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Let $_2F_1(a,b;c,z)$ be the (Gauss) hypergeometric function, and $m$ and $n$ positive integers.

From a simple plot it looks like

$_2F_1(m+n,1,m+1,\frac{m}{m+n})>\frac{m}{n} \,_2F_1(m+n,1,n+1,\frac{n}{m+n})$

when $m<n$, but how do I prove this?

Hope this is not too trivial, I am not very familiar with such functions.

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1 Answer 1

up vote 3 down vote accepted

Using Euler's transformation of the Gauss's hypergeometric function: $$ {}_2 F_1\left(a,b; c; z\right) = (1-z)^{c-a-b} \cdot {}_2F_1\left(c-a,c-b; c; z\right) $$ for $b=1$, $a=m+n$ and $c=m+1$ we have a "closed form" expression: $$ \begin{eqnarray} {}_2 F_1\left(m+n,1; m+1; \frac{m}{m+n}\right) &=& \left(\frac{n}{n+m}\right)^{-n} {}_2F_1\left( 1-n, m; m+1; \frac{m}{m+n} \right) \\ &=& m \left(\frac{n+m}{n}\right)^{n} \cdot \sum_{k=0}^{n-1} \frac{1}{k+m} \frac{(1-n)_k}{k!} \left(\frac{m}{n+m}\right)^k \\ &=& m \left(\frac{n+m}{n}\right)^{n} \cdot \sum_{k=0}^{n-1} \frac{(-1)^k}{k+m} \binom{n-1}{k} \left(\frac{m}{n+m}\right)^k \\ &=& m \left(\frac{n+m}{n}\right)^{n} \cdot \left(\frac{n+m}{m}\right)^{m} \int_0^{\frac{m}{n+m}} \left(1-x\right)^{n-1} x^{m-1} \mathrm{d} x \end{eqnarray} $$ Thus: $$ \frac{1}{m} {}_2 F_1\left(m+n,1; m+1; \frac{m}{m+n}\right) - \frac{1}{n} {}_2 F_1\left(m+n,1; n+1; \frac{n}{m+n}\right) = \frac{(n+m)^{n+m}}{n^n m^m} \left( \int_0^{\frac{m}{n+m}} \left(1-x\right)^{n-1} x^{m-1} \mathrm{d} x - \int_0^{\frac{n}{n+m}} \left(1-x\right)^{m-1} x^{n-1} \mathrm{d} x \right) = \frac{(n+m)^{n+m}}{n^n m^m} \left( \int_0^{\frac{m}{n+m}} \left(1-x\right)^{n-1} x^{m-1} \mathrm{d} x - \int_{\frac{m}{n+m}}^1 \left(1-x\right)^{n-1} x^{m-1} \mathrm{d} x \right) = \\ \frac{(n+m)^{n+m}}{n^n m^m} \int_0^1 \operatorname{sgn}\left({\frac{m}{n+m}}-x\right)\left(1-x\right)^{n-1} x^{m-1} \mathrm{d} x $$ Interpreting $x$ as a beta random variable the last expression becomes: $$ \frac{(n+m)^{n+m}}{\Gamma(n+m)} \frac{n^n}{\Gamma(n)} \frac{m^m}{\Gamma(m)} \mathbb{E}\left( \operatorname{sgn}\left(\mu-X\right) \right) = \frac{(n+m)^{n+m}}{\Gamma(n+m)} \frac{n^n}{\Gamma(n)} \frac{m^m}{\Gamma(m)} \left( \mathbb{P}\left( X < \mu \right) - \mathbb{P}\left( X > \mu \right)\right) $$ where $\mu = \mathbb{E}(X) = \frac{m}{m+n}$, where $X$ follows $\mathrm{Be}(m,n)$. Given that $m<n$, random variable $X$ is positively skewed, and the expression is positive.

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