Two Monotonicity Results for Beta Distribution Functions

Abstract

Write $\mathrm{pbeta}(·,\alpha ,\beta )$ for the distribution function of the Beta distribution with parameters $\alpha$ and $\beta$. We show that $\alpha \mapsto pbeta(\alpha /(\alpha +\beta ),\alpha ,\beta )$ is decreasing and $\alpha \mapsto pbeta(\alpha /(\alpha +\beta ),\alpha +1,\beta )$ is increasing over the positive reals, with the common limit for $\alpha \to \infty$ expressible in terms of the Gamma distribution functions, and discuss implications for the distribution functions of the Gamma, Poisson and Binomial distributions.

1. Introduction

Write $\mathrm{Gamma}\left(\alpha \right)$ and $\mathrm{Beta}(\alpha ,\beta )$ for, respectively, the Gamma distribution with shape parameter $\alpha$ and rate/scale parameter 1 and the Beta distribution with parameters $\alpha$ and $\beta$, and write $pgamma(·,\alpha )$ and $pbeta(·,\alpha ,\beta )$ for the corresponding cumulative distribution functions.

If $X_{\alpha }\sim \mathrm{Gamma}\left(\alpha \right)$, $\mathbb{E}(X_{\alpha }/\alpha )=1$. Using the addition theorem for the Gamma distribution, as $\alpha \to \infty$, we have $X_{\alpha }/\alpha \to 1$ by the law of large numbers and $\mathbb{P}(X_{\alpha }/\alpha \le 1)=pgamma(\alpha ,\alpha )\to 1/2$ by the central limit theorem. On the other hand, using integration by parts, for $ϵ>0$ we have

$$\mathbb{P}(X_{\alpha }/\alpha >ϵ)={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_{ϵ\alpha }^{\infty }{t}^{\alpha -1}{e}^{-t}dt={\displaystyle \frac{1}{\Gamma (\alpha +1)}}{\int }_{ϵ\alpha }^{\infty }{t}^{\alpha }{e}^{-t}dt-{\displaystyle \frac{{\left(ϵ\alpha \right)}^{\alpha }{e}^{-ϵ\alpha }}{\Gamma (\alpha +1)}}$$

so that as $\alpha \to 0+$

$$\mathbb{P}(X_{\alpha }/\alpha >ϵ)\to {\displaystyle \frac{1}{\Gamma \left(1\right)}}{\int }_0^{\infty }{e}^{-x}dx-1=0$$

and hence, in particular, $\mathbb{P}(X_{\alpha }/\alpha \le 1)=pgamma(\alpha ,\alpha )\to 1$.

Write

$${\pi }_{\alpha }=\mathbb{P}(X_{\alpha }/\alpha \le 1),X_{\alpha }\sim \mathrm{Gamma}\left(\alpha \right).$$

Ref. [1] shows that $\alpha \mapsto {\pi }_{\alpha }=\mathrm{pgamma}(\alpha ,\alpha )$ decreases monotonically from 1 to 1/2 as $\alpha$ varies from 0 to ∞. In this paper, we show that, quite remarkably, monotonicity continues to hold if 1 is replaced by the normalized Gamma distributed random variable $X_{\beta }/\beta$ independent of $X_{\alpha }/\alpha$. We also establish a second, related monotonicity result. These results are formulated and discussed in Section 2. Section 3 gives the proofs.

2. Results

Theorem 1.

For $\alpha ,\beta >0$, let $X_{\alpha }\sim \mathrm{Gamma}\left(\alpha \right)$ be independent from $X_{\beta }\sim \mathrm{Gamma}\left(\beta \right)$ and let $Y_{\alpha ,\beta }\sim \mathrm{Beta}(\alpha ,\beta )$ so that $\mathbb{E}\left(Y_{\alpha ,\beta }\right)=\alpha /(\alpha +\beta )$. Write

$$p_{\alpha ,\beta }=\mathbb{P}(X_{\alpha }/\alpha \le X_{\beta }/\beta ).$$

Then

$$p_{\alpha ,\beta }=\mathbb{P}(Y_{\alpha ,\beta }\le \mathbb{E}\left(Y_{\alpha ,\beta }\right))=pbeta\left({\displaystyle \frac{\alpha }{\alpha +\beta }},\alpha ,\beta \right),$$

the function $\alpha \mapsto p_{\alpha ,\beta }$ is decreasing for $\alpha >0$, with limits 1 for $\alpha \to 0+$ and $1-pgamma(\beta ,\beta )<1/2$ for $\alpha \to \infty$, and the function $\beta \mapsto p_{\alpha ,\beta }$ is increasing for $\beta >0$, with limits 0 for $\beta \to 0+$ and $pgamma(\alpha ,\alpha )>1/2$ as $\beta \to \infty$.

The above can also be formulated in terms of the Beta prime and F distributions, noting that trivially

$$p_{\alpha ,\beta }=\mathbb{P}\left({\displaystyle \frac{X_{\alpha }}{X_{\beta }}}\le {\displaystyle \frac{\alpha }{\beta }}\right)=\mathbb{P}\left({\displaystyle \frac{X_{\alpha }/\alpha }{X_{\beta }/\beta }}\le 1\right)$$

where $X_{\alpha }/X_{\beta }$ has a Beta prime distribution with parameters $\alpha$ and $\beta$, and $(X_{\alpha }/\alpha )/(X_{\beta }/\beta )$ has an F distribution with parameters $\alpha /2$ and $\beta /2$. Note also that for $\beta \to \infty$, $X_{\beta }/\beta \to 1$ in probability and, thus, $p_{\alpha ,\beta }\to \mathbb{P}(X_{\alpha }\le \alpha )=\mathrm{pgamma}(\alpha ,\alpha )$, so Theorem 1 implies the result of [1].

Refs. [2,3] show that $pbeta(\alpha /(\alpha +\beta ),\alpha +1,\beta )<1/2$ for, respectively, all positive integers or all positive reals $\alpha ,\beta$. The following substantially improves these results:

Theorem 2.

Let $\beta >0$. The function

$$\alpha \mapsto {\tilde{p}}_{\alpha ,\beta }=pbeta\left({\displaystyle \frac{\alpha }{\alpha +\beta }},\alpha +1,\beta \right)$$

is increasing for $\alpha >0$, with limits 0 for $\alpha \to 0+$ and $1-pgamma(\beta ,\beta )$ for $\alpha \to \infty$.

If we write $pbinom(·,n,p)$ for the cumulative distribution function of the Binomial distribution with parameters n and p, then for integer $0\le k\le n$ we have

$$pbinom(k,n,p)=pbeta(1-p,n-k,k+1).$$

Using Theorem 1 with $\alpha =n-k$, $\beta =k+1$ and $1-p=\alpha /(\alpha +\beta )=(n-k)/(n+1)$, we obtain for integer $0\le k<n$

$$pbinom\left(k,n,{\displaystyle \frac{k+1}{n+1}}\right)\ge 1-pgamma(k,k).$$

Similarly, using Theorem 2 with $\alpha +1=n-k$, $\beta =k+1$ and $1-p=\alpha /(\alpha +\beta )=(n-k-1)/n$, we obtain for integer $0\le k<n-1$

$$pbinom\left(k,n,{\displaystyle \frac{k+1}{n}}\right)\le 1-pgamma(k,k).$$

If $X_{\alpha +1}\sim \mathrm{Gamma}(\alpha +1)$ is independent from $X_{\beta }\sim \mathrm{Gamma}\left(\beta \right)$, $X_{\alpha +1}/(X_{\alpha +1}+X_{\beta })\sim \mathrm{Beta}(\alpha +1,\beta )$, so that

$$pbeta\left({\displaystyle \frac{\alpha }{\alpha +\beta }},\alpha +1,\beta \right)=\mathbb{P}\left({\displaystyle \frac{X_{\alpha +1}}{X_{\alpha +1}+X_{\beta }}}\le {\displaystyle \frac{\alpha }{\alpha +\beta }}\right)=\mathbb{P}\left(X_{\alpha +1}\le \alpha {\displaystyle \frac{X_{\beta }}{\beta }}\right).$$

For $\beta \to \infty$, $X_{\beta }/\beta \to 1$ in probability, so Theorem 2 yields the following:

Corollary 1.

The function $\alpha \mapsto \mathrm{pgamma}(\alpha ,\alpha +1)$ is increasing for $\alpha >0$, with limits of 0 for $\alpha \to 0+$ and $1/2$ for $\alpha \to \infty$.

In combination with [1] (or using Theorem 1 with $\beta \to \infty$), we, thus, find that the median $m_{\alpha }$ of $\mathrm{Gamma}\left(\alpha \right)$ satisfies $\alpha -1<m_{\alpha }<\alpha$, where the lower bound is worse than the sharp lower bound $m_{\alpha }>\alpha -1/3$ of [4].

If we write $\mathrm{ppois}(·,\lambda )$ for the cumulative distribution function of the Poisson distribution with parameter $\lambda$, then for the integer $n\ge 0$ we have

$$\mathrm{ppois}(n,\lambda )=1-\mathrm{pgamma}(\lambda ,n+1).$$

From Corollary 1, we, thus, find that $n\mapsto \mathrm{ppois}(n,n)=1-\mathrm{pgamma}(n,n+1)$ is decreasing from 1 to 1/2 as n varies over the non-negative integers, nicely extending the bounds of [5] in the integer case. We note that $\lambda \mapsto \mathrm{ppois}(\lambda ,\lambda )$ is not monotone: it jumps at the positive integers, and for $n\le \lambda <n+1$

$$\mathrm{ppois}(\lambda ,\lambda )=\mathrm{ppois}(n,\lambda )=1-\mathrm{pgamma}(\lambda ,n+1)$$

which decreases from $\mathrm{ppois}(n,n)$ to $\mathrm{ppois}(n+1,n)=1-\mathrm{pgamma}(n+1,n+1)$, where we just obtained that the former upper envelope sequence is decreasing in n, and based on the result of [1], the latter lower envelope sequence is increasing in n (with common limit 1/2 as $n\to \infty$).

Clearly, Theorem 2 is equivalent to

$$\alpha \mapsto pbeta\left({\displaystyle \frac{\alpha -1}{\alpha +\beta -1}},\alpha ,\beta \right)$$

increasing for $\alpha >1$ (it is zero for $0<\alpha \le 1$), where, in fact, $(\alpha -1)/(\alpha +\beta -1)$ is the harmonic mean of the Beta distribution with parameters $\alpha$ and $\beta$. Thus, if for $\alpha >max(u,0)$ and $\beta >0$, we write

$$p_{\alpha ,\beta ,u}=pbeta\left({\displaystyle \frac{\alpha -u}{\alpha +\beta -u}},\alpha ,\beta \right),$$

and our results say that $\alpha \mapsto p_{\alpha ,\beta ,0}$ is decreasing, and $\alpha \mapsto p_{\alpha ,\beta ,1}$ is increasing, and we can ask about the monotonicity for other values of u. Numerical experiments suggest that for $\beta >1$, $\alpha \mapsto p_{\alpha ,\beta ,u}$ is decreasing iff $u\le 1/3$ and increasing iff $u\ge 0.5$, but we have thus far been unable to obtain rigorous monotonicity results for $u\notin \{0,1\}$. We also note also that for $u=1$, we find that for $\alpha >1$, the median $m_{\alpha ,\beta }$ of the Beta distribution with parameters $\alpha$ and $\beta$ satisfies $m_{\alpha ,\beta }>(\alpha -1)/(\alpha +\beta -1)$, which for $\alpha <\beta$ is worse than the lower bound $m_{\alpha ,\beta }>(\alpha -1)/(\alpha +\beta -2)$ of [6]. This suggests the importance of more generally investigating the monotonicity of $\alpha \mapsto p_{\alpha ,\beta ,u,v}=pbeta((\alpha -u)/(\alpha +\beta -v),\alpha ,\beta )$. For $v=2u$, this includes the bounds in [6] and the approximations suggested in [7], and it conveniently allows using $p_{\alpha ,\beta ,u,2u}=1-p_{\beta ,\alpha ,u,2u}$ to obtain monotonicity in $\beta$ from the monotonicity in $\alpha$.

3. Proofs

We first establish several lemmas. We write that

$$g_{\alpha ,\beta }={\left.{\displaystyle \frac{x^{\alpha }{(1-x)}^{\beta }}{\alpha B(\alpha ,\beta )}}\right|}_{x={\displaystyle \frac{\alpha }{\alpha +\beta }}}={\displaystyle \frac{{\alpha }^{\alpha -1}{\beta }^{\beta }}{B(\alpha ,\beta ){(\alpha +\beta )}^{\alpha +\beta }}}$$

Lemma 1.

Let $\alpha ,\beta >0$. Then,

$$pbeta\left({\displaystyle \frac{\alpha }{\alpha +\beta }},\alpha ,\beta \right)=pbeta\left({\displaystyle \frac{\alpha }{\alpha +\beta }},\alpha +1,\beta \right)+g_{\alpha ,\beta }.$$

Proof. 

Immediate from Equation 8.17.20 (http://dlmf.nist.gov/8.17.E20, accessed on 28 October 2024) in [8]. □

Lemma 2.

Let $\beta >0$. Then, $\alpha \mapsto g_{\alpha ,\beta }$ is decreasing and positive for $\alpha >0$.

Proof. 

Positivity is trivial. As

$$g_{\alpha ,\beta }={\displaystyle \frac{{\beta }^{\beta }}{\Gamma \left(\beta \right)}}{\displaystyle \frac{\Gamma (\alpha +\beta )}{\Gamma \left(\alpha \right)}}{\displaystyle \frac{{\alpha }^{\alpha -1}}{{(\alpha +\beta )}^{\alpha +\beta }}},$$

we have

$${\displaystyle \frac{\partial log\left(g_{\alpha ,\beta }\right)}{\partial \alpha }}=\psi (\alpha +\beta )-\psi \left(\alpha \right)+log\left(\alpha \right)-{\displaystyle \frac{1}{\alpha }}-log(\alpha +\beta ),$$

where $\psi \left(z\right)={(log(\Gamma \left(z\right)))}^{\prime }={\Gamma }^{\prime }\left(z\right)/\Gamma \left(z\right)$ is the psi (or digamma) function (e.g., Equation 5.2.2 (https://dlmf.nist.gov/5.2.E2, accessed on 28 October 2024) in [8]). Using Equation 5.9.13 (http://dlmf.nist.gov/5.9.E13, accessed on 28 October 2024) in [8], for $\Re \left(z\right)>0$, we find that

$$\psi \left(z\right)-log\left(z\right)={\int }_0^{\infty }\left({\displaystyle \frac{1}{t}}-{\displaystyle \frac{1}{1-e^{-t}}}\right)e^{-zt}dt,$$

so that

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\partial log\left(g_{\alpha ,\beta }\right)}{\partial \alpha }}& =& {\int }_0^{\infty }\left({\displaystyle \frac{1}{t}}-{\displaystyle \frac{1}{1-e^{-t}}}\right)(e^{-(\alpha +\beta )t}-e^{-\alpha t})dt-{\displaystyle \frac{1}{\alpha }}\hfill \\ & =& {\int }_0^{\infty }\left({\displaystyle \frac{1}{t}}-{\displaystyle \frac{1}{1-e^{-t}}}\right)(e^{-\beta t}-1)e^{-\alpha t}dt-{\int }_0^{\infty }{e}^{-\alpha t}dt\hfill \\ & =& {\int }_0^{\infty }\left(\left({\displaystyle \frac{1}{t}}-{\displaystyle \frac{1}{1-e^{-t}}}\right)(e^{-\beta t}-1)-1\right)e^{-\alpha t}dt\hfill \\ & =& {\int }_0^{\infty }(k\left(t\right)(1-e^{-\beta t})-1)e^{-\alpha t}dt,\hfill \end{array}$$

where

$$k\left(t\right)={\displaystyle \frac{1}{1-e^{-t}}}-{\displaystyle \frac{1}{t}}$$

has the derivative

$$k^{\prime }\left(t\right)=-{\displaystyle \frac{e^{-t}}{{(1-e^{-t})}^2}}+{\displaystyle \frac{1}{t^2}}={\displaystyle \frac{-t^2{e}^{-t}+{(1-e^{-t})}^2}{t^2{(1-e^{-t})}^2}}$$

with

$$-t^2{e}^{-t}+{(1-e^{-t})}^2=1-2e^{-t}+e^{-2t}-t^2{e}^{-t}=e^{-t}(e^t-(2+t^2)+e^{-t})>0$$

for $t>0$. Hence, for all $t>0$, k is increasing, from which first $k\left(t\right)<k(\infty )=1$ and then $k\left(t\right)(1-e^{-\beta t})-1<1-1=0$ are derived. This in turn shows that $log\left(g_{\alpha ,\beta }\right)$ and, hence, $g_{\alpha ,\beta }$ are decreasing for $\alpha >0$. □

Lemma 3.

Let $\alpha ,\beta >0$. Then,

$$\begin{array}{c}{\displaystyle pbeta\left(\frac{\alpha +1}{\alpha +\beta +1},\alpha +1,\beta \right)-pbeta\left(\frac{\alpha }{\alpha +\beta },\alpha +1,\beta \right)}\hfill \\ \hspace{1em}={\displaystyle \frac{1}{B(\alpha +1,\beta )}}{\int }_0^1{\displaystyle \frac{{(\alpha +v)}^{\alpha }{\beta }^{\beta }}{{(\alpha +\beta +v)}^{\alpha +\beta +1}}}dv\hfill \\ \hspace{1em}=g_{\alpha +1,\beta }{\int }_0^1{\left({\displaystyle \frac{\alpha +v}{\alpha +1}}\right)}^{\alpha }{\left({\displaystyle \frac{\alpha +\beta +1}{\alpha +\beta +v}}\right)}^{\alpha +\beta +1}dv\hfill \\ \hspace{1em}=g_{\alpha ,\beta }{\int }_0^1{\left({\displaystyle \frac{\alpha +v}{\alpha }}\right)}^{\alpha }{\left({\displaystyle \frac{\alpha +\beta }{\alpha +\beta +v}}\right)}^{\alpha +\beta +1}dv.\hfill \end{array}$$

Proof. 

We have

$$\begin{array}{c}{\displaystyle pbeta\left(\frac{\alpha +1}{\alpha +\beta +1},\alpha +1,\beta \right)-pbeta\left(\frac{\alpha }{\alpha +\beta },\alpha +1,\beta \right)}\hfill \\ \hspace{1em}={\displaystyle \frac{1}{B(\alpha +1,\beta )}}{\int }_{{\displaystyle \frac{\alpha }{\alpha +\beta }}}^{{\displaystyle \frac{\alpha +1}{\alpha +1+\beta }}}{t}^{\alpha }{(1-t)}^{\beta -1}dt.\hfill \end{array}$$

Substituting $t=u/(1+u)$ so that $1-t=1/(1+u)$ and $dt=du/{(1+u)}^2$ and then $u=(\alpha +v)/\beta$,

$$\begin{array}{ccc}\hfill {\int }_{{\displaystyle \frac{\alpha }{\alpha +\beta }}}^{{\displaystyle \frac{\alpha +1}{\alpha +1+\beta }}}{t}^{\alpha }{(1-t)}^{\beta -1}dt& =& {\int }_{{\displaystyle \frac{\alpha }{\beta }}}^{{\displaystyle \frac{\alpha +1}{\beta }}}{\displaystyle \frac{u^{\alpha }}{{(1+u)}^{\alpha +\beta +1}}}du\hfill \\ & =& {\displaystyle \frac{1}{\beta }}{\int }_0^1{\left({\displaystyle \frac{\alpha +v}{\beta }}\right)}^{\alpha }{\left({\displaystyle \frac{\beta }{\beta +\alpha +v}}\right)}^{\alpha +\beta +1}\hfill \\ & =& {\int }_0^1{\displaystyle \frac{{(\alpha +v)}^{\alpha }{\beta }^{\beta }}{{(\alpha +\beta +v)}^{\alpha +\beta +1}}}dv\hfill \end{array}$$

from which the first equality follows. The second is immediate, and the third obtained from

$$g_{\alpha ,\beta }={\displaystyle \frac{{\alpha }^{\alpha }{\beta }^{\beta }}{{(\alpha +\beta )}^{\alpha +\beta }}}{\displaystyle \frac{1}{\alpha B(\alpha ,\beta )}}={\displaystyle \frac{{\alpha }^{\alpha }{\beta }^{\beta }}{{(\alpha +\beta )}^{\alpha +\beta +1}}}{\displaystyle \frac{1}{B(\alpha +1,\beta )}}.$$

□

Write

$$h_{\alpha ,\beta }=1-{\int }_0^1{\left({\displaystyle \frac{\alpha +v}{\alpha }}\right)}^{\alpha }{\left({\displaystyle \frac{\alpha +\beta }{\alpha +\beta +v}}\right)}^{\alpha +\beta +1}dv.$$

Lemma 4.

Let $\beta >0$. Then, $\alpha \mapsto h_{\alpha ,\beta }$ is decreasing and positive for $\alpha >0$.

Proof. 

We write

$$h_{\alpha ,\beta }=1-{\int }_0^1{k}_{\alpha ,\beta }\left(v\right)dv,k_{\alpha ,\beta }\left(v\right)={\left({\displaystyle \frac{\alpha +v}{\alpha }}\right)}^{\alpha }{\left({\displaystyle \frac{\alpha +\beta }{\alpha +\beta +v}}\right)}^{\alpha +\beta +1}.$$

Then,

$$log\left(k_{\alpha ,\beta }\left(v\right)\right)=\alpha log{\displaystyle \frac{\alpha +v}{\alpha }}+(\alpha +\beta +1)log{\displaystyle \frac{\alpha +\beta }{\alpha +\beta +v}}$$

and, hence,

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\partial log\left(k_{\alpha ,\beta }\left(v\right)\right)}{\partial \alpha }}& =& log{\displaystyle \frac{\alpha +v}{\alpha }}+\alpha \left({\displaystyle \frac{1}{\alpha +v}}-{\displaystyle \frac{1}{\alpha }}\right)\hfill \\ & & +log{\displaystyle \frac{\alpha +\beta }{\alpha +\beta +v}}+(\alpha +\beta +1)\left({\displaystyle \frac{1}{\alpha +\beta }}-{\displaystyle \frac{1}{\alpha +\beta +v}}\right).\hfill \end{array}$$

As a function of v, this has the derivative

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\partial }{\partial v}}{\displaystyle \frac{\partial log\left(k_{\alpha ,\beta }\left(v\right)\right)}{\partial \alpha }}& =& {\displaystyle \frac{1}{\alpha +v}}-{\displaystyle \frac{\alpha }{{(\alpha +v)}^2}}-{\displaystyle \frac{1}{\alpha +\beta +v}}+{\displaystyle \frac{\alpha +\beta +1}{{(\alpha +\beta +v)}^2}}\hfill \\ & =& {\displaystyle \frac{v}{{(\alpha +v)}^2}}+{\displaystyle \frac{1-v}{{(\alpha +\beta +v)}^2}}\hfill \end{array}$$

which is positive for $0<v<1$. Hence, for $\alpha ,\beta >0$ and $0<v<1$, so

$${\displaystyle \frac{\partial log\left(k_{\alpha ,\beta }\left(v\right)\right)}{\partial \alpha }}>{\left.{\displaystyle \frac{\partial log\left(k_{\alpha ,\beta }\left(v\right)\right)}{\partial \alpha }}\right|}_{v=0}=0$$

from which we infer that for $\beta >0$ and $0<v<1$, $\alpha \mapsto k_{\alpha ,\beta }\left(v\right)$ is increasing for $\alpha >0$, which in turn yields that $\alpha \mapsto h_{\alpha ,\beta }$ is decreasing for $\alpha >0$.

For $\alpha \to \infty$,

$$k_{\alpha ,\beta }\left(v\right)={\displaystyle \frac{{(1+v/\alpha )}^{\alpha }}{{(1+v/(\alpha +\beta ))}^{\alpha +\beta +1}}}\to {\displaystyle \frac{e^v}{e^v}}=1$$

and, hence, $h_{\alpha ,\beta }\to h_{\infty ,\beta }=1-{\int }_0^{1}1dv=0$. Thus, for all $\alpha >0$, $h_{\alpha ,\beta }>h_{\infty ,\beta }=0$, thus completing the proof. □

Lemma 5.

Let $\alpha ,\beta >0$. Then,

$$p_{\alpha ,\beta }=p_{\alpha +1,\beta }+g_{\alpha ,\beta }{h}_{\alpha ,\beta }$$

and

$$p_{\alpha ,\beta }=p_{\infty ,\beta }+\sum _{n=0}^{\infty }{g}_{\alpha +n,\beta }{h}_{\alpha +n,\beta }.$$

Proof. 

Using Lemmas 1 and 3,

$$\begin{array}{ccc}\hfill p_{\alpha ,\beta }& =& pbeta\left({\displaystyle \frac{\alpha }{\alpha +\beta }},\alpha ,\beta \right)\hfill \\ & =& pbeta\left({\displaystyle \frac{\alpha }{\alpha +\beta }},\alpha +1,\beta \right)+g_{\alpha ,\beta }\hfill \\ & =& pbeta\left({\displaystyle \frac{\alpha +1}{\alpha +\beta +1}},\alpha +1,\beta \right)+g_{\alpha ,\beta }\hfill \\ & & -\left(pbeta\left({\displaystyle \frac{\alpha +1}{\alpha +\beta +1}},\alpha +1,\beta \right)-pbeta\left({\displaystyle \frac{\alpha }{\alpha +\beta }},\alpha +1,\beta \right)\right)\hfill \\ & =& p_{\alpha +1,\beta }+g_{\alpha ,\beta }\left(1-{\int }_0^1{\left({\displaystyle \frac{\alpha +v}{\alpha }}\right)}^{\alpha }{\left({\displaystyle \frac{\alpha +\beta }{\alpha +\beta +v}}\right)}^{\alpha +\beta +1}dv\right)\hfill \\ & =& p_{\alpha +1,\beta }+g_{\alpha ,\beta }{h}_{\alpha ,\beta },\hfill \end{array}$$

from which

$$p_{\alpha ,\beta }=p_{\infty ,\beta }+\sum _{n=0}^{\infty }(p_{\alpha +n,\beta }-p_{\alpha +n+1,\beta })=p_{\infty ,\beta }+\sum _{n=0}^{\infty }{g}_{\alpha +n,\beta }{h}_{\alpha +n,\beta }$$

as asserted. □

Proof of Theorem 1.

As $x\mapsto t\left(x\right)=x/(1+x)$ increases monotonically from 0 to 1 as x varies from 0 to ∞, and $t(u/v)=u/(u+v)$,

$$p_{\alpha ,\beta }=\mathbb{P}(t(X_{\alpha }/X_{\beta })\le t(\alpha /\beta ))=\mathbb{P}\left({\displaystyle \frac{X_{\alpha }}{X_{\alpha }+X_{\beta }}}\le {\displaystyle \frac{\alpha }{\alpha +\beta }}\right)$$

where $Y_{\alpha ,\beta }=X_{\alpha }/(X_{\alpha }+X_{\beta })\sim \mathrm{Beta}(\alpha ,\beta )$ has the mean $\mathbb{E}\left(Y_{\alpha ,\beta }\right)=\alpha /(\alpha +\beta )$, so that $p_{\alpha ,\beta }=pbeta(\alpha /(\alpha +\beta ),\alpha ,\beta )$.

Combining Lemmas 2, 4 and 5, we see that $\alpha \mapsto p_{\alpha ,\beta }$ is decreasing for $\alpha >0$. To determine the limits, remember that if $X_{\alpha }\sim \mathrm{Gamma}\left(\alpha \right)$, then $X_{\alpha }/\alpha$ goes to 0 in probability as $\alpha \to 0+$ and to 1 as $\alpha \to \infty$. Hence,

$$p_{\alpha ,\beta }=\mathbb{P}\left({\displaystyle \frac{X_{\alpha }}{\alpha }}\le {\displaystyle \frac{X_{\beta }}{\beta }}\right)$$

goes to $\mathbb{P}(0\le X_{\beta }/\beta )=1$ as $\alpha \to 0+$ and to $\mathbb{P}(1\le X_{\beta }/\beta )=1-pgamma(\beta ,\beta )$ as $\alpha \to \infty$.

Finally, $p_{\beta ,\alpha }=1-p_{\alpha ,\beta }$, thus completing the proof. □

We write

$${\tilde{h}}_{\alpha ,\beta }={\int }_0^1{\left({\displaystyle \frac{\alpha +v}{\alpha +1}}\right)}^{\alpha }{\left({\displaystyle \frac{\alpha +\beta +1}{\alpha +\beta +v}}\right)}^{\alpha +\beta +1}dv-1.$$

Lemma 6.

Let $\beta >0$. Then, $\alpha \mapsto {\tilde{h}}_{\alpha ,\beta }$ is decreasing and positive for $\alpha >0$.

Proof. 

This parallels the proof of Lemma 4. We write

$${\tilde{h}}_{\alpha ,\beta }={\int }_0^1{\tilde{k}}_{\alpha ,\beta }\left(v\right)dv-1,{\tilde{k}}_{\alpha ,\beta }\left(v\right)={\left({\displaystyle \frac{\alpha +v}{\alpha +1}}\right)}^{\alpha }{\left({\displaystyle \frac{\alpha +\beta +1}{\alpha +\beta +v}}\right)}^{\alpha +\beta +1}.$$

Then,

$$log\left({\tilde{k}}_{\alpha ,\beta }\left(v\right)\right)=\alpha log{\displaystyle \frac{\alpha +v}{\alpha +1}}+(\alpha +\beta +1)log{\displaystyle \frac{\alpha +\beta +1}{\alpha +\beta +v}}$$

and, hence,

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\partial log\left({\tilde{k}}_{\alpha ,\beta }\left(v\right)\right)}{\partial \alpha }}& =& log{\displaystyle \frac{\alpha +v}{\alpha +1}}+\alpha \left({\displaystyle \frac{1}{\alpha +v}}-{\displaystyle \frac{1}{\alpha +1}}\right)\hfill \\ & & +log{\displaystyle \frac{\alpha +\beta +1}{\alpha +\beta +v}}+(\alpha +\beta +1)\left({\displaystyle \frac{1}{\alpha +\beta +1}}-{\displaystyle \frac{1}{\alpha +\beta +v}}\right).\hfill \end{array}$$

As a function of v, this (again) has the derivative

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\partial }{\partial v}}{\displaystyle \frac{\partial log\left({\tilde{k}}_{\alpha ,\beta }\left(v\right)\right)}{\partial \alpha }}& =& {\displaystyle \frac{1}{\alpha +v}}-{\displaystyle \frac{\alpha }{{(\alpha +v)}^2}}-{\displaystyle \frac{1}{\alpha +\beta +v}}+{\displaystyle \frac{\alpha +\beta +1}{{(\alpha +\beta +v)}^2}}\hfill \\ & =& {\displaystyle \frac{v}{{(\alpha +v)}^2}}+{\displaystyle \frac{1-v}{{(\alpha +\beta +v)}^2}}\hfill \end{array}$$

which is positive for $0<v<1$. Hence, for $\alpha ,\beta >0$ and $0<v<1$,

$${\displaystyle \frac{\partial log\left({\tilde{k}}_{\alpha ,\beta }\left(v\right)\right)}{\partial \alpha }}<{\left.{\displaystyle \frac{\partial log\left({\tilde{k}}_{\alpha ,\beta }\left(v\right)\right)}{\partial \alpha }}\right|}_{v=1}=0$$

from which we infer that for $\beta >0$ and $0<v<1$, $\alpha \mapsto {\tilde{k}}_{\alpha ,\beta }\left(v\right)$ is decreasing for $\alpha >0$, which in turn implies that $\alpha \mapsto {\tilde{h}}_{\alpha ,\beta }$ is decreasing for $\alpha >0$. Finally, for $\alpha \to \infty$,

$${\tilde{k}}_{\alpha ,\beta }\left(v\right)={\left({\displaystyle \frac{1+v/\alpha }{1+1/\alpha }}\right)}^{\alpha }{\left({\displaystyle \frac{1+1/(\alpha +\beta )}{1+v/(\alpha +\beta )}}\right)}^{\alpha +\beta +1}\to {\displaystyle \frac{e^v}{e}}{\displaystyle \frac{e}{e^v}}=1$$

and, hence, ${\tilde{h}}_{\alpha ,\beta }\to {\tilde{h}}_{\infty ,\beta }={\int }_0^{1}1dv-1=0$. Thus, for all $\alpha >0$, ${\tilde{h}}_{\alpha ,\beta }>{\tilde{h}}_{\infty ,\beta }=0$, thus completing the proof. □

Lemma 7.

Let $\alpha ,\beta >0$. Then,

$${\tilde{p}}_{\alpha ,\beta }={\tilde{p}}_{\alpha +1,\beta }-g_{\alpha +1,\beta }{\tilde{h}}_{\alpha ,\beta }$$

and

$${\tilde{p}}_{\alpha ,\beta }={\tilde{p}}_{\infty ,\beta }-\sum _{n=0}^{\infty }{g}_{\alpha +n+1,\beta }{\tilde{h}}_{\alpha +n,\beta }.$$

Proof. 

Using Lemma 1 (with $\alpha +1$ instead of $\alpha$) and Lemma 3, we have

$$\begin{array}{ccc}\hfill {\tilde{p}}_{\alpha ,\beta }& =& pbeta\left({\displaystyle \frac{\alpha }{\alpha +\beta }},\alpha +1,\beta \right)\hfill \\ & =& pbeta\left({\displaystyle \frac{\alpha +1}{\alpha +1+\beta }},\alpha +1,\beta \right)\hfill \\ & & -\left(pbeta\left({\displaystyle \frac{\alpha +1}{\alpha +\beta +1}},\alpha +1,\beta \right)-pbeta\left({\displaystyle \frac{\alpha }{\alpha +\beta }},\alpha +1,\beta \right)\right)\hfill \\ & =& pbeta\left({\displaystyle \frac{\alpha +1}{\alpha +1+\beta }},\alpha +2,\beta \right)+g_{\alpha +1,\beta }\hfill \\ & & -g_{\alpha +1,\beta }{\int }_0^1{\left({\displaystyle \frac{\alpha +v}{\alpha +1}}\right)}^{\alpha }{\left({\displaystyle \frac{\alpha +\beta +1}{\alpha +\beta +v}}\right)}^{\alpha +\beta +1}dv\hfill \\ & =& {\tilde{p}}_{\alpha +1,\beta }-g_{\alpha +1,\beta }{\tilde{h}}_{\alpha ,\beta }.\hfill \end{array}$$

The second assertion again follows by taking telescope sums. □

Proof of Theorem 2.

Combining Lemmas 2, 6 and 7, we see that $\alpha \mapsto {\tilde{p}}_{\alpha ,\beta }$ is increasing for $\alpha >0$. The limits are immediate from (the proof of) Theorem 1. □