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I have modified the original question.


Does beta distribution function $$f(x,\alpha) = \frac{[x^a(1-x)^b]^\alpha}{B(a\alpha+1,b\alpha+1)}$$ where $B$ is the beta function, approach delta function $\delta(x-a)$ on $[0,1]$ as $\alpha\rightarrow\infty$, in distribution, for fixed positive $(a,b) \ni a+b=1$?

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    $\begingroup$ Let at least one of $\alpha, \beta \to \infty$. e.g. let $\alpha=\beta$ and let them both $\to \infty$. What happens to the variance? What happens to the distribution function? $\endgroup$
    – Glen_b
    Commented Apr 24, 2014 at 1:54
  • $\begingroup$ @Glen_b: You get a delta function centered at one end. $\endgroup$
    – Hans
    Commented Apr 24, 2014 at 2:00
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    $\begingroup$ If you hold one of $\alpha$ or $\beta$ constant while the other goes to $\infty$, then it will be centered at one end. In my e.g. there, $\alpha=\beta$, so the mean is at 0.5 (and it's symmetric). If they both go to infinity such that $\alpha/\beta=k$, then the delta function it approaches won't be at one end, but at a value that's an obvious function of $k$. $\endgroup$
    – Glen_b
    Commented Apr 24, 2014 at 2:05
  • $\begingroup$ @Glen_b: You edited your first comment after I have posted my reply. Your original comment was letting one of $\alpha, \beta \rightarrow\infty$ and hold the other constant. What you suggest in your second comment and the edited first comment is what I had in mind when posting my question. I did not say it explicitly so as not to interfere with possible other ideas from people. However, the question remains, does the limit function exist and if it does, what is it, a step function? $\endgroup$
    – Hans
    Commented Apr 24, 2014 at 2:17
  • $\begingroup$ actually my original comment said "at least one". If the density approaches a delta function, what's the cdf? $\endgroup$
    – Glen_b
    Commented Apr 24, 2014 at 3:49

1 Answer 1

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I come to the solution of my own question. The answer is in the affirmative. I prove it in two ways.

Proof 1:

The mean $E[x]=\frac{a\alpha+1}{\alpha+2}\rightarrow a$ and the variance $\text{var}[x] = \frac{(a\alpha+1)(b\alpha+1)}{(\alpha+2)^2(\alpha+3)}\rightarrow 0$, as $\alpha\rightarrow 0$.

By Chebyshev's inequality, together with boundedness of $f(x,\alpha)$, it can be shown $f(x,\alpha)$ uniformly approaches $0$ for $|x-a|>\epsilon$, for arbitrary fixed positive $\epsilon$.

Proof 2:

Use Stirling's approximation to write the asymptotics of Beta function $$B(u,v) \sim \sqrt{2\pi}\frac{u^{u-\frac{1}{2}}v^{v-\frac{1}{2}}}{(u+v)^{u+v-\frac{1}{2}}},$$ for large $u$ and $v$.

$$f(x,\alpha) \sim \sqrt{\frac{\alpha+2}{2\pi ab}}g(x)^\alpha$$ where $$g(x) := \Big(\frac{x}{a}\Big)^a\Big(\frac{1-x}{b}\Big)^b.$$ By the strict concavity of the natural logarithmic function, so long as $x\ne a$,
$$\ln g(x)<\ln\Big(a\frac{x}{a}+b\frac{1-x}{b}\Big)=0.$$


I will fill in the detail and finish the derivation later.

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    $\begingroup$ I would like to suggest the essence of the situation is conveyed by a generalization of the problem. Let $h:[0,1]\to\mathbb{R}^{+}$ be any bounded measurable non-negative function with a unique maximum at, say, $a$ and suppose $h$ is continuous at $a$. For $\lambda\gt 0$ define $h_\lambda(x)=C_\lambda h(x)^\lambda$ where $C_\lambda$ makes $\int_0^1 h_\lambda(x)dx=1$. If $f$ is any continuous function on $[0,1]$ and $\epsilon\gt 0$, then ${\lim}_{\lambda\to\infty}\int_{h(x)\le h(a)-\epsilon}h_\lambda(x)f(x)dx = 0.$ This readily implies the limit of $h_\lambda$ is $\delta_{a}$. $\endgroup$
    – whuber
    Commented Apr 24, 2014 at 20:39
  • $\begingroup$ @whuber: I like your generalization very much. I am having trouble though to argue $h(x)$ away from $a$ is either less than $1$ or $h(x)^\lambda$ is suppressed enough by $C_\lambda$ to vanish. Perhaps this is not the right approach. How would you prove it? $\endgroup$
    – Hans
    Commented Apr 24, 2014 at 21:10
  • $\begingroup$ Renormalize $h$ at the outset so that $h(a)=1+\epsilon.$ When $h(x)\lt 1$ $$\log(h(x)^\lambda)=\lambda\log(h(x))\lt \lambda\log(1)$$ is negative, showing $h(x)^\lambda\to 0$. A similar calculation shows $h(x)^\lambda(x)\to\infty$ when $h(x)\gt 1$. The integral therefore is concentrated within the region where $h(x)\gt 1.$ For sufficiently small $\epsilon$ this is a neighborhood of $a$. The normalized integral is bounded below by the minimum of $f$ in a slightly larger neighborhood. Now appeal to the continuity of $f$. $\endgroup$
    – whuber
    Commented Apr 24, 2014 at 21:23
  • $\begingroup$ @whuber: Clever renormalization trick for every given $\epsilon$. The difficult technical part for me now is to argue the measure of set of $h(x)>1$ away from an interval of $a$ approaches zero faster than the value $h(x)^\lambda$ of the set. Would you mind writing out the complete proof as an answer? I will accept yours as THE answer. Just a small nitpicking, there is no need to invoke the logarithm to prove $h(x)^\lambda\rightarrow 0$, for $h(x)<1$ as $\lambda\rightarrow\infty$, right? $\endgroup$
    – Hans
    Commented Apr 24, 2014 at 21:43
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    $\begingroup$ The generalized question is posed in math.stackexchange.com/q/1274804/64809 and is elegantly proved by math.stackexchange.com/a/1275075/64809. $\endgroup$
    – Hans
    Commented Dec 16, 2015 at 1:23

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