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(Asking a final time 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 $f(x)=g(x^2)$, 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 $\phi=\pi/2$.

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$\newcommand{\ep}{\varepsilon}$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. $\quad\Box$

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  • $\begingroup$ Damn. That function apparently with appropriate exponent seems to work for any finite number of derivative conditions. I guess it requires absolutely monotonic or Bernstein functions. I'll work on it before addressing it here again. $\endgroup$
    – japalmer
    Commented Aug 15, 2023 at 16:19
  • $\begingroup$ Actually not absolutely monotonic, as $|x|^p$ works for $p>2$. Maybe just univalent derivatives of all orders. $\endgroup$
    – japalmer
    Commented Aug 15, 2023 at 16:21
  • $\begingroup$ Last resort conjecture: $g$ is a Bernstein function in the concave case, or $g$ is an $m$th integral of a Bernstein function, or a scale mixture of $\cosh(x)$ in the convex case. $\endgroup$
    – japalmer
    Commented Aug 15, 2023 at 17:05
  • $\begingroup$ @japalmer : This will probably be hard to prove for a general class of functions. You may want to try different ones using my Mathematica notebook. $\endgroup$ Commented Aug 15, 2023 at 20:35
  • $\begingroup$ I think it will be possible using the Levy-Khinchine representation of Bernstein functions. The $m$th integral of the Levy-Khinchine kernel can be calculated as well. Proving that these kernels have monotone likelihood ratio, and then that the scale mixing density also has MLRP should do it. $\endgroup$
    – japalmer
    Commented Aug 16, 2023 at 3:26

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