Asymptotics of incomplete Beta function $B_{1/2}(y+1,y)$ when $y\to\infty$

Question

My question concerns the behavior of the incomplete Beta function

$$B_{1/2}(y+1,y)=\int_0^{1/2}x^y (1-x)^{y-1}dx$$

in the large $y$ limit. I have been looking everywhere, but I can't find anything. I was thinking that I can get a good approximation by letting $(1-x)^{y-1}\approx x^y$, thus yielding a solution of the form $\approx 1/(2^{2y+1})$. However, is there any reference that talks about the above function in the large argument limit?

Answer 1

This is straightforward steepest descent (or Laplace's method). Let $f(x):=\ln(x)+\ln(1-x),$ and $g(x)=(1-x)^{-1}$ so that the integral becomes:

$$\int_0^{1/2}g(x)e^{yf(x)}dx.$$

On $[0,1/2]$, $f$ achieves its maximum at $x_0=1/2$. Furthermore, $f''(x)=-1/x^2-1/(1-x)^2$, and $f''(x_0)<0$.

Then Laplace's method gives:

$$\int_0^{1/2}g(x)e^{yf(x)}dx\sim \sqrt{\frac{2\pi}{y\,|f''(x_0)|}}\,g(x_0)\,e^{yf(x_0)}$$ in the sense that the ratio of the LHS and RHS converges to $1$ when $y\to+\infty$.

Can you wrap it up from here? Note the extra $1/\sqrt{y}$ that's going to contribute.

Answer 2

I'll play around and see if anything interesting happens.

$\begin{array}\\ B_{1/2}(y+1,y) &=\int_0^{1/2}x^y (1-x)^{y-1}dx\\ &\ge\int_0^{1/2}x^y (\frac12)^{y-1}dx\\ &=\dfrac1{2^{y-1}}\int_0^{1/2}x^ydx\\ &=\dfrac1{2^{y-1}}\dfrac{x^{y+1}}{y+1}\big|_0^{1/2}\\ &=\dfrac1{2^{2y}(y+1)}\\ \end{array} $

and

$\begin{array}\\ B_{1/2}(y+1,y) &=\int_0^{1/2}x^y (1-x)^{y-1}dx\\ &\le\int_0^{1/2}(\frac12)^y (1-x)^{y-1}dx\\ &=\dfrac1{2^y}\int_0^{1/2} (1-x)^{y-1}dx\\ &=\dfrac1{2^y}\int_{1/2}^1 x^{y-1}dx\\ &=\dfrac1{2^y}\dfrac{x^y}{y}\big|_{1/2}^1\\ &=\dfrac1{2^y}\dfrac{1-(1/2)^y}{y}\\ \end{array} $

I'll leave it at these bounds. Don't know which is best.

This has an obvious generalization to $B_{z}(y+1,y) =\int_0^{z}x^y (1-x)^{y-1}dx $.

Answer 3

This is a very empirical approach of the problem since based on numerical simulations.

Computing the values of $B_{1/2}(y+1,y)$ for $0\leq y\leq 5000$, what we can notice is that the logarithm varies as a linear function of $y$. For this range, I obtained $$\log\left(B_{1/2}(y+1,y)\right)\approx -1.38659 y-3.1602$$ which seems to be a very good fit (as shown below) $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & -3.16020 & 6.70 \times 10^{-3} & \{-3.17333,-3.14706\} \\ b & -1.38659 & 2.32 \times 10^{-6} & \{-1.38659,-1.38658\} \\ \end{array}$$ Extrapolated to $n=10000$, this simple model leads, for the logarithm, to a value equal to $-13869.0$ while the exact value should be $\approx -13867.7$ which does not seems to be too bad.

You could notice that $$\log\left(\frac 1 {2^{2y+1}}\right)\approx -1.38629 y-0.693147$$ showing almost the same slope as in the regression (and this is almost $-2\log(2)$).

Finally, considering $$F(k)=\int_0^{\infty } \left(B_{\frac{1}{2}}(y+1,y)-\frac k {2^{2y+1}} \right){}^2 \, dy$$ the minimum of the function is obtained for $k\approx 1.21973$.