Let $f(x) = \log(\cosh(x))$, and define the kernel density:

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

Show that for all $r>0$, $p_r(\phi;\theta)$ has a decreasing likelihood ratio. I.e., for $0\le\theta_1 < \theta_2\le\pi/2$, the function:

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

is decreasing on $[0,\pi/2]$, and:

$$ \frac{\partial}{\partial \phi} \frac{\partial}{\partial \theta}\hspace{2pt} \log p_r(\phi;\theta) \le 0,\quad 0 \le \phi,\theta\le \mbox{$\large\frac{\pi}{2}$}$$

If we define: $$g_1\hspace{-1pt}(r,x) = rx\tanh(rx)-r^2(1-x^2)\big(1-\tanh^2(rx)\big) $$ and: $$g_2\hspace{-0.5pt}(r,x) = r^2(1-x^2)\tanh^2(rx) $$

then the result will be true if, for $r>0$ and $0\le v < u \le 1$:

$$\big(g_1\hspace{-1pt}(r,u)+g_1\hspace{-1pt}(r,v)\big)\Big(\hspace{-1pt}\log\hspace{-1pt}\big(\hspace{-1.5pt}\cosh(ru)\big)-\log\hspace{-1pt}\big(\hspace{-1.5pt}\cosh(rv)\big)\hspace{-1pt}\Big) +g_2\hspace{-0.5pt}(r,u)-g_2\hspace{-0.5pt}(r,v) \;<\; 0$$

This function is actually monotonic in $u$ and $v$. Since it is 0 when $u=v$ it suffices to show that the derivative with respect to $v$ is non-negative. This results in an expression in which the multiplier of $\log(\cosh(u))-\log(\cosh(v))$ is non-negative. The bound:

$$\log(\cosh(ru))-\log(\cosh(rv)) \ge \sqrt{r^2u^2+1}-\sqrt{r^2v^2+1}$$

can then be used to eliminate the $\log(\cosh())$ resulting in a function with only algebraic and $\tanh$ terms that is positive for $u,v,r>0$. However it is not clear how to show this.

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 variation diminishing property of MLR densities.