Bounds for the beta CDF

Question

This question is closely related to a previous question that I asked here: An inequality involving the beta distribution

Let $a,b$ be strictly positive integers, and let $F_{a,b}(x)$ denote the CDF for a Beta distribution with parameters $a$ and $b$: $$F_{a,b}(x) = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\int_0^x \frac{ t^{a-1} (1-t)^{b-1} }{B(a,b)} dt $$ I am trying to understand the behavior of the function $$g(x)=\int_0^x\sqrt{F_{a,b}(t)} ~dt$$ Since there is not an algebraic expression for the integrand, I am wondering if there are any existing bounds for $F_{a,b}(\cdot)$ that would help me bound $g(x)$, for all $a$, $b$, and $x$? In my previous question, it was answered that $$\frac{b}{a+b}\leq g(1)\leq\frac{2b}{a+b}$$
As one example, an easy observation is that if we fix $b=pa$ for some constant $p$ and let $a$ become large, then $F_{a,b}$ looks like a step function that jumps from $0$ to $1$ at the point $t = a/(a+b)$, and my integral inherits this same behavior.

Answer 1

If $\min(a,b)\ge100$, then the mean of the beta distribution will be $$m=\frac a{a+b},$$ and for its standard deviation we will have
\begin{equation} s\approx\sqrt{\frac{ab}{(a+b)^3}}=m\sqrt{\frac b{a(a+b)}}<m\sqrt{\frac1a}\le\frac m{10} \tag{0} \end{equation} and similarly $$s\approx\sqrt{\frac{ab}{(a+b)^3}}=(1-m)\sqrt{\frac a{b(a+b)}}<(1-m)\sqrt{\frac1b}\le\frac{1-m}{10}.$$ So, the beta distribution will be approximated well by $N(m,s^2)$, with $s\lesssim\min(\frac m{10},\frac{1-m}{10})<\frac1{20}$, which latter is much less than $1$. So, for $x\in(0,1)$ and with $\Phi$ denoting the standard normal cdf, the integral in question, $g(x)$, will be approximated well by \begin{equation} \int_0^x\sqrt{\Phi\Big(\frac{t-m}s\Big)}\,dt =s \int_{-m/s}^{(x-m)/s}\sqrt{\Phi(z)}\,dz \approx s\psi\Big(\frac{x-m}s\Big), \tag{1} \end{equation} where \begin{equation} \psi(u):=\int_{-\infty}^u\sqrt{\Phi(z)}\,dz. \end{equation} The approximate equality in (1) holds because, by (0), $m/s$ is much greater than $1$.

So, to approximate $g(x)$, you only need to use the values of one function, $\psi$, of one real variable.

For the error $E(x):=g(x)-s\psi\big(\frac{x-m}s\big)$ of this approximation, here is the graph $\{(t,E(t))\colon0<t<1\}$ for $(a,b)=(100,500)$: