a question on a central limit theorem for independent random variables (r.v.) here, let $Y_j=-\log(1-V_j)$, where $V_j\sim\mbox{beta}(1-\sigma,j\sigma)$, $j\in\mathbb{N}^*$, $\sigma\in(0,1)$. The shifted sums $S_n=\sum_{j=1}^{n}Y_j -\frac{1-\sigma}{\sigma}\log n$ have moment generating functions (MGF) which admit a simple limit when $n\rightarrow \infty$: $$\mathbb{E}\left(e^{\lambda S_n}\right)\rightarrow M(\lambda)=\frac{\Gamma(1-\lambda/\sigma)}{\sigma^\lambda \Gamma(1-\lambda)}.$$

I'm trying to work out to the r.v. $S$ that admits $M(\lambda)$ as a MGF. The existence of $S$ is by the Kolmogorov three-series theorem, which ensures a.s. convergence. Note that $S$ is infinitely divisible since it is the limit of the infinitely divisible $Y_j$s.

In the expression of $M(\lambda)$, $\sigma^\lambda\Gamma(1-\lambda)$, resp. $\Gamma(1-\lambda/\sigma)$, is the MGF of a Gumbel r.v. shifted by $\log(\sigma)$, resp. Gumbel r.v. rescaled by $\sigma$. Though I don't see how to make use of this ratio since the inverse of an MGF isn't an MGF.

  • $\begingroup$ How did you derive this moment generating function? $\endgroup$ Commented Oct 16, 2021 at 20:49
  • $\begingroup$ @SextusEmpiricus I've wondered, too, about the derivation of $M(\lambda)$ but I was able to obtain the same result with Mathematica. Are you looking for a derivation or questioning the result (or both) ? I note that the user hasn't asked another question since 2016 so I suspect he won't be answering. $\endgroup$
    – JimB
    Commented Oct 17, 2021 at 3:46
  • $\begingroup$ @JimB it is a bit of both. The characteristic function did not seem intuitive to me. It needs to involve some product, but while there are some infinite product representations of the gamma function, I did not see how it would get together. $\endgroup$ Commented Oct 17, 2021 at 9:54
  • $\begingroup$ @JimB I am currently working it out by hand here stats.stackexchange.com/q/548613 but it is not giving me much intuition. It is not yet finished but there are already many steps involved. (I was hoping that it was gonna be something simple) $\endgroup$ Commented Oct 17, 2021 at 15:07
  • $\begingroup$ What I found helpful to get the result is to look for the log MGF: much easier to deal with limits of sums rather than limits of products, IMO. Not active in the last 5 years, but still alive :-) $\endgroup$
    – julyan
    Commented Oct 18, 2021 at 9:52

2 Answers 2


Solution for the case $\sigma = 1/k$ with $k \in \mathbb{N}$

The answer of JimB for the case of $\sigma = 1/2$ can be adjusted in order to get to an expression for the cases $\sigma = 1/k$ where $k$ is an integer with $k \geq 2$.

In terms of $k$, the moment generating function is

$$ M(\lambda)=\frac{k^\lambda\Gamma(1-k \lambda)}{ \Gamma(1-\lambda)} = \frac{k^\lambda\Gamma(-k \lambda)\cdot(-k\lambda)}{ \Gamma(-\lambda)\cdot(-\lambda)} = \frac{k^{\lambda+1}\Gamma(-k \lambda)}{ \Gamma(-\lambda)} $$

For the derivation of the end result we will be using Gauss's multiplication formula and the generalized Gumbel distribution.

Gauss's multiplication formula

With Gauss's multiplication formula, we can write a gamma function with a parameter $k\lambda$ in terms of gamma functions with parameter $\lambda$

$$\Gamma(k\lambda) = k^{k\lambda -\frac{1}{2}} (2\pi)^{\frac{1-k}{2}} \prod_{j=0}^{k-1} \Gamma\left(\frac{j}{k} + \lambda \right)$$

And when we take the $j=0$ term in the product to the left hand side and inverse the parameter (use $-\lambda$ instead of $\lambda$)

$$\frac{\Gamma(-k\lambda)}{\Gamma(-\lambda)} = k^{-k\lambda -\frac{1}{2}} (2\pi)^{\frac{1-k}{2}} \prod_{j=1}^{k-1} \Gamma\left( \frac{j}{k} - \lambda \right)$$

If we substitute this into the moment generating function then

$$M(\lambda) = k^{\frac{1}{2}} (2\pi)^{\frac{1-k}{2}} \prod_{j=1}^{k-1} \left( \frac{1}{k}\right)^{\lambda} \Gamma\left(\frac{j}{k} - \lambda \right) $$

Generalized Gumbel distribution

The expression in the product is like the generalized Gumbel distribution

$$f(x) = \frac{b^a}{\Gamma(a)} e^{-(ax+be^{-x})}$$

with moment generating function

$$M(\lambda;a,b) = \frac{1}{\Gamma(a)}b^\lambda\Gamma(a-\lambda)$$

A description of this generalized Gumbel distribution occurs in Ahuja, J. C., and Stanley W. Nash. "The generalized Gompertz-Verhulst family of distributions." Sankhyā: The Indian Journal of Statistics, Series A (1967): 141-156.

Sum of generalized Gumbel distributions

The term $$\prod_{j=1}^{k-1} \left( \frac{1}{k}\right)^{\lambda} \Gamma\left(\frac{j}{k} - \lambda \right)$$ relates to a product of moment generating functions of the generalized Gumbel distribution with $a = j/k$ and $b=\frac{1}{k}$.

This product of moment generating functions of the gumbel distribution will involve a constant based on a product of gamma functions which we can simplify with the multiplication formula

$$\prod_{j=1}^{k-1} {\Gamma(j/k+z)} = (2\pi)^{\frac{ k-1}{2}} k^{1/2-kz} \frac{\Gamma(kz)}{\Gamma(z)} $$

setting $z=0$

$$\prod_{j=1}^{k-1} {\Gamma(j/k)} = (2\pi)^{\frac{ k-1}{2}} k^{1/2}$$

And the inverse

$$\prod_{j=1}^{k-1} \frac{1} {\Gamma(j/k)} = (2\pi)^{\frac{1-k}{2}} k^{-1/2}$$

this equals $1/k$ times the constant $k^{\frac{1}{2}} (2\pi)^{\frac{1-k}{2}} $ that we have in the moment generating function that we are looking for.

So we can write

$$ M(\lambda)=\frac{k^\lambda\Gamma(1-k \lambda)}{ \Gamma(1-\lambda)} = \frac{1}{k} \prod_{j=1}^{k-1} \frac{1} {\Gamma(j/k)} \left( \frac{1}{k}\right)^{\lambda} \Gamma\left(\frac{j}{k} - \lambda \right) $$

The product on the right-hand side is the product of the moment generating function of generalized Gumbel distributions with $a = j/k$ and $b=\frac{1}{k}$. The factor $1/k$ relates to a translation with a factor $-\log(k)$.

  • $\begingroup$ I know one is not supposed to use comments this way but: Very nice! $\endgroup$
    – JimB
    Commented Oct 20, 2021 at 14:10

There is an explicit density for $S_n$ when $\sigma=1/2$ and a limiting density of

$$\frac{e^{-\frac{z}{2}-\frac{e^{-z}}{2}}}{\sqrt{2 \pi }}$$

(And there might be a general density for other values of $\sigma$.) I have to confess ignorance of what extreme value distribution has the above density. It's similar to a Gumbel distribution not quite.

This is done with a brute force approach where the pdf of $\sum_{i=1}^n Y_i$ is found followed by the density of $S_n=\sum_{i=1}^n Y_i - \log(n)$ (again, just for $\sigma=1/2$). Then the limit of the density of $S_n$ is found along with the limiting moment generating function.

First, the distribution of $Y_i$ (using Mathematica throughout).

pdf[j_, y_] := Simplify[PDF[TransformedDistribution[-Log[1 - x],
  x \[Distributed] BetaDistribution[1 - 1/2, j /2]], y] // 
  TrigToExp, Assumptions -> y > 0]

So the pdf for $Y_1$ and $Y_2$ are

pdf[1, y1]

$$\frac{1}{\pi \sqrt{e^{y_1}-1}}$$

pdf[2, y2]

$$\frac{e^{-\frac{y_2}{2}}}{2 \sqrt{e^{y_2}-1}}$$

The distribution of the sums of the $Y$'s are constructed sequentially:

pdfSum[1] = pdf[1, y1]
pdfSum[2] = Integrate[pdfSum[1] pdf[2, y2 - y1], {y1, 0, y2}, Assumptions -> y2 > 0]
pdfSum[3] = Integrate[pdfSum[2] pdf[3, y3 - y2], {y2, 0, y3}, Assumptions -> y3 > 0]
pdfSum[4] = Integrate[pdfSum[3] pdf[4, y4 - y3], {y3, 0, y4}, Assumptions -> y4 > 0]
pdfSum[5] = Integrate[pdfSum[4] pdf[5, y5 - y4], {y4, 0, y5}, Assumptions -> y5 > 0]
pdfSum[6] = Integrate[pdfSum[5] pdf[6, y6 - y5], {y5, 0, y6}, Assumptions -> y6 > 0]
pdfSum[7] = Integrate[pdfSum[6] pdf[7, y7 - y6], {y6, 0, y7}, Assumptions -> y7 > 0]
pdfSum[8] = Integrate[pdfSum[7] pdf[8, y8 - y7], {y7, 0, y8}, Assumptions -> y8 > 0]

Resulting pdf's

We see the pattern and the pdf for a general $n$ is

$$\frac{\Gamma \left(\frac{n+1}{2}\right) \exp \left(\frac{1}{2} (-(n-1)) z\right) (\exp (z)-1)^{\frac{n-2}{2}}}{\sqrt{\pi } \Gamma \left(\frac{n}{2}\right)}$$

So the pdf for $S_n$ is that of the above but including the shift of $\log(n)$:

pdfSn[n_] := Gamma[(1 + n)/2]/(Sqrt[\[Pi]] Gamma[n/2])*
   Exp[-(n - 1) (z + Log[n])/2] (-1 + Exp[z + Log[n]])^((n - 2)/2)

Taking the limit of this function we have

pdfS = Limit[pdfSn[n], n -> \[Infinity], Assumptions -> z \[Element] Reals]

$$\frac{e^{-\frac{z}{2}-\frac{e^{-z}}{2}}}{\sqrt{2 \pi }}$$

The moment generating function associated with this pdf is

mgf = Integrate[Exp[\[Lambda] z] pdfS, {z, -\[Infinity], \[Infinity]},
   Assumptions -> Re[\[Lambda]] < 1/2]

$$\frac{2^{-\lambda } \Gamma \left(\frac{1}{2}-\lambda \right)}{\sqrt{\pi }}$$

This doesn't look exactly like the mgf in the OP's question when $\sigma=1/2$ but Mathematica declares them identical with

Gamma[1 - \[Lambda]/(1/2)]/((1/2)^\[Lambda] Gamma[1 - \[Lambda]]) // FunctionExpand

$$\frac{2^{-\lambda } \Gamma \left(\frac{1}{2}-\lambda \right)}{\sqrt{\pi }}$$

Another check involves extracting several moments which gives identical results. Here is for the first moment:

D[mgf, {\[Lambda], 1}] /. \[Lambda] -> 0 // FunctionExpand // FullSimplify
D[Gamma[1 - \[Lambda]/(1/2)]/((1/2)^\[Lambda] Gamma[1 - \[Lambda]]), {\[Lambda], 1}] /. \[Lambda] -> 0

Both give $\gamma +\log (2)$ (where $\gamma$ is Euler's constant).

For the 17th moment:

D[mgf, {\[Lambda], 17}] /. \[Lambda] -> 0 // N
D[Gamma[1 - \[Lambda]/(1/2)]/((1/2)^\[Lambda] Gamma[1 - \[Lambda]]), {\[Lambda], 17}] /. \[Lambda] -> 0 // N

Both give 3.71979*10^19.

There might be some simple way to insert $\sigma$ into the limiting pdf but I haven't played with that yet.

  • 1
    $\begingroup$ This distribution looks like a type of generalized Gumbel distribution described in this answer. It has the probability density function $$f(x) = \frac{b^a}{\Gamma(a)} e^{-(ax+be^{-x})}$$ and has the moment generating function $$M(t;a,b) = b^t \frac{\Gamma(a-t)}{\Gamma(a)}$$ But this does not have this gamma function (with a dependency on $t$) in the denominator. $\endgroup$ Commented Oct 19, 2021 at 7:04
  • 1
    $\begingroup$ An interpretation of the case $a=b$ is an m-th order distribution (and was described by Gumbel). $\endgroup$ Commented Oct 19, 2021 at 7:13
  • $\begingroup$ The equivalence for $\sigma=1/2$ can possibly be shown with Legendre's multiplication formula en.m.wikipedia.org/wiki/… And there are similar forms that will work for $\sigma = 1/k$ $\endgroup$ Commented Oct 19, 2021 at 9:16
  • $\begingroup$ Like this $$\Gamma(z+0.5) = \frac{\Gamma(2z)}{\Gamma(z)} \sqrt{\pi}2^{1-2z} = \frac{\Gamma(1+2z)}{\Gamma(1+z)} \sqrt{\pi}2^{-2z} $$ $\endgroup$ Commented Oct 19, 2021 at 10:02
  • $\begingroup$ Thanks for link to the generalize Gumbel distribution. Unfortunately it appears that isn't the desired distribution as I think that if it were, we would need $a=b=\sigma$ and that doesn't produce the same moments with values of $\sigma\neq1/2$. $\endgroup$
    – JimB
    Commented Oct 19, 2021 at 15:18

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