# Proving Bernoulli is the limit of Beta

It's clear to me by inspection that if we fix $\beta = \frac{1-\mu}{\mu} \alpha$ (thereby fixing the mean) and let $\alpha \rightarrow 0$, the Beta distribution approaches a Bernoulli($\mu$) distribution.

For example:

par(mfrow = c(1, 2),
oma = c(0, 0, 1.5, 0))
xx = seq(0, 1, length.out = 1000)
mus = c(.2, .7)
for (ii in 1:2) {
mu = mus[ii]
matplot(xx, sapply(10^(-1:-5), function(al)
pbeta(xx, al, (1-mu)/mu * al)),
type = 'l', lty = ii,
main = paste('Mean:', mu),
ylab = 'Cumulative Probability', xlab = 'x')
}
title('Beta Approaches Bernoulli', outer = TRUE) That is, if $X \sim B(\alpha, \frac{1-\mu}{\mu} \alpha)$, its CDF $F_X(x; \alpha, \mu)$, satisfies

$$\lim_{\alpha \rightarrow 0} F_X(x; \alpha, \mu) = 1 - \mu = F_Y(x) \quad \forall x \in (0, 1)$$

Where $Y \sim$ Bernoulli($\mu$), and the convergence is non-uniform. I tried to do a more formal proof of this but couldn't make any progress, even in simple cases (like $\mu = \frac12$). The Wikipedia page on Beta makes reference to the $\mu = \frac12$ case several times without proof. Perhaps I'm missing something easy about doing the integral, or some fact about the Beta/incomplete Beta function. Any help would be appreciated!

• My answer at stats.stackexchange.com/a/237849/919 implies this, because it shows that the limit probabilities will be concentrated at $0$ and $1$. See the second pair of graphics there for an illustration. – whuber Mar 27 '17 at 18:48

## 1 Answer

The analysis can be a little messy, especially if carried out with full and elementary rigor, but the idea is simple and easily grasped. Focus on small regions very close to $0$ and $1$. As $\alpha$ and $\beta$ approach $0$, almost all the probability of a Beta$(\alpha,\beta)$ distribution becomes located within these regions. By shrinking the sizes of the regions, we see that the limiting distribution if one exists can only be a Bernoulli distribution. We can create a limiting distribution only by making the ratio $\alpha:\beta$ approach a constant, exactly as described in the question.

The nice thing about this analysis is that looking at relative areas obviates any need to consider the behavior of the normalizing constant, a Beta function $B(\alpha,\beta)$. This is a considerable simplification. (Its avoidance of the Beta function is similar in spirit to my analysis of Beta distribution quantiles at Do two quantiles of a beta distribution determine its parameters?)

A further feature of this analysis is approximating the incomplete Beta function by simple integrals of the form $\int t^c\mathrm{d}t$ for constants $c\gt -1$. This reduces everything to the most elementary operations of Calculus and algebraic inequalities.

The Beta PDF is proportional to $$f(x)=x^{\alpha-1}(1-x)^{\beta-1}.$$ Consider small $\epsilon\gt 0$ and examine the contributions to the area under $f$ within the three intervals $(0,\epsilon]$, $(\epsilon, 1-\epsilon)$, and $[1-\epsilon, 1)$ as $\alpha$ and $\beta$ grow small (but remain positive).

Eventually both $\alpha$ and $\beta$ will both be less than $1$: $f$ will therefore have poles at both $0$ and $1$, looking like this: The graph of $f$ is the upper blue line. Compared to it are the graphs of $x^{\alpha-1}$ (red curve, with a pole only at $0$) and $(1-x)^{\beta-1}$ (gold curve, with a pole only at $1$).

What happens to the three areas under $f$, relative to each other, in the limit?

As a matter of notation, write $$F(x) = \int_0^x f(t)\mathrm{d}t = \int_0^x t^{\alpha-1}(1-t)^{\beta-1}\mathrm{d}t$$for the area under the graph of $f$ between $0$ and $x$. I am asking about the relative sizes of $F(\epsilon)$, $F(1-\epsilon)-F(\epsilon)$, and $F(1)-F(1-\epsilon)$.

Let's estimate these areas one at a time, always assuming $0 \lt \alpha \lt 1$ and $0\lt \beta \lt 1,$ $0\lt x \lt 1,$ and $0\lt \epsilon \lt 1/2$. Under these assumptions

$$x^{\alpha-1} \gt 1;\quad(1-x)^{\beta-1}\gt 1,$$

$x\to x^{\alpha-1}$ (red) is a decreasing function in $x,$ and $x\to (1-x)^{\beta-1}$ (gold) is an increasing function.

1. On the left, it looks like the blue and red curves draw close. Indeed, for $0\lt x \lt \epsilon$, the foregoing inequalities yield the bounds $$x^{\alpha-1} \lt x^{\alpha-1}(1-x)^{\beta-1} \lt x^{\alpha-1}(1-\epsilon)^{\beta-1}.$$Integrating each between $0$ and $\epsilon$ is simple and squeezes $F(\epsilon)$ between two close bounds, $$\frac{\epsilon^\alpha}{\alpha} \lt F(\epsilon) \lt (1-\epsilon)^{\beta-1} \frac{\epsilon^\alpha}{\alpha}.\tag{1}$$

2. The same analysis applies to the right hand side, yielding a similar result.

3. Because $f$ is concave, on the middle interval $[\epsilon, 1-\epsilon]$ it attains its extreme values at the endpoints. Consequently the area is less than that of the trapezoid spanned by those points: \eqalign{ F(1-\epsilon) - F(\epsilon) &\lt \frac{1}{2}\left(f(\epsilon) + f(1-\epsilon)\right)(1-\epsilon - \epsilon)\\ &= \frac{1-2\epsilon}{2}\left(\epsilon^{\alpha-1}(1-\epsilon)^{\beta-1} + (1-\epsilon)^{\alpha-1}\epsilon^{\beta-1}\right)).\tag{2} }

Although this threatens to get messy, let's temporarily fix $\epsilon$ and consider what happens to the ratio $(F(1-\epsilon)-F(\epsilon)):F(\epsilon)$ as $\alpha$ and $\beta$ approach $0$. In expressions $(1)$ and $(2)$, both $(1-\epsilon)^{\alpha-1}$ and $(1-\epsilon)^{\beta-1}$ will approach $(1-\epsilon)^0=1$. Thus, the only terms that matter in the limit are

$$\frac{F(1-\epsilon)-F(\epsilon)}{F(\epsilon)} \approx \frac{(\epsilon^{\alpha-1} + \epsilon^{\beta-1})/2}{\epsilon^\alpha / \alpha} = \frac{\alpha}{2\epsilon} + \frac{\alpha}{2\epsilon^{\alpha-\beta}} \approx \frac{\alpha}{\epsilon}\tag{3}$$

because $\alpha-\beta \approx 0$. Consequently, since $\alpha\to 0$, eventually the middle area is inconsequential compared to the left area.

The same argument shows that eventually the middle area is close to $\beta/\epsilon$ times the right area, which also becomes inconsequential. This shows that

$(*)$ No matter what $0\lt \epsilon\lt 1/2$ may be, if we take both $\alpha$ and $\beta$ to be sufficiently small, then essentially all the area under $f$ is concentrated within the left interval $(0,\epsilon)$ and the right interval $(1-\epsilon, 1)$.

The rest is easy: the mean will be very close to the area near the right pole (proof: underestimate it by replacing $xf(x)$ by $0f(x)$ in the integrals over the left and middle intervals and by $(1-\epsilon)f(x)$ in the right interval, then overestimate it by replacing $xf(x)$ by $\epsilon f(x)$ at the left, $(1-\epsilon)f(x)$ in the middle, and $f(x)$ at the right. Both expressions closely approximate $F(1)-F(1-\epsilon)$.) But, by $(3),$ the relative areas are approximately

$$\frac{F(1)-F(1-\epsilon)}{F(\epsilon)} \approx \frac{\epsilon/\beta}{\epsilon/\alpha} = \frac{\alpha}{\beta}.$$

By keeping the mean constant, this ratio remains constant, allowing us to add one more observation to $(*)$:

$(**)$ If we let $\alpha\to 0$ and $\beta\to 0$ in such a way that $\alpha/\beta$ approaches a limiting constant $\lambda$, then eventually the ratio of the area at the right to the area at the left will be arbitrarily close to $\lambda$, too.

Now contemplate $\epsilon$ shrinking to zero. The result is that the limiting distribution exists and it must have all its probability concentrated around the values $0$ and $1$: this is the class of Bernoulli distributions. $(**)$ pins down which one: since the Bernoulli$(p)$ distribution, whose mean is $p,$ assigns probability $p$ to $1$ and probability $1-p$ to $0$, the ratio $p/(1-p)$ must be the limiting ratio $\lambda.$

In the terminology of the question, $$\lambda = \alpha / \left(\frac{1-\mu}{\mu}\alpha\right) = \frac{\mu}{1-\mu} = \frac{p}{1-p},$$

as claimed.