# Proving monotonicity of this ratio of Hypergeometric functions

Let $$n\in\Bbb N$$, $$\omega=0,1,\dots,n$$, and $$\nu,z>0$$. We define $$\tilde g_{n,\omega}(z,\nu):=\frac{z^{n-\omega}\partial_z^n z^\omega {_1F_0}(1;-;z)_\nu}{{_1F_0}(1;-;z)_\nu},$$ where $${_1F_0}(1;-;z)_\nu:=\frac{1-z^{\nu+1}}{1-z}=(\nu+1)z^{\nu+1}\mathbf F(1,\nu+2,2,1-z),$$ which is a continuous interpolation of the truncated geometric series and $$\mathbf F(a,b,c,z)=F(a,b,c,z)/\Gamma(c)$$ is the regularized Gauss hypergeometric function.

Conjecture: Under the specified restrictions on the parameters, $$|\tilde g_{n,\omega}(z,\nu)|$$ is nondecreasing in $$z$$ and is strictly increasing in $$z$$ when $$n-\omega-\nu\notin\Bbb N$$.

I am seeking to solve this conjecture but have been unsuccessful so far and am turning to the SE math community for help in finishing the proof. See below for what I have tried as a potential path forward.

• I think numerical evidence would disprove the conjecture, rather than just provide evidence that it is false, right? Dec 27 '19 at 17:02
• @mathworker21 Yes. I meant to say I couldn't find a set of parameters that yielded a (numerical) counterexample. Derivative appears to always be positive. Dec 27 '19 at 18:23

Let, $${_1F_0}(1;-;z)_\nu:=\frac{1-z^{\nu+1}}{1-z}=(\nu+1)z^{\nu+1}\mathbf F(1,\nu+2,2,1-z),$$ where $$\mathbf F(a,b,c,z)=F(a,b,c,z)/\Gamma(c)$$ is the regularized Gauss hypergeometric function. Then for $$n\in\Bbb N$$, $$\omega=0,1,\dots,n$$, $$z>0$$, and $$\nu>0$$ $$\tilde g_{n,\omega}(z,\nu):=\frac{z^{n-\omega}\partial_z^n z^\omega {_1F_0}(1;-;z)_\nu}{{_1F_0}(1;-;z)_\nu},$$ is nondecreasing in $$z$$ when $$n-\omega-\nu\in\Bbb N$$ and strictly increasing in $$z$$ otherwise.
Using differentiation formulas for the hypergeometric function $$\tilde g$$ can be expressed in closed-form as $$\tilde g_{n,\omega}(z,\nu)=\frac{n!(\omega+\nu+1)^{(n+1)}}{\nu+1}\frac{\mathbf F(1,\omega+\nu+2;n+2;1-z)}{\mathbf F(1,\nu+2,2,1-z)}.$$ where $$(a)^{(n)}=\Gamma(a+1)/\Gamma(a-n+1)$$ is the falling factorial. It is trivial to show that $$\tilde g_{n,\omega}(z,\nu)$$ is nondecreasing for the special case $$n-\omega-\nu\in\Bbb N$$ since $$(\omega+\nu+1)^{(n+1)}=0$$ under such conditions.
Now assuming $$n-\omega-\nu\notin\Bbb N$$ the leading constant term is nonzero. Furthermore, by way of the integral representation $$\mathbf F(a,b;c;z)=\int_0^1\frac{t^{b-1}(1-t)^{c-b-1}(1-zt)^{-a}}{\Gamma(b)\Gamma(c-b)}\,\mathrm dt,\quad |\operatorname{ph}(1-z)|<\pi,\ \Re c>\Re b>0,$$ each hypergeometric term can be shown to be strictly positive over the entire parameter space such that $$|\tilde g_{n,\omega}(z,\nu)|=\left\lvert\frac{n!(\omega+\nu+1)^{(n+1)}}{\nu+1}\right\rvert\frac{\mathbf F(1,\omega+\nu+2;n+2;1-z)}{\mathbf F(1,\nu+2,2,1-z)}.$$ For a strictly positive function $$f(z)$$, $$\operatorname{sgn}(\partial_z\log f)=\operatorname{sgn}(\partial_z f)$$; thus, we begin by evaluating the logarithmic derivative of $$|\tilde g_{n,\omega}|$$ yielding $$\partial_z\log |\tilde g_{n,\omega}(z,\nu)|=\Psi_{0,0}(z,\nu)-\Psi_{n,\omega}(z,\nu),$$ where $$\Psi_{n,\omega}(z,\nu)=(\omega+\nu+2) \frac{\mathbf F(2,\omega+\nu+3;n+3;1-z)}{\mathbf F(1,\omega+\nu+2;n+2;1-z)}.$$ For brevity, let $$\beta=n+1$$ and $$\gamma=\omega+\nu+2$$, then \begin{aligned} \Psi_{n,\omega}(z,\nu)% &=\gamma\frac{\mathbf F(2,\gamma+1;\beta+2;1-z)}{\mathbf F(1,\gamma;\beta+1;1-z)}\\ &=\frac{\gamma}{z}\int_0^1\frac{zx}{1+(z-1)x}\frac{(1-x)^{\beta-1}(1+(z-1)x)^{-\gamma}}{\operatorname B(1,\beta)F(1,\gamma,\beta+1,-(z-1))}\,\mathrm dx\\ &=\frac{\gamma}{z}\,\mathsf E\left(\frac{zX}{1+(z-1)X}\right), \end{aligned} where $$X\sim\operatorname{GH}(1,\beta,\gamma,z-1)$$ is distributed according to the Gauss hypergeometric distribution $$f(x;\alpha,\beta,\gamma,\xi)=\frac{x^{\alpha-1}(1-x)^{\beta-1}(1+\xi x)^{-\gamma}}{\operatorname B(\alpha,\beta)F(\alpha,\gamma;\alpha+\beta;-\xi)},\quad 0 as defined for $$\alpha,\beta>0$$, $$\gamma\in\Bbb R$$, and $$\xi>-1$$. For mathematical convenience define $$W=zX(1+(z-1)X)^{-1}$$ such that $$\Psi_{n,\omega}(z,\nu)=\gamma/z\,\mathsf E W$$. Since $$W$$ is a monotone increasing transformation of $$X$$ for all $$z>0$$ we have $$F_W(w)=\mathsf P(zX(1+(z-1)X)^{-1}\leq w)= \mathsf P(X\leq (z(w^{-1}-1)+1)^{-1}),$$ which after a significant amount of work yields $$F_W(w)=1-\frac{\operatorname B_{(1-z)(1-w)}(\beta,\gamma-\beta)}{\operatorname B_{1-z}(\beta,\gamma-\beta)},\quad 0 where $$\operatorname B_s(\alpha,\beta)$$ is the incomplete beta function. Differentiating the derived cdf w.r.t. $$w$$ then provides the density in the form $$f_W(w)=\frac{(1-w)^{\beta-1}(1-(1-z)(1-w))^{\gamma-\beta-1}}{(1-z)^{-\beta}\operatorname B_{1-z}(\beta,\gamma-\beta)}.$$ Now let $$\delta>0$$ and consider the family of density functions $$f_\beta=\{f_W(w|\beta):\beta\geq 1\}$$. We have for the likelihood ratio $$\frac{f_{\beta+\delta}}{f_\beta}(w)=\frac{(1-z)^\delta\operatorname B_{1-z}(\beta,\gamma-\beta)}{\operatorname B_{1-z}(\beta+\delta,\gamma-\beta-\delta)}\left(\frac{1-w}{1-(1-w)(1-z)}\right)^\delta:= C (h(w,z))^\delta,$$ where the constant $$C$$ is easily shown to be strictly positive. Differentiating the likelihood ratio w.r.t. $$w$$ yields $$\partial_w\frac{f_{\beta+\delta}}{f_\beta}(w)=-\delta C\frac{(1-w)^{\delta-1}}{(1-(1-w)(1-z))^{\delta+1}},$$ which is strictly negative on $$w\in(0,1)$$; thus, the family of densities $$f_\beta$$ admits a strictly decreasing monotone likelihood ratio. It follows that $$\partial_w\frac{f_{\beta+\delta}}{f_\beta}(w)<0\implies F_{\beta}(w)\mathsf E W_{\beta+\delta},$$ which upon recalling $$\beta=n+1$$ shows $$\partial_n\mathsf E W<0\implies \operatorname{sgn}(\partial_n\Psi_{n,\omega})=\operatorname{sgn}(\gamma/z\,\partial_n\mathsf E W)=-\operatorname{sgn}(\gamma).$$ With this result, suppose we impose the constraint $$\nu>-2\implies\gamma>0$$. It follows that $$\Psi_{n,\omega}(z,\nu)>0$$ and $$\partial_n\Psi_{n,\omega}(z,\nu)<0$$; hence, for $$\nu>-2$$ we have proven $$\Psi_{n,\omega}(z,\nu)<\Psi_{\omega,\omega}(z,\nu).$$ Next, we consider the behavior of $$\Psi_{\omega,\omega}(z,\nu)$$ in $$\omega$$. To aid in the following calculations we introduce the operators \begin{aligned} \mathcal A_1^k F(a_1,a_2;a_3;s) &=F(a_1+k,a_2;a_3;s),\\ \mathcal A_2^k F(a_1,a_2;a_3;s) &=F(a_1,a_2+k;a_3;s),\\ \mathcal A_3^k F(a_1,a_2;a_3;s) &=F(a_1,a_2;a_3+k;s), \end{aligned} for which $$\mathcal A_i^0=\mathcal I$$ is the identity. In terms of these operators we then have $$\Psi_{\omega,\omega}(z,\nu)% =\frac{\frac{\omega+\nu+2}{\omega+2}\mathcal A_1\mathcal A_2\mathcal A_3F(1,\omega+\nu+2;\omega+2;1-z)}{F(1,\omega+\nu+2;\omega+2;1-z)},$$ which upon application of the identities Eqs. $$10$$,$$13$$ $$\begin{gather*} \mathcal A_1=\mathcal I+\frac{a_2}{a_3}s\mathcal A_1\mathcal A_2\mathcal A_3,\\ \mathcal A_1^{-1}=\frac{a_1(s-1)}{a_1-a_3}\mathcal A_1+\frac{2a_1+(a_2-a_1)s-a_3}{a_1-a_3}\mathcal I, \end{gather*}$$ permits us to write $$\Psi_{\omega,\omega}(z,\nu)% =\frac{1}{1-z}\left(\frac{\omega+1}{z F(1,\omega+\nu+2;\omega+2;1-z)}+\frac{1}{z}\left((\omega+\nu+1)(1-z)-\omega\right)-1\right).$$ Introducing the relations Eq. $$07.23.03.0122.01$$. $$F(1,\beta;\gamma;s)=(\gamma-1)s^{1-\gamma}(1-s)^{-(\beta-\gamma+1)}\operatorname B_s(\gamma-1,\beta-\gamma+1).$$ and Eq. $$06.19.20.0003.01$$ $$\partial_\alpha\operatorname B_s(\alpha,\beta)=\operatorname B_s(\alpha,\beta)\log s-\frac{s^\alpha}{\alpha^2}{_3F_2}(1-\beta,\alpha,\alpha;\alpha+1,\alpha+1;s).$$ we are able to derive after a significant amount of work $$\partial_\omega\Psi_{\omega,\omega}(z,\nu)=\frac{1}{1-z}\left(\frac{{_3F_2}(\nu,\omega+1,\omega+1;\omega+2,\omega+2;1-z)}{z^{\nu+2}F(1,\omega+\nu+2;\omega+2;1-z)^2}-1\right).$$ Using the integral representation of the generalized hypergeometric function in Eq. $$16.5.2$$ this result is equivalent to the expected value $$\partial_\omega\Psi_{\omega,\omega}(z,\nu)=\frac{1}{1-z}\mathsf E\left(\frac{h(X)}{h(1)}-1\right),$$ where $$h(X)=(1-(1-z)X)F(1,\omega+\nu+2;\omega+2;(1-z)X)$$ and $$X\sim\operatorname{GH}(\omega+1,1,-\nu,z-1)$$. Upon evaluating the derivative $$\partial_X h(X)=(1-z)\frac{\nu}{\omega+2}F(2,\omega+\nu+2;\omega+3;(1-z)X),$$ and noting that $$0, $$\forall X\in(0,1)$$ we observe that if $$\nu>0$$ then \begin{aligned} \partial_X h(X)% \begin{cases} >0, &01. \end{cases}% &\implies h(X)/h(1) \begin{cases} <1, &01, &z>1. \end{cases}\\ &\implies\frac{(h(X)/h(1)-1)}{1-z}<0,\quad \forall z\in\Bbb R^+\setminus\{1\}\\ &\implies\partial_\omega\Psi_{\omega,\omega}(z,\nu)<0,\quad \forall z\in\Bbb R^+\setminus\{1\}. \end{aligned} Similarly, at the boundary point $$z=1$$ we find $$\Psi_{\omega,\omega}(1,\nu)=1+\frac{\nu}{\omega+2}\implies \partial_\omega \Psi_{\omega,\omega}(1,\nu)=-\frac{\nu}{(\omega+2)^2}<0.$$ Consequently, if $$\nu>0$$, then $$\Psi_{\omega,\omega}(z,\nu)<0$$ and \begin{aligned} \Psi_{n,\omega}(z,\nu)<\Psi_{\omega,\omega}(z,\nu)<\Psi_{0,0}(z,\nu)% &\implies \partial_z\log |\tilde g_{n,\omega}(z,\nu)|>0\\ &\implies \partial_z|\tilde g_{n,\omega}(z,\nu)|>0,\\ \end{aligned} which completes the proof. $$\quad\quad\square$$