In order to prove the following Probability result:
"Find a family of random variables $X$, having pdf $f$, such that $X$ and $Y=f(X)$ have the same distribution."
Let's call $F(x)$ the CDF of the random variable $X$ and $f(x)$ its PDF. Furthermore, the random variable $Y$ has CDF $K(y)$ and PDF $k(y)$.
I know that $y$ belongs to a set of values in which it can be considered valid the relation $y=f(x)$.
I'd like to "investigate" when the random variables $X$ and $Y$ have the same distribution, in the case of $f$ stictly increasing for $x\leq m$ and stricly decreasing for $x \geq m$ (where $m$ is the mode of the distribution).
So we have to prove when $F(y)=K(y)$.
From a previous valid result, I know that, under these assumptions, the CDF of the random variable $Y$ is:
$K(y)=2 F(f^{-1}(y))=2F(x)$ or, equivalently, $K(y)=2(1-F(f^{-1}(y))=2-2F(x)$.
In order to find the explicit expression of the PDF $f$, using the relation $F(y)=K(y)$ (our thesis), I should prove that:
- in the case $f$ strictly increasing for $x \leq m$,
$$K(y)=2F(x)=F(y) \Leftrightarrow 2F(x)=F(f(x)) \Leftrightarrow 2F(x)=F(F'(x)).$$
So the first functional differential equation that should be solved is:
$$2F(x)=F(F'(x))$$ in order to find the expression of $F$ and then obtain $f$, derivating $F$.
- in the case of $f$ strictly decreasing for $x \geq m$
$K(y)=2-F(x)=F(y) \Leftrightarrow 2-F(x)=F(f(x)) \Leftrightarrow 2-F(x)=F(F'(x))$.
The second functional differential equation to be solved is the following one: $$F(F'(x))+F(x)=2.$$
The solutions of the aforesaid equations should be in a certain way "symmetrical".
I should solve those functional differential equations, but I have NO IDEA on HOW to solve them because I've never met this kind of equations before in my career.
Could you help me, please?