Let $f_X(x)$ be the probability density function of a real random variable $X$, $X\geq0$. Define $Y=\mathrm{ln}(X+1)$ and denote the PDF of $Y$ as $f_Y(y)$. I'm curious when will $f_X=f_Y$?
I've tried to use the formula for functions of random variables. The distribution $f_X$ seems need to satisfy: (call $f_X=f$ to save the subscript)
$$
f(x) = f(e^x-1)e^x \quad\quad x\geq0
$$
And the antiderivative $F$ of $f$ will satisfy:
$$
F(x)=F(e^x-1) \quad\quad x\geq0
$$
I was unable to solve the above equations. Could you please help me? I apologize if I made mistakes on the problem statement or the derivation.
Note that $\ln(x+1) \le x$, with equality only at $x=0$. For any $y > 0$, $$\mathbb P(X < y) = \mathbb P(\ln(X+1) < \ln(y+1)) = \mathbb P(X < \ln(y+1))$$ which says $\mathbb P(\ln(y+1) \le X < y) = 0$. From that you can conclude that $X = 0$ with probability $1$.