Concavity of hypergeometric function ratio

Question

I would like to show that the function, $$ f(x) = \frac{{}_2\mathrm{F}_1\big(\frac{1}{2},\frac{1}{2};c+1\,;x\big)}{{}_2\mathrm{F}_1\big(\frac{1}{2},\frac{1}{2};c\,;x\big)} $$ is concave for $0 < x < 1$, where $c\ge\frac{1}{2}$.

Edit: from a comment, it seems that the stronger statement is true that: $$ g(x) = 1 - \frac{{}_2\mathrm{F}_1\big(\frac{1}{2},\frac{1}{2};c+1\,;x\big)}{{}_2\mathrm{F}_1\big(\frac{1}{2},\frac{1}{2};c\,;x\big)} $$ is absolutely monotonic on $0<x<1$, with $g(0)=0$ and $g(1) = \min\big(1,(2c-1)^{-2}\big)$.

Edit (2): I found this paper https://dmkrp.wordpress.com/wp-content/uploads/2022/10/mathematics-10-03903-v2.pdf which gives an integral representation for $F(a+n_1,b+n_2;c+m;x)/F(a,b;c;x)$. In the present case the integral representation of $f(x)$ is a constant minus a scale mixture of $(1-x)^{-1}$, which is absolutely monotonic on $(0,1)$, showing that $g(x)$ is also absolutely monotonic.

Concavity of $f$ (or convexity of $g$) would complete a proof that the correlation coefficient distribution with Normal predictors has monotone likelihood ratio.

Answer 1

According to the paper cited in Edit (2), when $a,b>0$ and $c>\max(a,b)$, we have the integral representation: $$ f(x) = \frac{{}_2\mathrm{F}_1(a,b;c+1;x)}{{}_2\mathrm{F}_1(a,b;c;x)} = c_1 - c_2\int_0^1 \frac{t^{a+b-1}(1-t)^{c-a-b}}{(1-xt)\,|{}_2\mathrm{F}_1(a,b;c;t^{-1})|^2}\,dt $$ where $c_1=\frac{c}{c-\min(a,b)}>0$ and $c_2 = \frac{c\,\Gamma(c)^2}{\Gamma(a)\Gamma(b)\Gamma(c-a+1)\Gamma(c-b+1)}>0$. Since $(1-x)^{-1}$ is absolutely monotonic on $(0,1)$, we have that $1-f(x)$ is absolutely monotonic on $(0,1)$ (and has positive Maclaurin series coefficients). This implies concavity of $f(x)$ on $(0,1)$.

Using this result, we can show that $\log {}_2\mathrm{F}_1(a,b;c;x)$ and its derivative have decreasing (and positive) Maclaurin coefficients when $c>\max(a,b)$. We have, $$ \frac{d}{dx} \log {}_2\mathrm{F}_1(a,b;c;x) = \frac{ab}{c}\frac{{}_2\mathrm{F}_1(a+1,b+1;c+1;x)}{{}_2\mathrm{F}_1(a,b;c;x)} $$ Since, $$ 1-(1-x)\frac{{}_2\mathrm{F}_1(a+1,b+1;c+1;x)}{{}_2\mathrm{F}_1(a,b;c;x)} = 1-\frac{{}_2\mathrm{F}_1(c-a,c-b;c+1;x)}{{}_2\mathrm{F}_1(c-a,c-b;c;x)} $$ has positive Maclaurin coefficients, we have that the Maclaurin coefficients of $\frac{d}{dx}\log {}_2\mathrm{F}_1(a,b;c;x)$ are decreasing, and thus the coefficients of $\log {}_2\mathrm{F}_1(a,b;c;x)$ are decreasing.

The fact that the Maclaurin coefficients of $\log {}_2\mathrm{F}_1(a,b;c;x)$ are positive follows from the integral representation, $$ \frac{{}_2\mathrm{F}_1(a+1,b+1;c+1;x)}{{}_2\mathrm{F}_1(a,b;c;x)} = c_3\int_0^1 \frac{t^{a+b}(1-t)^{c-a-b-1}}{(1-xt)\,|{}_2\mathrm{F}_1(a,b;c;t^{-1})|^2}\,dt $$ where $c_3 = \frac{c\,\Gamma(c)^2}{\Gamma(a+1)\Gamma(b+1)\Gamma(c-a)\Gamma(c-b)}>0$.