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I am trying to determine the limiting form of a beta distribution as its range expands under isoparametric constraints on its first two moments....

For reference

$$X_A \sim \operatorname{Beta}(0,A,\alpha,\beta) = \frac{1}{A} \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\left(\frac x A\right)^{\alpha-1} \left(\frac{A-x} A\right)^{\beta-1} \text{ for } x \in [0,A]$$

$$\operatorname{E}[X_A] = A\frac \alpha {\alpha+\beta}, \quad \operatorname{Var}(X_A) = A^2\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}$$

Problem Statement

Let $\mu,\sigma^2 \in (0,\infty)$ be fixed parameters.

Find $g(x) := \LARGE \{\normalsize\lim\limits_{A\rightarrow\infty} \operatorname{Beta}(x;0,A,\alpha,\beta):\int_0^A xg(x) \, dx = E[X_A] = \mu$ and $\int_0^{A} (x-\mu_A)^2g(x) \, dx = \operatorname{Var}[X_A] = \sigma^2\LARGE\}$

In words, we want the function that results from taking the limit of the beta distribution as its range approaches $\infty$ holding the mean and variance constant.

Attempts Thus Far

I tried an approach that quickly led to an algebraic mess, so I'm hoping the more mathematically experienced folks can provide additional, and perhaps more elegant, approaches.

Given the expected value equation, I solved for $\beta$ in terms of $\alpha, \mu$ and A to get: $\beta = \alpha \frac{A-\mu_A}{\mu}$

Plugging this into the variance formula, I got a big mess:

$$\frac{\sigma^2}{A^2} = \frac{\alpha^2(A-\mu)}{\mu \alpha^2 (1+\frac{A-\mu}{\mu})^2(\alpha+\alpha \frac{A-\mu}{\mu}+1)}$$

Next Steps

I need help doing two things:

1: Simplify, if possible, the formulas for $\alpha,\beta$ in terms of just $A, \mu, \sigma^2$.

2: Plug into beta equation and take the limit as $A \rightarrow \infty$. Hopefully step 1 will show that the form reduces to something, but I am at a loss right now.

Thanks for any hints, suggestions, links, or partial solutions that you can provide :)!

Numerical Study

Started with a beta with mean 50, variance 600 and right tail at 200 (short beta). Compared to the gamma with equivalent mean and variance as well as to a beta with same mean and variance but tail extending out to 4.29E+11 (long beta). Note that the long beta falls direclty on the gamma, with essentially zero numerical error!

Numerical results for Beta mean = 50, variance = 600 vs equivalent gamma

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  • $\begingroup$ AS $A$ gets larger, the only way to maintain the mean and variance is to have a thin right tail, creating a highly right-skewed distribution. A very limited experiment on my part indicated the parameters approaching zero. $\endgroup$
    – soakley
    Oct 30, 2013 at 14:03
  • $\begingroup$ I did a quick study as well. I found that while holding the mean and variance constant while the tail grows, alpha appears to approach $\frac{mean}{2}$ while beta grows without bound. $\endgroup$
    – user76844
    Oct 30, 2013 at 14:56
  • $\begingroup$ Ok - did a comparison study in excel. It definitely appears to converge to the gamma. See attached chart. $\endgroup$
    – user76844
    Oct 30, 2013 at 15:06
  • $\begingroup$ Ah, yes, after correcting a bug I am seeing the gamma connection you found. Very interesting! $\endgroup$
    – soakley
    Oct 30, 2013 at 16:50

1 Answer 1

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Fix the mean to $m$ and the variance to $v$. Since $A\to\infty$ while $E[X_A]=m$ stays constant, $\alpha\ll\beta$ hence $\beta\sim A\alpha/m$. Thus, $v\sim A^2\alpha/\beta^2$, that is, $\beta\sim A\sqrt{\alpha/v}$. Equating the two equivalents of $\beta$, one sees that $\alpha\to a$ with $a=m^2/v$ is mandatory. The density of $X_A$ is $$ f_A(x)=B(\alpha,\beta)A^{-\alpha}x^{\alpha-1}(1-x/A)^{\beta-1}\mathbf 1_{0\lt x\lt A}. $$ When $A\to\infty$ and $\beta\to\infty$ in the prescribed regime, for every fixed $x$, one sees that $\mathbf 1_{0\lt x\lt A}\to\mathbf 1_{x\gt 0}$, $B(\alpha,\beta)\sim\beta^\alpha/\Gamma(\alpha)$, and $(1-x/A)^{\beta-1}\sim\mathrm e^{-x\beta/A}\sim\mathrm e^{-xm/v}$, hence $$ f_A(x)\sim\beta^\alpha\Gamma(\alpha)^{-1}A^{-\alpha}x^{\alpha-1}\mathrm e^{-xm/v}\mathbf 1_{x\gt 0}, $$ thus, $f_A(x)\to f(x)$ where $$ f(x)=c^a\Gamma(a)^{-1}x^{a-1}\mathrm e^{-cx}\mathbf 1_{x\gt 0}, $$ with $c=m/v$. This is the gamma density with parameters $(a,c)=(m^2/v,m/v)$, or, equivalently, the gamma density with mean $m$ and variance $v$.

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    $\begingroup$ Excellent! Thanks for the derivation :-) That confirms my intuition that the Gamma was a limiting function for the Beta. This completes a triad of distributions: Beta $\rightarrow$ Gamma $\rightarrow$ Normal as you move from compact, to positive unbounded, to unbounded support. $\endgroup$
    – user76844
    Oct 31, 2013 at 12:30
  • $\begingroup$ BTW - is this result cited anywhere, or is it common knowledge among mathematicians? OR...did you derive this de novo? $\endgroup$
    – user76844
    Oct 31, 2013 at 12:33
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    $\begingroup$ Deriving it from scratch seemed simpler--but surely this is well known. $\endgroup$
    – Did
    Oct 31, 2013 at 12:57
  • $\begingroup$ Thanks Did. Much appreciated! $\endgroup$
    – user76844
    Oct 31, 2013 at 12:59

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