Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.