Monotone likelihood ratio of a family of densities with convexity property

Question

(Asking again in a new question because the previous version had insufficient conditions, as pointed out in the answer there.)

Define the densities:

$$p(\phi;\theta,r) = \Big(f\big(r\cos(\phi-\theta)\big) - f\big(r\cos(\phi+\theta)\big)\Big)\hspace{0.5pt} \frac{\sin(2\phi)}{\sin(2\theta)}, \quad 0 \le \phi,\theta\le \pi/2,\,r>0$$

where $f(x)=g(x^2)$, and the following hold: $g$ is twice-differentiable, increasing, and convex or concave on $(0,\infty)$, $g''(x)$ is non-decreasing on $(0,\infty)$, and $f'(0^+) < \infty$. Remarkably, the area of these densities is independent of $\theta$ whenever $g$ is increasing and concave or convex (without requiring the monotonicity of $g''(x)$), which can be shown using an integral representation. For $f(x)=|x|$, we have $\int_0^{\pi/2} p(\phi;\theta,r)d\phi=\frac{2}{3}r$.

Show that for all $r$, $p(\phi;\theta,r)$ has a monotone likelihood ratio (decreasing for concave $g$, increasing for convex $g$). I.e., for $0\le\theta_1 < \theta_2\le\pi/2$:

$$ h(\phi) = \frac{f\big(r\cos(\phi-\theta_2)\big) - f\big(r\cos(\phi+\theta_2)\big)}{f\big(r\cos(\phi-\theta_1)\big) - f\big(r\cos(\phi+\theta_1)\big)}$$

is monotonic on $[0,\pi/2]$.

Examples of functions are $f(x) = |x|^p$, $1\le p<2$, or for $p>2$. (For $p=2$, $p(\phi;\theta) = \sin^2(2\phi)$.) And the function $f(x) = \log( \cosh(x))$, which is concave in $x^2$, i.e. $g(x) = \log(\cosh(\sqrt{x}))$ is concave on $[0,\infty)$, and twice differentiable at $x=0$, unlike $|x|^p$, $p<2$.

An answer showing the this holds for either convex or concave $g$ is acceptable. This question can be considered as only asking for one of them.

This result is important to prove uniqueness of stable optima in the unmixing and deconvolution of linear mixtures of independent random variables with strongly sub- and super-gaussian densities, using the Karlin-Rubin theorem. The result can be proved for $f(x)=x^4$ by simplifying the derivative expression. This corresponds to using kurtosis as the the cost function, and the uniqueness result is already known in this case. For $f(x) = |x|$, the likelihood ratio is non-increasing, constant around $\phi=0$ and $\phi=\pi/2$.

Answer 1

$\newcommand{\ep}{\varepsilon}$The "convex" part of this conjecture is not true in general.

Indeed, suppose it is true. Then (letting $x:=\phi$, $t:=\theta_1$, and $\theta_2\downarrow\theta_1=t$) we see that for any strictly increasing convex smooth function $g$ with $g'''\ge0$ and all $x$ and $t$ in $(0,\pi/2)$ we would have $h_2(g;x,t):=\partial_x\partial_t\,\ln(g(\cos^2(x-t))-g(\cos^2(x+t))\ge0$. (Note that for all $x$ and $t$ in $(0,\pi/2)$ we have $\cos^2(x-t)-\cos^2(x+t)=\sin2x\,\sin2t>0$, so that $h_2(g;x,t)$ is well defined.) For $c$ and $c_*$ in $[0,\infty)$ and real $\ep>0$, let $g(c):=g_{c_*,\ep}(c):=(\sqrt{(c-c_*)^2+\ep^2}+c-c_*)^2$.

Then the function $g$ is strictly increasing, convex, and smooth on $\mathbb R$, and $g'''>0$. However, $h_2(g;x,t)=-461586.955\ldots\not\ge0$ if $c_*=\frac{526}{1000}$, $\ep=\frac1{1000}$, $x=\frac{812}{1000}$, and $t=\frac{157}{100}$. So, the "convex" part of your conjecture is not true in general. $\quad\Box$