1
$\begingroup$

In the article https://en.wikipedia.org/wiki/Beta_distribution#Symmetric_(α_=_β) it is said that a Beta distribution with parameters $\alpha = \beta \to 0$ has a Bernoulli distribution with probability $p=0.5$ at $0$ and $1$.

Formally, does this mean that the sequence $\text{Beta}(1/n, 1/n)$ with $n \in \mathbb{N}$ converge to $\text{Bernoulli}(0.5)$ and what kind of convergence it is (distribution, a. e.)?

How to prove this fact?

Beta density function when $\alpha = \beta$ is:

$$f_X(x)=\frac{x^{\alpha -1} (1-x)^{\beta -1}}{B(\alpha, \beta)} = \frac{\Gamma(2\alpha)\;x^{\alpha -1} (1-x)^{\alpha -1}}{\Gamma(\alpha)^2}$$

then I need to find

$$\lim_{\alpha^+ \to 0}\frac{\Gamma(2\alpha)\;x^{\alpha -1} (1-x)^{\alpha -1}}{\Gamma(\alpha)^2}$$

where $\lim_{\alpha^+ \to 0}$ is the limit from the right since the parameters for the Beta distribution are real positive numbers. We know that $\lim_{\alpha^+ \to 0} \Gamma(\alpha) = +\infty$ and for every $x \in \mathbb{R}$ $x^{-1}$ and $(1-x)^{-1}$ is another real number, then we have a limit of the form $\infty / \infty$ and we can use L'Hopital, but I dont know how to use it with the $\Gamma$ function and how to conclude that the new function is a density function of Bernoulli distribution.

$\endgroup$
  • $\begingroup$ Try checking the definition of the derivative of the gamma function when applying L'hopital.. $\endgroup$ – venrey Sep 5 '18 at 3:51
  • $\begingroup$ Alternatively, you can appeal to the good old 'approximation-to-the-identity' argument. Notice that the measure concentrates near both $0$ and $1$ as $\alpha\to0$. $\endgroup$ – Sangchul Lee Sep 5 '18 at 4:45
  • 2
    $\begingroup$ Yes the convergence in distribution holds but showing that it does by studying the pointwise limit of the PDF is hopeless since the limit distribution has no PDF. Instead, one could show the pointwise convergence of the CDF (very tedious) or, better, that, for every $x\in(0,1)$, $$P(\tfrac12(1-x)\leqslant X_\alpha\leqslant\tfrac12(1+x))\to0$$ while $$P(X_\alpha\leqslant\tfrac12(1-x))\to\tfrac12\qquad P(X_\alpha\geqslant\tfrac12(1+x))\to\tfrac12$$ $\endgroup$ – Did Sep 5 '18 at 5:56
1
$\begingroup$

I find a theorem that I think can answer all the questions

Theorem: If $X$ is a random variable determined by its moments and $\{ X_n \}_{n \in \mathbb{N}}$ is a sequence of random variables such that $E[X_n^k] \to E[X^k]$ then $X_n \to X$ in distribution.

Bernoulli distribution is determined by its moments (we can verify with Carleman's condition). The moments of a Bernoulli distribution $X$ with parameter $p$ are $E(X^k)=p$ and the moments for a $Y \sim \text{Beta}(\alpha, \beta)$ are:

$$ E(Y^k) = \frac{\alpha(\alpha+1) \cdots (\alpha+k-1)}{(\alpha+\beta)(\alpha+\beta+1) \cdots (\alpha+\beta+k-1) }$$

then if $\alpha = \beta = 1/n \to 0$, $n \in \mathbb{N}$, and $X_n = \text{Beta}(1/n,1/n)$ we have:

$$ lim_{n \to \infty}E(X_n^k)=\lim_{a^+ \to 0}\frac{\alpha(\alpha+1) \cdots (\alpha+k-1)}{(2\alpha)(2\alpha+1) \cdots (2\alpha+k-1) } = \lim_{a^+ \to 0}\frac{(\alpha+1) \cdots (\alpha+k-1)}{(2)(2\alpha+1) \cdots (2\alpha+k-1) } $$ $$=\frac{(k-1)!}{2(k-1)! }= \frac{1}{2} $$

By the theorem, we have $\{X_n\}$ converges to Bernoulli with parameter $p=1/2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.