A note on the asymptotics for incomplete Betafunctions

Abstract

We determine the asymptotic behaviour of certain incomplete Betafunctions.

1 Introdution and results

For integers $k,\ell \ge 1$ define

$$\begin{aligned} P_{k,\ell } = \frac{(k+\ell )!}{(k-1)!\ell !}\int _{k/(k+\ell )}^1 t^{k-1}(1-t)^\ell \;dt. \end{aligned}$$

By repeated partial integration we obtain the representation

$$\begin{aligned} P_{k, \ell } = \left( \frac{\ell }{\ell +k}\right) ^{k+\ell }\sum _{\nu =0}^{k-1} \left( {\begin{array}{c}k+\ell \\ \nu \end{array}}\right) \left( \frac{k}{\ell }\right) ^\nu . \end{aligned}$$

Vietoris[5] showed that $P_{k,\ell }\le \frac{1}{2}$. Alzer and Kwong [1] proved that $P_{k,\ell }\ge \frac{1}{4}$ holds for all $k, \ell$. Interest in bounds for $P_{k,\ell }$ stems from application to statistics, see [4].

In this note we consider the asymptotic behaviour of $P_{k,\ell }$. We prove the following.

Theorem 1

We have

$$\begin{aligned} P_{k, l} = e^{-k}\sum _{\nu =0}^{k-1}\frac{k^\nu }{\nu !} + \mathcal {O}\left( \frac{k^2}{\ell }\right) , \end{aligned}$$

(1)

$$\begin{aligned} P_{k,\ell } = 1-e^{-\ell }\sum _{\nu =0}^{\ell }\frac{\ell ^\nu }{\nu !} + \mathcal {O}\left( \frac{\ell ^2}{k}\right) \end{aligned}$$

(2)

and

$$\begin{aligned} P_{k, \ell } = \frac{1}{2}+\mathcal {O}\left( \frac{1}{\sqrt{\min (k, \ell )}}\right) . \end{aligned}$$

(3)

Note that the first two estimates are good if one of the parameters $k, \ell$ is rather small, whereas the third one gives information in the general case.

Comparing (1) and (2) with Vietoris’ result that $P_{k,\ell }\le \frac{1}{2}$ we see that

$$\begin{aligned} \sum _{\nu =0}^{\ell -1}\frac{\ell ^\nu }{\nu !}<\frac{1}{2}e^\ell < \sum _{\nu =0}^{\ell }\frac{\ell ^\nu }{\nu !}. \end{aligned}$$

Note that equality is impossible as e is transcendental. A more precise version of this inequality has been asked by Ramanujan (Question 294) and was answered by Karamata [2]. For a detailed discussion of this result, Uhlmann’s inequalities [4] and Vietoris bound we refer the reader to the historical notes by Vietoris [6].

From Theorem 1 we deduce the following.

Corollary 1

The only accumulation points of the set $\{P_{k,\ell }:k, \ell \in \mathbb {N}\}$ are $\frac{1}{2}$ and the real numbers $e^{-k}\sum _{\nu =1}^{k-1}\frac{k^\nu }{\nu !}$ and $1-e^{-\ell }\sum _{\nu =1}^{\ell }\frac{\ell ^\nu }{\nu !}$.

Proof

Suppose that $(k_n), (\ell _n)$ are integer sequences such that the pairs $(k_n, \ell _n)$ are all distinct and that the sequence $(P_{k_n, \ell _n})$ converges. If both $k_n, \ell _n$ tend to infinity, then by (3) we have that $P_{k_n, \ell _n}$ tends to $\frac{1}{2}$. If one of these sequences does not tend to infinity, then we can pass to a subsequence and assume that $k_n$ or $\ell _n$ is constant. Then our claim follows from (1) or (2). $\square$

Together with Vietoris bound we obtain the following.

Corollary 2

We have

$$\begin{aligned} \lim _{\ell \rightarrow \infty } P_{k, \ell } \ge \frac{1}{2} - \frac{1}{\sqrt{2\pi k}}. \end{aligned}$$

Proof

We have

$$\begin{aligned} \lim _{\ell \rightarrow \infty }\left( P_{k, \ell }+P_{\ell , k}\right) = 1-\frac{k^k}{e^k k!}, \end{aligned}$$

using Stirling’s formula in the form $k!=\left( \frac{k}{e}\right) ^k\sqrt{2\pi k}e^{\theta /12 k}$ with $0\le \theta \le 1$, and $P_{\ell , k}\le \frac{1}{2}$, we conclude

$$\begin{aligned} \lim _{\ell \rightarrow \infty } P_{k, \ell } \ge \frac{1}{2}-\frac{k^k}{e^k k!}\ge \frac{1}{2}-\frac{1}{\sqrt{2\pi k}}. \end{aligned}$$

$\square$

2 Preliminary estimates

For the proof of (3) we use another representation of $P_{k,\ell }$, which is due to Raab [3].

Theorem 2

We have $P_{k,\ell }=U_{k,\ell } V_{k,\ell }$, where

$$\begin{aligned} U_{k,\ell } = \exp \int _0^\infty \frac{1}{t}\left( \frac{1}{e^t-1}-\frac{1}{t}+\frac{1}{2}\right) \left( e^{-(k+\ell )t}-e^{-kt}-e^{-\ell t}\right) \;dt, \end{aligned}$$

$$\begin{aligned} V_{k, \ell } = \frac{\sqrt{\ell }}{2\pi }\sum _{\nu =1}^\infty \frac{c_\nu (k/\ell )}{\sqrt{\nu }(\nu +\ell )}, \end{aligned}$$

and

$$\begin{aligned} c_\nu (x) = \exp \int _0^\infty \frac{1}{t}\left( \frac{1}{e^t-1}-\frac{1}{t}+\frac{1}{2}\right) \left( e^{-\nu (1+x)t}-e^{-\nu xt}-e^{-\nu t}\right) \;dt. \end{aligned}$$

We first compute the occurring integrals.

Lemma 1

We have for $x\ge 1$

$$\begin{aligned} \int _0^\infty \frac{1}{t}\left( \frac{1}{e^t-1}-\frac{1}{t}+\frac{1}{2}\right) e^{-xt}\;dt = \frac{1}{12x} + \mathcal {O}\left( \frac{1}{x^2}\right) \end{aligned}$$

and

$$\begin{aligned} \int _0^\infty \frac{1}{t}\left( \frac{1}{e^t-1}-\frac{1}{t}+\frac{1}{2}\right) e^{-xt}\;dt = \frac{1}{2}\log x+ C + \mathcal {O}(x) \end{aligned}$$

for $0<x\le 1$, where C is some constant.

Proof

The series expansion of $e^x$ yields for $t\rightarrow 0$

$$\begin{aligned} \frac{1}{t}\left( \frac{1}{e^t-1}-\frac{1}{t}+\frac{1}{2}\right) = \frac{1}{t^2}\left( \frac{1}{1+t/2+t^2/6+\mathcal {O}(t^3)}-1+\frac{t}{2}\right) = 1/12 + \mathcal {O}(t). \end{aligned}$$

For $t\rightarrow \infty$ this expression tends to 0, in particular, it is bounded for all positive t. Hence for $x\rightarrow \infty$ the integral in question becomes

$$\begin{aligned} \frac{1}{12}\int _0^\infty e^{-xt}\;dt + \mathcal {O}\left( \int _0^\infty te^{-xt}\;dt\right) = \frac{1}{12 x} + \mathcal {O}\left( \frac{1}{x^2}\right) . \end{aligned}$$

For $x\rightarrow 0$ we use $e^{-xt}=1-xt + \mathcal {O}(x^2t^2)$ and obtain

$$\begin{aligned} \int _0^1 \frac{1}{t}\left( \frac{1}{e^t-1}-\frac{1}{t}+\frac{1}{2}\right) e^{-xt}\;dt = \int _0^1 \frac{1}{t}\left( \frac{1}{e^t-1}-\frac{1}{t}+\frac{1}{2}\right) \;dt + \mathcal {O}(x) =C_1+\mathcal {O}(x) \end{aligned}$$

and

$$\begin{aligned} \int _1^\infty \frac{1}{t(e^t-1)} e^{-xt}\;dt = \int _1^\infty \frac{1}{t(e^t-1)} \;dt -x \int _1^\infty \frac{1}{e^t-1} \;dt + \mathcal {O}(x^2). \end{aligned}$$

As a function of x, the integral $\int _1^{\infty } \frac{e^{-xt}}{t^2}\;dt$ defines a function that is differentiable from the right in 0 and has bounded second derivative in $(0,\infty )$, hence, for $x\ge 0$ this integral is $1+\mathcal {O}(x)$.

Finally

$$\begin{aligned} \int _1^\infty \frac{e^{-xt}}{2t}\;dt&= \int _x^\infty \frac{e^{-s}}{2s}\;ds =\\ \int _x^1\frac{dt}{2t}&+ \int _0^1\frac{e^{-t}-1}{2t}\;dt +\int _1^\infty \frac{e^{-t}}{2t}\;dt+\int _0^x\frac{1-e^{-t}}{2t}\;dt\\&= -\frac{1}{2}\log x + C_3 + \mathcal {O}(x). \end{aligned}$$

Combining these estimates our claim follows. $\square$

From this we obtain

Lemma 2

We have

$$\begin{aligned} U_{k, \ell } =1-\frac{1}{12}\left( \frac{1}{k}+\frac{1}{\ell }-\frac{1}{k+\ell }\right) +\mathcal {O}\left( \frac{1}{\min (k, \ell )^2}\right) . \end{aligned}$$

Proof

From Lemma 1 we obtain

$$\begin{aligned} U_{k, \ell } = \exp \left( \frac{1}{12}\left( \frac{1}{k+\ell }-\frac{1}{k}-\frac{1}{\ell }\right) +\mathcal {O}\left( \frac{1}{\min (k, \ell )^2}\right) \right) , \end{aligned}$$

inserting the Taylor series for $\exp$ and using the fact that $k, \ell \ge 1$ our claim follows. $\square$

Next we compute $c_\nu (x)$.

Lemma 3

If $\nu x\ge 1$, then

$$\begin{aligned} c_\nu (x)=1+\frac{1}{12\nu (1+x)}-\frac{1}{12\nu x}+\frac{1}{12\nu } + \mathcal {O}\left( \frac{1}{\nu ^2 \min (1,x^2)}\right) . \end{aligned}$$

If $\nu x\le 1$, then

$$\begin{aligned} c_\nu (x) = K\sqrt{\nu x} + \mathcal {O}\left( \sqrt{\frac{x}{\nu }}+(\nu x)^{3/2}\right) \end{aligned}$$

for some constant K.

Proof

If $\nu x>1$, then we apply Lemma 1 to obtain

$$\begin{aligned} c_\nu (x)= & {} \exp \left( \frac{1}{12\nu (x+1)}-\frac{1}{12\nu x}+\frac{1}{12\nu } +\mathcal {O}\left( \frac{1}{\nu ^2\min (1, x^2)}\right) \right) \\= & {} 1 + \frac{1}{12\nu (x+1)}-\frac{1}{12\nu x}+\frac{1}{12\nu }+\mathcal {O} \left( \frac{1}{\nu ^2\min (1, x^2)}\right) . \end{aligned}$$

If $\nu x<1$, then $\nu (x+1)\ge 1$, and we obtain

$$\begin{aligned} c_\nu (x)= & {} \exp \left( \frac{1}{12\nu (x+1)}+\frac{1}{2}\log (\nu x)+C +\frac{1}{12\nu }+\mathcal {O}\left( \frac{1}{\nu ^2}+\nu x\right) \right) \\= & {} K\sqrt{\nu x} + \mathcal {O}\left( \sqrt{\frac{x}{\nu }}+(\nu x)^{3/2}\right) . \end{aligned}$$

$\square$

We now compute $V_{k,\ell }$.

Lemma 4

We have

$$\begin{aligned} V_{k,\ell } = \frac{1}{2} + \mathcal {O}\left( \frac{1}{\sqrt{k}}+\frac{1}{\sqrt{\ell }}\right) \end{aligned}$$

Proof

We have

$$\begin{aligned} \sum _{\nu \le \ell /k}\frac{c_\nu (k/\ell )}{\sqrt{\nu }(\nu +\ell )}\ll \sum _{\nu \le \ell /k}\frac{\sqrt{\nu k/\ell }}{\sqrt{\nu }(\nu +\ell )}\ll \frac{1}{\sqrt{\ell k}} \end{aligned}$$

thus,

$$\begin{aligned} V_{k,\ell }= & {} \frac{\sqrt{\ell }}{2\pi }\sum _{\nu>\ell /k}\frac{1}{\sqrt{\nu } (\nu +\ell )}\left( 1+\frac{1}{12\nu \frac{k+\ell }{\ell }} - \frac{1}{12\nu \frac{k}{\ell }} + \frac{1}{12\nu }+\mathcal {O}\left( \frac{1}{\nu ^2 \min (1, k^2/\ell ^2)}\right) \right) \\&+\, \mathcal {O}\left( \frac{1}{\sqrt{k}}\right) \\= & {} \frac{\sqrt{\ell }}{2\pi }\sum _{\nu>\ell /k}\frac{1}{\sqrt{\nu } (\nu +\ell )}\left( 1 - \frac{1}{12\nu \frac{k}{\ell }}+\mathcal {O} \left( \frac{1}{\nu ^2\min (1, k^2/\ell ^2)}\right) \right) \\&+\, \mathcal {O} \left( \frac{1}{\sqrt{k}}+\sum _{\nu>\ell /k}\frac{1}{\nu ^{3/2}\sqrt{\ell }}\right) \\= & {} \frac{\sqrt{\ell }}{2\pi }\sum _{\nu >\ell /k}\frac{1}{\sqrt{\nu }(\nu +\ell )} \left( 1 - \frac{1}{12\nu \frac{k}{\ell }}+\mathcal {O}\left( \frac{1}{\nu ^2 \min (1, k^2/\ell ^2)}\right) \right) \\&+\, \mathcal {O}\left( \frac{1}{\sqrt{k}} +\frac{1}{\sqrt{\ell }}\right) \end{aligned}$$

If $k>\ell$ we obtain

$$\begin{aligned} V_{k,\ell }= & {} \frac{\sqrt{\ell }}{2\pi }\sum _{\nu =1}^\infty \frac{1}{\sqrt{\nu }(\nu +\ell )} + \mathcal {O}\left( \frac{1}{\sqrt{k}} +\frac{1}{\sqrt{\ell }}+\sqrt{\ell }\sum _{\nu =1}^\infty \frac{1}{\nu ^{3/2}(\nu +\ell )}\right) \\= & {} \frac{1}{2\pi }\int _0^\infty \frac{dt}{\sqrt{t}(t+1)} + \mathcal {O} \left( \frac{1}{\sqrt{k}}+\frac{1}{\sqrt{\ell }}\right) \\= & {} \frac{1}{2} + \mathcal {O}\left( \frac{1}{\sqrt{\ell }}\right) . \end{aligned}$$

We have

$$\begin{aligned} \frac{\sqrt{\ell }}{2\pi }&\sum _{\nu>\ell }\frac{1}{\sqrt{\nu }(\nu +\ell )}\left( 1 - \frac{1}{12\nu \frac{k}{\ell }}+\mathcal {O}\left( \frac{1}{\nu ^2\min (1, k^2/\ell ^2)}\right) \right) = \frac{\sqrt{\ell }}{2\pi }\sum _{\nu >\ell }\frac{1}{\sqrt{\nu }(\nu +\ell )}\\&+\,\mathcal {O}\left( \frac{1}{\sqrt{k}}\right) , \end{aligned}$$

and for $\ell /k\le N\le \ell$

$$\begin{aligned} \sum _{N\le \nu \le 2N}\frac{1}{\nu ^{3/2}(\nu +\ell )} \le \frac{1}{\sqrt{N}\ell } \end{aligned}$$

as well as

$$\begin{aligned} \sum _{N\le \nu \le 2N}\frac{1}{\nu ^{5/2}(\nu +\ell )} \le \frac{1}{N^{3/2}\ell } \end{aligned}$$

and therefore

$$\begin{aligned} \sqrt{\ell }\sum _{\nu \ge \ell /k}\frac{1}{\sqrt{\nu }(\nu +\ell )} \cdot \frac{1}{\nu \frac{k}{\ell }} = \frac{\ell ^{3/2}}{k}\sum _{\nu \ge \ell /k}\frac{1}{\nu ^{3/2}(\nu +\ell )} =\mathcal {O}\left( \frac{1}{\sqrt{k}}\right) \end{aligned}$$

and

$$\begin{aligned} \sqrt{\ell }\sum _{\nu \ge \ell /k} \frac{1}{\sqrt{\nu }(\nu +\ell )}\cdot \frac{1}{\nu ^2\min (k^2/\ell ^2)} = \frac{\ell ^{5/2}}{k^2}\sum _{\nu \ge \ell /k}\frac{1}{\nu ^{5/2}(\nu +\ell )} =\mathcal {O}\left( \frac{1}{\sqrt{k}}\right) . \end{aligned}$$

Using these estimates we obtain

$$\begin{aligned} \frac{\sqrt{\ell }}{2\pi }&\sum _{\ell /k\le \nu \le \ell }\frac{1}{\sqrt{\nu }(\nu +\ell )} \left( 1 - \frac{1}{12\nu \frac{k}{\ell }}+\mathcal {O}\left( \frac{1}{\nu ^2 \min (1, k^2/\ell ^2)}\right) \right) =\\ \frac{\sqrt{\ell }}{2\pi }&\sum _{\ell /k\le \nu \le \ell }\frac{1}{\sqrt{\nu }(\nu +\ell )} +\mathcal {O}\left( \frac{1}{\sqrt{k}}\right) . \end{aligned}$$

For $k\le \ell$ we obtain

$$\begin{aligned} V_{k,\ell }= & {} \frac{\sqrt{\ell }}{2\pi }\sum _{\nu \ge \ell /k} \frac{1}{\sqrt{\nu }(\nu +\ell )} +\mathcal {O}\left( \frac{1}{\sqrt{k}}\right) \\= & {} \frac{1}{2\pi }\int _{1/k}^\infty \frac{dt}{\sqrt{t}(t+1)} +\mathcal {O}\left( \frac{1}{\sqrt{k}}\right) \\= & {} \frac{1}{2\pi }\int _0^\infty \frac{dt}{\sqrt{t}(t+1)} +\mathcal {O}\left( \frac{1}{\sqrt{k}}\right) \\= & {} \frac{1}{2}+\mathcal {O}\left( \frac{1}{\sqrt{k}}\right) , \end{aligned}$$

and the proof is complete. $\square$

3 Proof of Theorem 1

We first prove (1). Note that this inequality is only interesting if the error term is o(1), in particular we may assume that $k^2<\ell$. Under this assumption we have

$$\begin{aligned} P_{k, \ell }= & {} \left( \frac{\ell }{\ell +k}\right) ^{k+\ell }\sum _{\nu =0}^{k-1} \left( {\begin{array}{c}k+\ell \\ \nu \end{array}}\right) \left( \frac{k}{\ell }\right) ^\nu \\= & {} \sum _{\nu =0}^{k-1}\left( \frac{\ell }{\ell +k}\right) ^{k+\ell -\nu } \frac{k^\nu }{\nu !} \exp \left( \sum _{\mu =1}^\nu \log \frac{k+\ell -\mu }{k+\ell }\right) \\= & {} \sum _{\nu =0}^{k-1}\left( \frac{\ell }{\ell +k}\right) ^{k+\ell -\nu } \frac{k^\nu }{\nu !} \exp \left( -\sum _{\mu =1}^\nu \frac{\mu }{k+\ell }+ \mathcal {O} \left( \frac{\mu ^2}{(k+\ell )^2}\right) \right) \\= & {} \sum _{\nu =0}^{k-1}\left( \frac{\ell }{\ell +k}\right) ^{k+\ell -\nu } \frac{k^\nu }{\nu !} \exp \left( \mathcal {O}\left( \frac{\nu ^2}{\ell }\right) \right) \\= & {} \sum _{\nu =0}^{k-1}\left( \frac{\ell }{\ell +k}\right) ^{k+\ell } \frac{k^\nu }{\nu !} \exp \left( \mathcal {O}\left( \frac{k^2}{\ell }\right) \right) \\= & {} \exp \left( (k+\ell )\log \frac{\ell }{\ell +k}\right) \sum _{\nu =0}^{k-1} \frac{k^\nu }{\nu !} \exp \left( \mathcal {O}\left( \frac{k^2}{\ell }\right) \right) \\= & {} \exp \left( -k+\mathcal {O}\left( \frac{k^2}{\ell }\right) \right) \sum _{\nu =0}^{k-1}\frac{k^\nu }{\nu !}\\= & {} e^{-k}\sum _{\nu =0}^{k-1}\frac{k^\nu }{\nu !}+\mathcal {O}\left( \frac{k^2}{\ell }\right) . \end{aligned}$$

The proof of (2) is quite similar. Analogously to the previous case we may assume that $\ell ^2<k$.

$$\begin{aligned} P_{k, \ell }= & {} \left( \frac{\ell }{\ell +k}\right) ^{k+\ell }\left( \left( \frac{\ell +k}{\ell }\right) ^{k+\ell } - \sum _{\nu =k}^{k+\ell } \left( {\begin{array}{c}k+\ell \\ \nu \end{array}}\right) \left( \frac{k}{\ell }\right) ^\nu \right) \\= & {} 1-\left( \frac{\ell }{\ell +k}\right) ^{k+\ell } \sum _{\mu =0}^{\ell } \left( {\begin{array}{c}k+\ell \\ \mu \end{array}}\right) \left( \frac{k}{\ell }\right) ^{k+\ell -\mu }\\= & {} 1-\left( \frac{k}{\ell +k}\right) ^{k+\ell } \sum _{\mu =0}^{\ell } \left( {\begin{array}{c}k+\ell \\ \mu \end{array}}\right) \left( \frac{\ell }{k}\right) ^{\mu }\\= & {} 1-\exp \left( -\ell +\mathcal {O}\left( \frac{\ell ^2}{k}\right) \right) \sum _{\mu =0}^{\ell }\frac{\ell ^\mu }{\mu !}\left( \frac{k+\ell }{k}\right) ^{\mu }\\= & {} 1-e^{-\ell }\sum _{\mu =0}^{\ell }\frac{\ell ^\mu }{\mu !}. \end{aligned}$$

Finally (3) follows from Theorem 2, Lemma 2 and Lemma 4.