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Is there a closed form for the $n$'th cumulant of the Beta distribution, as a function of the parameters $\alpha$, $\beta$?

The cumulant generating function of the Beta distribution is the logarithm of a confluent hypergeometric function:

$$\log(\mathrm{E}(\mathrm{e}^{tX}))=\log{_1F_1(\alpha,\alpha+\beta ;t)},$$

where $X$ is a random variable following a Beta distribution with parameters $\alpha$, $\beta$ and pdf $f(x)$:

$$f(x)=\frac{x^{\alpha-1}(1-x)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}.$$

So, equivalently, I am asking for a general closed form expression of the coefficient $\kappa_n$ in:

$$\log{_1F_1(\alpha,\alpha+\beta ;t)}=\sum_{n=1}^\infty \kappa_n \frac{t^n}{n!}.$$

Note: By closed form I mean an expression involving no derivatives or integrals, and no infinite sums or products. It can contain finite sums and finite products.

a06e
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1 Answers1

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The $n$-th raw moment of the Beta distribution is given by:

$$E[X^n]= \mu_n' = \frac{B(\alpha+n,\beta)}{B(\alpha,\beta)}$$

Thus, to obtain a closed form solution for the $n$-th cumulant, all we need is a method to express the $n$-th cumulant $\kappa_n$ in terms of the raw moments $\mu_i'$ (which we know), and this can be done as follows:

$$\kappa_n = (-1)^{n-1}\left| \begin{array}{cccc} \mu_1' & 1 & 0& 0 & \cdots & 0 \\ \mu_2' & \binom{1}{0}\mu_1' & 1& 0 & \cdots & 0 \\ \mu_3' & \binom{2}{0}\mu_2' & \binom{2}{1}\mu_1'& 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & & 1 \\ \mu_n' & \binom{n-1}{0}\mu_{n-1}' & \binom{n-1}{1}\mu_{n-2}' &\dots &\dots & \binom{n-1}{n-2}\mu_1' \\ \end{array} \right|$$

This has "no derivatives or integrals, and no infinite sums or products" :)

Example: $n=2$

With $n = 2$, the formula is:

$$\kappa_2 = (-1)^{1}\left| \begin{array}{cccc} \mu_1' & 1 \\ \mu_2' &\binom{1}{0}\mu_1' \\ \end{array} \right| = \mu_2' - \mu_1'^2$$

which is the formula for variance, and so must give the correct solution.

Example: $n=3$

With $n = 3$, the formula is:

$$\kappa_3 = (-1)^{2}\left| \begin{array}{cccc} \mu_1' & 1 &0 \\ \mu_2' &\binom{1}{0}\mu_1' &1 \\ \mu_3' &\binom{2}{0}\mu_2' &\binom{2}{1}\mu_1' \\ \end{array} \right| = 2\mu_1'^3 - 3\mu_1' \mu_2' + \mu_3'$$

Automated form

The above was provided because the OP sought a 'closed-form' solution. But, instead of manually deriving each of these cumulant to raw moment conversions, there exist automated ways of doing this that make it much easier. For example, here are the first 5 such conversions using the CumulantToRaw function in mathStatica:

Even with $n = 20$, the conversion takes less than 0.15 seconds to compute.

Glorfindel
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wolfies
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    Are you sure? Something seems to be wrong... The variance of the Beta distribution (a.k.a., the 2nd cumulant) is

    $$\frac{\alpha\beta}{(\alpha+\beta)^2(1+\alpha+\beta)}$$

    but your formula with $n=2$ gives:

    $$\frac{\alpha(1+\alpha)}{(\alpha+\beta)(1+\alpha+\beta)}$$

     – a06e Dec 09 '13 at 17:43
  • I think there's something wrong again. Putting $n=2$ in your formula gives

    $$\frac{6\alpha\beta(\alpha^3 + \alpha^2(1-2\beta) + \beta^2(1+\beta) - 2\alpha\beta(2+\beta))}{(\alpha+\beta)^4(1+\alpha+\beta)^2(2+\alpha+\beta)(3+ \alpha +\beta)}$$

    and again this isn't equivalent to the variance in my previous comment.

     – a06e Dec 09 '13 at 20:15
  • It appears you have made a mistake in your calculations. It is working fine. Please see detailed examples above. – wolfies Dec 10 '13 at 10:29
  • I found my mistake... Sorry about that! I tried some more values of $n$ up to 20 and your formula gives the correct value (as computed using Cumulant in Mathematica). – a06e Dec 10 '13 at 13:55
  • Commercial software not attributed, again. I thought you were already told to behave? – Did Feb 16 '14 at 07:51
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    Stalking again, Didier? – wolfies Feb 16 '14 at 08:07