TITLE: An inequality involving the beta distribution
QUESTION [7 upvotes]: Let $F$ be the CDF for a Beta distribution, $$F(x)=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\int_{0}^{x}t^{a-1}(1-t)^{b-1}\,dt$$ with $a,b\geq 1$.  Is it true that $$\frac{b}{a+b} \leq\int_0^1\sqrt{F(x)}dx \leq \frac{2b}{a+b}\,?$$  Numerical simulations very strongly suggest that both inequalities hold, but I have no idea how to prove it.

REPLY [8 votes]: You understand that it may be a good time to quit when you start forgetting what used to be your favorite tricks.
Let $u=\frac ab>0$. Let $F_s$ be the CDF corresponding to $a'=u(1+s),b'=1+s$, so $F_s'(t)=c_s t^{u-1}[t^u(1-t)]^s$. Notice that $F_s(0)=0$, $F_s(1)=1$ and $\int_0^1 F_s=\frac{b}{a+b}$ for all $s$. Now, when $s>0$ (we are interested in $s=b-1$), we have $F_s<F_0$ slightly to the right of $0$ and $F_s>F_0$ slightly to the left of $1$ (the order of tangency is greater). Thus, the graphs of $F_s$ and $F_0$ intersect at least once on $(0,1)$.
If they intersect more than once on $(0,1)$, then, by Rolle, the difference of the derivatives $F_s'-F_0'$ has at least three zeroes on $(0,1)$ but that difference is $t^{u-1}\{c_s[t^u(1-t)]^s-c_0\}$, which, clearly, has at most two roots on $(0,1)$ (the function $t\mapsto t^u(1-t)$ is unimodal). Thus we have $F_s\le F_0$ on $[0,x_s]$, $F_s\ge F_0$ on $[x_s,1]$ and $F_s(x_s)=F_0(x_s)=Y$ for some $x_s\in(0,1)$.
Then, since both $F_s,F_0$ are increasing and $\Phi(v)=\sqrt v$ is concave, we get
$$
\Phi(F_s)-\Phi(F_0)\le \Phi'(Y)(F_s-F_0)
$$
on $[0,1]$ (consider $[0,x_s]$ and $[x_s,1]$ separately),
so, since $F_s$ and $F_0$ have the same integral, we get $\int_0^1\Phi(F_s)\le\int_0^1 \Phi(F_0)$.
Now it remains to evaluate $\int_0^1\sqrt{F_0}=\int_0^1 x^{u/2}\,dx=\frac{2}{u+2}=\frac{2b}{a+2b}$.