Assume we know the distributions of $X $and $h(X)$, how to find $h$?

Question

Let $X$ be a random variable in $\mathbb{R}$

Let $h:\mathbb{R} \rightarrow \mathbb{R}$ be an unknown function

Assume we know the distribution of $X $ and the distribution of $h(X)$

Can one find $h$ ?

Answer 1

No, we cannot. Consider if $X$ and $h(X)$ are both uniform on $[0,1]$. Then we might have $h(x)=x$, we might have $h(x)=1-x$, and we can't tell which. If we allow discontinuous $h$, then there are many functions we could have, starting with $$ h(x)=\cases{x+\frac12&if $x\leq\frac12$\\1-x&if $x>\frac12$} $$ and only getting uglier from there.

Answer 2

As made clear by the answer of Arthur $h$ cannot always be found. Extra conditions are needed.

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First a note that will be used below.

For a CDF $F$ we can define $\Phi_F:(0,1)\to\mathbb R$ by stating:$$\Phi_F(u):=\inf\{x\mid F(x)\geq u\}$$

Then $\Phi_F(F(x))\leq x$ for every $x\in\mathbb R$ and in the special case that $F$ is strictly increasing we find: $$\Phi_F\circ F=\mathsf{id}_{\mathbb R}\tag1$$

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Let it be that $h$ and $F_X$ are both strictly increasing and let it be that the range of $h$ is not bounded below and is not bounded above. Then $F_{h(X)}$ is also strictly increasing.

Further we have: $$F_{h(X)}(h(x))=P(h(X)\leq h(x))=P(X\leq x)=F_X(x)$$

or shortly:$$F_{h(X)}\circ h=F_X$$

Since $F_{h(X)}$ is strictly increasing then we can apply $(1)$ and find:$$h=\Phi_{F_{h(X)}}\circ F_X$$

I do not exclude that the conditions can be weakened.