(Asking a final time in a new question because the [previous version](https://mathoverflow.net/q/452519/36721) had insufficient conditions, as pointed out in the [answer there](https://mathoverflow.net/a/452659/36721).)


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 $f(x)=g(x^2)$, with $g'(x)>0$, $g''(x)>0$, $g'''(x)> 0$, and $g'''(x)$ monotonic on $(0,\infty)$. Remarkably, the area of these densities is independent of $\theta$ whenever $g$ is increasing and convex, which can be shown using an integral representation.


Show that for all $r$, $p(\phi;\theta,r)$ has a monotone likelihood ratio. 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 $f(x) = |x|^p$ for $p>2$ and $f(x)=\cosh(x)$.


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 $\phi=\pi/2$.



Sign Regularity of a Density Kernel with Convexity Properties
$\newcommand{\ep}{\varepsilon}$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 $g'''>0$ and $g''''>0$ 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
$g(c):=g\_{c\_\*,\ep}(c):=(\sqrt{(c-c\_\*)^2+\ep^2}+c-c\_\*)^3$.


Then the function $g$ is strictly increasing, convex, and smooth on $\mathbb R$, and $g'''>0$ and $g''''>0$. However, $h\_2(g;x,t)=-132194.575\ldots\not\ge0$ if $c\_\*=\frac{5}{10}$, $\ep=\frac1{1000}$, $x=\frac{118}{100}$, and $t=\frac{39}{100}$. So, your conjecture is not true in general. $\quad\Box$


