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Define the family of densities:

$$p(\phi;\theta) = \Big(f\big(\hspace{-1pt}\cos(\phi-\theta)\big) - f\big(\hspace{-1pt}\cos(\phi+\theta)\big)\Big)\hspace{0.5pt} \frac{\sin(2\phi)}{\sin(2\theta)}, \quad 0 \le \phi,\theta\le \pi/2$$

where $f(x)=g(x^2)$ with $g$ non-negative, increasing, and convex or concave, on $[0,\infty)$. (Interestingly these densities indeed have the same measure for all $\theta$ whenever $g$ is concave or convex, which can be shown by an integral representation.)

Edit: As shown in the answer, these conditions are not sufficient. I believe it is true though if additionally $g^{(3)}(x) \ge 0$, for $x>0$.

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

$$ h(\phi) = \frac{f\big(\hspace{-1pt}\cos(\phi-\theta_2)\big) - f\big(\hspace{-1pt}\cos(\phi+\theta_2)\big)}{f\big(\hspace{-1pt}\cos(\phi-\theta_1)\big) - f\big(\hspace{-1pt}\cos(\phi+\theta_1)\big)}$$

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. $\log(\cosh(\sqrt{x}))$ is concave on $[0,\infty)$), and twice differentiable (unlike $|x|^p$, $p<2$).

This result is important to prove uniqueness of stable optima in blind source separation and deconvolution with strongly sub- and super-gaussian input 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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  • $\begingroup$ What do you mean by "decreasing or increasing depending on the concavity or convexity of $g$"? Decreasing for concave $g$ and increasing for convex $g$? $\endgroup$ Commented Aug 11, 2023 at 20:01
  • $\begingroup$ @Iosif Edited to clarify. $\endgroup$
    – japalmer
    Commented Aug 12, 2023 at 0:12

1 Answer 1

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$\newcommand{\ep}{\varepsilon}$This conjecture is not true in general.

Indeed, suppose the "convex" part of your conjecture 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 $g$ 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 real $\ep>0$ and real $u$, let $u_{+;\ep}:=\frac12(u+\sqrt{\ep^2+u^2})$, an "$\ep$-smoothed" version of $u_+:=\max(0,u)$. For $c$ and $c_*$ in $[0,\infty)$, let $g_{c_*,\ep}(c):=(c-c_*)_{+;\ep}$.

Then the function $g_{c_*,\ep}$ is strictly increasing, convex, and smooth on $[0,\infty)$. However, $h_2(g;x,t)=-44051.358\ldots\not\ge0$ if $g=g_{c_*,\ep}$, $c_*=\frac12$, $\ep=\frac1{1000}$, $x=\frac{39}{100}$, and $t=\frac{118}{100}$. So, the "convex" part of your conjecture is not true in general.


Suppose now the "concave" part of your conjecture is true. Then for any strictly increasing concave smooth function $g$ and all $x$ and $t$ in $(0,\pi/2)$ we would have $h_2(g;x,t)\le0$.

For $c$ and $c_*$ in $[0,\infty)$, let $G_{c_*,\ep}(c):=c-\sqrt{\ep^2+(c-c_*)^2}$.

Then the function $G_{c_*,\ep}$ is strictly increasing, concave, and smooth on $[0,\infty)$. However, $h_2(G;x,t)=32614.565\ldots\not\le0$ if $G=G_{c_*,\ep}$, $c_*=\frac12$, $\ep=\frac1{1000}$, and $x=\frac{39}{100}=t$. So, the "concave" part of your conjecture is not true in general either. $\quad\Box$

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  • $\begingroup$ Wow, that's not good. I thought it applied in the convex case as well. But I'm pretty sure I can show that it holds for concave $g$. Do you have a counter-example for that too? $\endgroup$
    – japalmer
    Commented Aug 13, 2023 at 16:27
  • $\begingroup$ I guess the inverse of that function also disproves the concave case. It must need a 3rd derivative condition. $\endgroup$
    – japalmer
    Commented Aug 13, 2023 at 16:50
  • $\begingroup$ @japalmer : The "concave" part of your conjecture has now considered as well. $\endgroup$ Commented Aug 13, 2023 at 16:58
  • $\begingroup$ Yes, I checked it by inverting the convex function. It's strange because I'm pretty sure I can prove that if $g$ is concave or convex and $a(\phi)$ is montonic, then $b(\theta) = \int_0^{\pi/2} a(\phi) p(\phi;\theta) d\phi$ is monotonic. (reversed for concave $g$). I thought this was a stronger property given a theorem by Karlin on TP(3) kernels having that property, which is stronger than the SR(2) property, which is equivalent to the monotone likelihood ratio. $\endgroup$
    – japalmer
    Commented Aug 14, 2023 at 1:38
  • $\begingroup$ So the new conjecture is that it holds with the additional condition that $g^{(3)}(x) \ge 0$ for $x > 0$. $\endgroup$
    – japalmer
    Commented Aug 14, 2023 at 1:58

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