A sum of logbeta distributed variables occurs in this question Distribution with a given moment generating function

Let, $X_j \sim Beta(j\sigma, 1-\sigma)$, $Y_j = -\log(X_j)$ and $S_n = \sum_{j=1}^n Y_j - \frac{1-\sigma}{\sigma}\log(n)$ then the moment generating function of $S_n$ approaches, for $n \to \infty$

$$E(e^{tS_n}) \to \frac{\Gamma(1-t/\sigma)}{\sigma^t\Gamma(1-t)}$$

How is this derived?

  • $\begingroup$ While going through some of my old questions, and trying to close whatever is reasonably answered, I encountered this problem. I had been able to come up with a solution, but, I find some steps not very intuitive. $\endgroup$ Feb 22, 2023 at 21:11

1 Answer 1


The expectation of $E(e^{tS_n})$ is similar to $E(Z_{n,t})$ where $Z_{n,t}$ is a product

$$Z_{n,t} = e^{tS_n} = e^{-t\sum_{j=1}^n \log X_j -t \frac{1-\sigma}{\sigma} \log(n)}= \left( n^{(1-\sigma)/\sigma} \prod_{j=1}^n X_j \right)^{-t}$$

And the expectation of a product of independent variables is the product of the expectation of the individual variables

$$E(Z_{n,t}) = n^{-t(1-\sigma)/\sigma} \prod_{j=1}^n E( X_j^{-t}) $$

The expectation $E(X_j^{-t})$ for a beta distribution is

$$E(X_j^{-t}) = \frac{1}{B(\alpha,\beta)} \int_0^1 x^{-t} x^{\alpha-1}(1-x)^{\beta-1} dx = \frac{B(\alpha-t,\beta)}{B(\alpha,\beta)}$$

We get to

$$E(e^{tS_n}) = n^{-t(1-\sigma)/\sigma} \prod_{j=1}^n \frac{B(j\sigma-t,1-\sigma)}{B(j\sigma,1-\sigma)} $$

We can rewrite the beta function in terms of gamma functions $B(x,y) = \Gamma(x)\Gamma(y)/\Gamma(x+y)$

$$\begin{array}{} E(e^{tS_n}) & = & n^{-t(1-\sigma)/\sigma} \prod_{j=1}^n \frac{\Gamma(j\sigma-t)}{\Gamma(j\sigma-t+1-\sigma)} \frac{\Gamma(j\sigma+1-\sigma)}{\Gamma(j\sigma)} \\ &=& n^{-t(1-\sigma)/\sigma} \prod_{j=1}^n \frac{\Gamma(j\sigma-t+1)}{\Gamma((j-1)\sigma-t+1)} \cdot \frac{\Gamma((j-1)\sigma+1)}{\Gamma(j\sigma+1)} \cdot \frac{j\sigma}{j\sigma-t} \end{array}$$

in the second line we rewrote the gamma function such that we get terms $\Gamma((j-1)\sigma+1)$ and $\Gamma(j\sigma+1)$ where the difference is only in $j$ versus $j-1$. This makes that many terms cancel because in the sum they appear both once in the numerator and once in the denominator.

$$\begin{array}{} E(e^{tS_n}) &=& n^{-t(1-\sigma)/\sigma} \frac{1}{\Gamma(1-t)} \cdot \frac{\Gamma(n\sigma + 1 -t)}{\Gamma(n\sigma +1)} \cdot \prod_{j=1}^n \frac{j\sigma}{j\sigma-t} \\ \end{array}$$

using $\Gamma(1+z) = \prod_{j=1}^{\infty} \frac{(1+1/j)^z}{1+z/j}$

$$\begin{array}{} E(e^{tS_n}) &=& n^{-t(1-\sigma)/\sigma} \frac{1}{\Gamma(1-t)} \cdot \frac{\Gamma(n\sigma + 1 -t)}{\Gamma(n\sigma +1)} \cdot \prod_{j=1}^n \frac{1}{1-\frac{t/\sigma}{j}} \\ &=& n^{-t(1-\sigma)/\sigma} \frac{\Gamma(1-t/\sigma)}{\Gamma(1-t)} \cdot \frac{\Gamma(n\sigma + 1 -t)}{\Gamma(n\sigma +1)} \cdot \prod_{j=1}^n \frac{1}{1-\frac{t/\sigma}{j}} \prod_{j=1}^\infty \frac{1-\frac{t/\sigma}{j}}{(1+1/j)^{-t/\sigma}} \\ &=& n^{-t(1-\sigma)/\sigma} \frac{\Gamma(1-t/\sigma)}{\Gamma(1-t)} \cdot \frac{\Gamma(n\sigma + 1 -t)}{\Gamma(n\sigma +1)} \cdot \prod_{j=1}^n \frac{1}{(1+1/j)^{-t/\sigma}} \prod_{j=n+1}^\infty \frac{1-\frac{t/\sigma}{j}}{(1+1/j)^{-t/\sigma}} \\ &=& n^{-t(1-\sigma)/\sigma} \frac{\Gamma(1-t/\sigma)}{\Gamma(1-t)} \cdot \frac{\Gamma(n\sigma + 1 -t)}{\Gamma(n\sigma +1)} \cdot (1+n)^{t/\sigma} \cdot \prod_{j=n+1}^\infty \frac{1-\frac{t/\sigma}{j}}{(1+1/j)^{-t/\sigma}}\\ &\approx& n^{-t(1-\sigma)/\sigma} \frac{\Gamma(1-t/\sigma)}{\Gamma(1-t)} \cdot (n\sigma + 1)^{-t} \cdot (1+n)^{t/\sigma} \cdot \prod_{j=n+1}^\infty \frac{1-\frac{t/\sigma}{j}}{(1+1/j)^{-t/\sigma}} \end{array}$$

in the limit for $n \to \infty$

$$\begin{array}{} \lim_{n \to \infty} E(e^{tS_n}) &=&n^{-t(1-\sigma)/\sigma} \cdot (1+n)^{t/\sigma} \cdot (n\sigma + 1)^{-t} \cdot \frac{\Gamma(1-t/\sigma)}{\Gamma(1-t)} \\ &=& \underbrace {\left( \frac{(1+1/n)^{1/\sigma} }{\sigma + 1/n}\right)^t }_{\to 1/\sigma^t} \cdot \frac{\Gamma(1-t/\sigma)}{\Gamma(1-t)} \\ &=& \frac{\Gamma(1-t/\sigma)}{\sigma^t\Gamma(1-t)} \end{array}$$

  • 1
    $\begingroup$ If you still need steps to get $1/\sigma^t$, then writing $\frac{n \left(\frac{1}{n}+1\right)^{1/\sigma }}{n \sigma +1}$ as $\frac{\left(\frac{1}{n}+1\right)^{1/\sigma }}{\frac{1}{n}+\sigma }$ and applying L'Hôpital's rule once will get a form that you can just plug in $\infty$ for $n$ to get the limit. $\endgroup$
    – JimB
    Oct 17, 2021 at 17:46
  • $\begingroup$ @JimB I got the last step. Or at least, I did it intuitively (not with L'Hopital's rule). The term $(1+1/n)^{1/\sigma}$ approaches $(1)^{1/\sigma}=1$ leaving $n/(n \sigma + 1) \approx n/(n \sigma) = 1/\sigma$. But indeed, it could be done more formally. $\endgroup$ Oct 17, 2021 at 19:01
  • $\begingroup$ I find the derivation still simple untill $$E(e^{tS_n}) = n^{-t(1-\sigma)/\sigma} \prod_{j=1}^n \frac{B(j\sigma-t,1-\sigma)}{B(j\sigma,1-\sigma)}$$ but after that I am applying some standard rules to get to the end result, but it is not intuitive anymore. $\endgroup$ Oct 17, 2021 at 19:13
  • 1
    $\begingroup$ Where you have "nominator" do you mean to say "numerator"? (I do this myself.) $\endgroup$
    – Glen_b
    Oct 17, 2021 at 23:17
  • $\begingroup$ As just commented here stats.stackexchange.com/questions/143027/… I suggest working wi the log MGF to help with limits. $\endgroup$
    – julyan
    Oct 18, 2021 at 9:59

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