Asymptotic behavior of the CDF of a $\operatorname{Beta}(n,n)$ at $x<1/2$.

Question

I know that the if $X_n \sim \operatorname{Beta}(n,n)$ (see here for a definition) then $\mathbb{P}_{X_n} \to \delta_{1/2}, n \to \infty$ weakly.

If $\varepsilon \in (0,1/2)$, I'm wondering about how fast $\mathbb{P}_{X_n}\big([0, \frac{1}{2}-\varepsilon]\big) \to 0$ as $n \to \infty$. As a power of $1/n$? Exponentially? I didn't find anything on wiki or elsewhere. Can someone point to a reference or give an answer?

Answer 1

Let $Y_1,\dots,Y_{2n-1}$ i.i.d. Bernoulli random variables of parameter $p = \frac{1}{2}+\varepsilon$.

Let $m \in \mathbb{N}$ and let $Z_1,\dots,Z_m$ be i.i.d. random variables uniformly distribuited on $[0,1]$. Let $Z^{(1)}_m,\dots, Z^{(m)}_m$ the order statistics of $Z_1,...,Z_m$. Then $$\forall j \in\{1,\dots,m\}, Z^{(j)}_m \sim \operatorname{Beta}(j,m+1-j).$$

So, if $j=n$ and $m=2n-1$ then \begin{align*} \mathbb{P}\bigg[X_n \le \frac{1}{2}-\varepsilon\bigg] &= \mathbb{P}\bigg[Z^{(n)}_{2n-1} \le \frac{1}{2}-\varepsilon\bigg] \\ &= \mathbb{P}\bigg[\#\bigg\{Z \in \big\{Z_1,\dots,Z_{2n-1} \big\} : Z\ge \frac{1}{2}-\varepsilon \bigg\} \le (2n-1)-n\bigg] \\ &= \mathbb{P}\bigg[\#\bigg\{Z \in \big\{Z_1,\dots,Z_{2n-1} \big\} : Z\ge \frac{1}{2}-\varepsilon \bigg\} \le n-1\bigg] \\ &= \mathbb{P}\bigg[\sum_{j=1}^{2n-1} Y_n\le n-1\bigg] \\ &= \mathbb{P}\bigg[\bigg(\frac{1}{2n-1}\sum_{j=1}^{2n-1} Y_n\bigg) - p\le \frac{n-1}{2n-1} -p\bigg] \\ &= \mathbb{P}\bigg[\bigg(\frac{1}{2n-1}\sum_{j=1}^{2n-1} Y_n\bigg) - p\le -\frac{1}{2(2n+1)} - \varepsilon\bigg] \\ &\le \exp\bigg(-2(2n-1)\bigg(\frac{1}{2(2n+1)}+\varepsilon\bigg)^2\bigg) \\ &< \exp(-2(2n-1)\varepsilon^2), \end{align*} where we have used Hoeffding's inequality.

It follows that $\mathbb{P}_{X_n}\big([0, \frac{1}{2}-\varepsilon]\big)$ decreases exponentially fast in $n$ as $n \to \infty$.

Answer 2

This answer is incomplete, but perhaps it may be useful. It sounds like you want the asymptotic behaviour of $P\left(X_n < \frac{1}{2}-\varepsilon\right)$, for some fixed $\varepsilon>0$, and $X_n\sim Beta\left(n,n\right)$? This is what is known as a "large deviations" result. For many distributions, such probabilities decay exponentially.

A standard approach is through the Gartner-Ellis theorem. Letting $\left\{Y_n\right\}$ be a sequence of random variables, not necessarily i.i.d., you let $M_n\left(t\right) = \log\mathbb{E}\left(e^{tY_n}\right)$ be the log-mgf of $Y_n$, and apply the scaling

$$M\left(t\right) = \lim_{n\rightarrow\infty} \frac{1}{n} M_n\left(nt\right).$$

If this limit exists, then the convex conjugate

$$R\left(s\right) = \sup_t s t - M\left(t\right)$$

produces results of the form

$$\lim_{n\rightarrow\infty} \frac{1}{n}\log P\left(Y_n \in E\right) = -\inf_{s\in E} R\left(s\right),$$

so, roughly speaking, $P\left(Y_n\in E\right)$ decays exponentially for some suitably chosen set $E$. In this case you would have $E = \left(0,\frac{1}{2}-\varepsilon\right)$. Unfortunately, these computations are not easy with the beta mgf, so I am not sure if this approach is tractable.