Monotone likelihood ratio of a family of densities with compact support

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 $$\phi=\pi/2$$.

• What do you mean by "decreasing or increasing depending on the concavity or convexity of $g$"? Decreasing for concave $g$ and increasing for convex $g$? Commented Aug 11, 2023 at 20:01
• @Iosif Edited to clarify. Commented Aug 12, 2023 at 0:12

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

• 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? Commented Aug 13, 2023 at 16:27
• I guess the inverse of that function also disproves the concave case. It must need a 3rd derivative condition. Commented Aug 13, 2023 at 16:50
• @japalmer : The "concave" part of your conjecture has now considered as well. Commented Aug 13, 2023 at 16:58
• 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. Commented Aug 14, 2023 at 1:38
• So the new conjecture is that it holds with the additional condition that $g^{(3)}(x) \ge 0$ for $x > 0$. Commented Aug 14, 2023 at 1:58