I'm looking for a continuous distribution for $x \in R$, such that after applying the uni-variate transformation for $y = \exp\{ax+b\} / (1 + \exp\{ax+b\})$, the resultant PDF $f_Y(y)$ becomes non-increasing in $y$.
I already tried a normal distribution for $x$, but $f_Y(y)$ turned out to be non-decreasing in $y$. Rather than a trial-and-error approach, is there any mechanism that I can narrow down my search for an appropriate distribution for $x$?
$$\frac{dy}{dx}=\frac{a\exp[ax+b]}{\left(1+\exp[ax+b]\right)^2}$$ $$f_Y(y)=f_X(x)\left|\frac{dx}{dy}\right|$$ $$\frac{df_Y(y)}{dy}=\frac{d}{dx}\left(f_X(x)\left|\frac{dx}{dy}\right|\right)\frac{dx}{dy}\le 0$$
$$\left[f'_X(x)\left|\frac{dx}{dy}\right|+f_X(x)\frac{d}{dx}\left|\frac{dx}{dy}\right|\right]\frac{dx}{dy} \le 0$$ $$\left[f'_X(x)\frac{\left(1+\exp[ax+b]\right)^2}{|a|\exp[ax+b]}+f_X(x)\frac{d}{dx}\frac{\left(1+\exp[ax+b]\right)^2}{|a|\exp[ax+b]}\right]\frac{\left(1+\exp[ax+b]\right)^2}{a\exp[ax+b]}\le 0$$ $$\left[f'_X(x)\frac{\left(1+\exp[ax+b]\right)^2}{\exp[ax+b]}+f_X(x)\frac{d}{dx}\frac{\left(1+\exp[ax+b]\right)^2}{\exp[ax+b]}\right]\frac{1}{a|a|}\le 0$$ $$\frac{1}{a|a|}\bigg\{f'_X(x)\frac{\left(1+\exp[ax+b]\right)^2}{\exp[ax+b]}$$ $$+f_X(x)\left[\frac{2(1+\exp[ax+b])a\exp[ax+b]}{\exp[ax+b]}-\frac{(1+\exp[ax+b])^2}{(\exp[ax+b])^2}\exp[ax+b]a\right]\bigg\}\le 0$$ $$\frac{1}{a}\bigg\{f'_X(x)\left(1+\exp[ax+b]\right)^2$$ $$+f_X(x)\left[2a\exp[ax+b]\left(1+\exp[ax+b]\right)-a\left(1+\exp[ax+b]\right)^2\right]\bigg\} \le 0$$ $$\frac{1}{a}\left\{f'_X(x)+f_X(x)\frac{a\exp[ax+b]-a}{1+\exp[ax+b]}\right\} \le 0$$ $$\frac{1}{a}\left[f'_X(x)+af_X(x)\tanh\frac{ax+b}{2}\right] \le 0$$ $$\frac{1}{a}\left[\frac{f'_X(x)}{f_X(x)}+a\tanh\frac{ax+b}{2}\right] \le 0$$ $$\frac{1}{a}\left[\frac{d}{dx}\ln f_X(x)+a\tanh\frac{ax+b}{2}\right] \le 0$$
Case 1: If $a>0$, $$\frac{d}{dx}\ln f_X(x)+a\tanh\frac{ax+b}{2} \le 0$$ $$\frac{d}{dx}\ln f_X(x) \le -a\tanh\frac{ax+b}{2}$$ $$\ln f_X(x)-\ln f_X(x_0) \le \int_{x_0}^x -a\tanh\frac{ax+b}{2} dx$$ $$=-2\int_{x_0}^x \tanh\frac{ax+b}{2}d\frac{ax+b}{2}$$ $$=-2\ln\frac{\cosh\frac{ax+b}{2}}{\cosh\frac{ax_0+b}{2}}$$ $$f_X(x) \le f_X(x_0)\frac{\cosh^2\frac{ax_0+b}{2}}{\cosh^2\frac{ax+b}{2}}$$
Case 2: If $a<0$, similarly, $$f_X(x) \ge f_X(x_0)\frac{\cosh^2\frac{ax_0+b}{2}}{\cosh^2\frac{ax+b}{2}}$$
If one chooses $x_0=-b/a$, then for $a>0$ $$f_X(x) \le f_X(-b/a)\cosh^{-2}\frac{ax+b}{2}$$ and for $a<0$, $$f_X(x) \ge f_X(-b/a)\cosh^{-2}\frac{ax+b}{2}$$