Limit of ratio of definite integrals, Conditional expectation of Beta distribution

Question

I'm trying to find the following limit:

$$\lim_{a,b \to \infty}\frac{\int_{\frac{a}{a+b}}^1 t^{a}(1-t)^{b-1} dt}{\int_{\frac{a}{a+b}}^1 t^{a-1}(1-t)^{b-1} dt}-\frac{a}{a+b}.$$

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I think the limit should exist, and should be $0$, but could not formally show that is the case.

And what if we have $a\rightarrow \infty, b>0$, or $b\rightarrow \infty, a>0$? Does the limit exist? Is it $0$?

Any suggestion helps.

Notice that the expression has the interpretation of $$\mathbb{E}\left[t-\frac{a}{a+b} \Big| a,b, \text{ and } t-\dfrac{a}{a+b}>0\right]$$ where $t|a,b\sim Beta(a,b)$. i.e., the conditional expectation of $t$ given $t$ is above the mean of the $Beta$ distribution.

May assume $a/(a+b)$ stays constant if necessary.

Answer 1

Concerning $$I=\dfrac{\int_{\frac{a}{a+b}}^1 t^{a}\,(1-t)^{b-1}\, dt}{\int_{\frac{a}{a+b}}^1 t^{a-1}\,(1-t)^{b-1}\, dt}=\frac{\text{num}}{\text{den}}$$ we have, after simplifications $$\frac{\text{num}}{ab}=\Gamma (a+2) \Gamma (b)-(a+1) \Gamma (a+b+1) B_{\frac{a}{a+b}}(a+1,b)$$ $$\frac{\text{den}}{a+b}=a (a+1) \Gamma (a+b)+\Gamma (a+2) \Gamma (b+1)+ $$ $$ \left(\frac{a^a b^b}{ (a+b)^{a+b}}-(a+b) B_{\frac{a}{a+b}}(a,b+1)\right)$$ Assuming $\frac a{a+b}=k$ that is to say $b=\frac{a}{k}-a$, I do not see what to do beside numerical experiment.

Trying for $a=10^4$ and a few $b$, some results $$\left( \begin{array}{cc} b & I-\frac a{a+b}\\ 1000 & 0.00217116 \\ 2000 & 0.00270260 \\ 3000 & 0.00293954 \\ 4000 & 0.00303978 \\ 5000 & 0.00306629 \\ 6000 & 0.00305040 \\ 7000 & 0.00300948 \\ 8000 & 0.00295378 \\ 9000 & 0.00288959 \\ 10000 & 0.00282091 \\ 11000 & 0.00275029 \\ 12000 & 0.00267938 \\ 13000 & 0.00260925 \\ 14000 & 0.00254059 \\ 15000 & 0.00247383 \\ 16000 & 0.00240921 \\ 17000 & 0.00234689 \\ 18000 & 0.00228690 \\ 19000 & 0.00222925 \\ 20000 & 0.00217391 \end{array} \right)$$

Answer 2

If $a/(a+b)=\tau\in(0,1)$ is fixed, then we let $a=\lambda\tau, b=\lambda(1-\tau)$ with $\lambda>0$ and ask for $$L=\lim_{\lambda\to\infty}\frac{\Phi(\lambda,\tau,0,-1)}{\Phi(\lambda,\tau,-1,-1)},\qquad\Phi(\lambda,\tau,\alpha,\beta)=\int_\tau^1 t^{\lambda\tau+\alpha}(1-t)^{\lambda(1-\tau)+\beta}\,dt.$$

The asymptotics of $\Phi$ as $\lambda\to\infty$ can be analysed using Laplace's method: the integral is of the form $I(\lambda)=\int_a^b f(t)e^{\lambda g(t)}dt$ with $g(t)$ having the maximum at $t=a$, $g'(a)=0$ and $g''(a)<0$; in this case, the method yields the asymptotics $I(\lambda)\asymp f(a)e^{\lambda g(a)}\sqrt{\pi/(-2g''(a)\lambda)}$ (notation: here $F(\lambda)\asymp G(\lambda)$ means $\lim\limits_{\lambda\to\infty}F(\lambda)/G(\lambda)=1$). In our case $$g(t)=\tau\log t+(1-\tau)\log(1-t),\\g'(\tau)=0,\qquad g''(\tau)=-\frac{1}{\tau(1-\tau)},\\\Phi(\lambda,\tau,\alpha,\beta)\asymp\tau^{\lambda\tau+\alpha+1/2}(1-\tau)^{\lambda\tau+\beta+1/2}\sqrt\frac{\pi}{2\lambda},$$ making $L=\tau$ obvious. The cases of fixed $a$ or $b$ could also be handled by (other variations of) this method, but there are also elementary means (integration by parts and $\int_\tau^1=\int_0^1-\int_0^\tau$).