Is any r.v. standard Gaussian up to reparametrization?

Question

I propose this exercise. Is it true, that given a real r.v. $X$ with a generic (continuous) p.d.f., there is always a real function $f$, s.t. $f(X) \sim N(0,1)$ where $N$ is a standard gaussian function ?

Than same question, in the multivariate case. For example, given $X_i,i=1...N$ r.v., is there always a non-linear transformation $Y_j=f_j(\{X_i\}_{i=1...N}),j=1,...,N$ such that the $Y_j$ are (multidimensional) standard Gaussian ?

This is a question I created during my self-study journey :)

PS: I think it is true but I wanted to check if I am wrong hence the question

Answer 1

If on its support $X$ has CDF strictly increasing $F$ of inverse $F^{-1}$, take $f(x)=\Phi^{-1}(F(x))$. Then$$P(f(X)\le y)=P(X\le F^{-1}(\Phi(y)))=\Phi(y)\implies f(X)\sim N(0,\,1).$$

Answer 2

1. It can not be true in general. For counterexample, let $X_1\sim U(0,1)$ distribution, and $X_2=X_1$. Then there can not be functions $f_1$ and $f_2$ such that $f_1(X_1)$ and $f_2(X_2)$ that are i.i.d standard normal. You can see that by taking two sets $A$ and $B$ such that $f_1^{-1}(A)$ and $f_2^{-1}(B)$ are disjoint and $\mathbb{P}(X_1 \in f_1^{-1}(A))\mathbb{P}(X_1 \in f_2^{-1}(B))>0$.

Details: First note that if $f_1$ and $f_2$ are equal almost everywhere then there is nothing to prove. So we assume that $f_1\neq f_2$ on a set of positive measure.

Now let me first show that: There are $A,B\subseteq \mathbb{R}$ such that $f_1^{-1}(A)$ and $f_2^{-1}(B)$ are disjoint and $\mathbb{P}(X_1 \in f_1^{-1}(A))\mathbb{P}(X_1 \in f_2^{-1}(B))>0$.

Take any two sets $A$ and $B$ with $\mathbb{P}(X_1 \in f_1^{-1}(A))=\mathbb{P}(X_1 \in f_2^{-1}(B))>0$, and write $A':=f_1^{-1}(A)\cap f_2^{-1}(B)^c$ and $B':=f_2^{-1}(B)\cap f_1^{-1}(A)^c$. Clearly $f_1(A')\cap f_2(B')$ is empty. Now, either the firt scenario happens, that is $\mathbb{P}(X_1 \in f_1^{-1}(A'))\mathbb{P}(X_1 \in f_1^{-1}(B'))>0$, in which case we are done or the second scenario happens: $\mathbb{P}(X_1 \in f_1^{-1}(A))=\mathbb{P}(X_1 \in f_2^{-1}(B)) = \mathbb{P}(X_1 \in f_1^{-1}(A) \cap f_2^{-1}(B))$. If the second scenario happens we choose another pair of $A,B$. If the second scenario occurs for all $A,B$ then $f_1=f_2$ almost surely, otherwise we will have the first scenario.

Coming to the main problem: after choosing such a pair $A,B$, assume that $f_1(X_1)$ and $f_2(X_2)$ are independent. Then using independence $$ \mathbb{P}(f_1(X_1) \in A, f_2(X_2) \in B) =\mathbb{P}(X_1 \in f_1^{-1}(A))\mathbb{P}(X_1 \in f_2^{-1}(B)), $$ but by our choice of $A,B$, the left side is $0$ and the right side is positive.

2. It is true if $X_1,X_2,\ldots,X_n$ are independent (but not necesserily identically distributed) random variables, and $X_i$ has strictly increasing CDF $F_i$ for $i=1,2,\ldots,n$. Then following J.G's answer take the $f_i(x)=\Phi^{-1}(F_i(x))$ and then

$$ \mathbb{P}(f_1(X_1)\leq y_1,\ldots,f_n(X_n)\leq y_n) = \prod_{i=1}^n\mathbb{P}(f_i(X_i)\leq y_i) = \prod_{i=1}^n\Phi(y_i). $$ Therefore $f_i(X_i)$'s are i.i.d normal.