Suppose $Z\sim N(0,1)$ (standard Gaussian) and $f: \mathbb{R} \to \mathbb{R}$ is a differentiable function such that $f(Z)\sim N(0,1)$. My question is whether there exists any such $f$ other than $f(x)=x$ or $f(x) =-x$. I know that there exists counter examples if we only assume continuity of $f$. For example, consider the continuous (non-differentiable) function $$f(x) = \Phi^{-1}(F_{Z^3-Z}(x^3-x)),$$ where $F_{Z^3-Z}(\cdot)$ is the CDF of the random variable $Z^3-Z$. The above function is non-differentiable at $x = \pm\sqrt{1/3}$ and $x= \pm \sqrt{4/3}$. Although, under continuity and invertibility assumption on $f$, I am able to show that $f(x)=x$ for $\forall x\in \mathbb{R}$ or $f(x)=-x$ for $\forall x\in \mathbb{R}$.

If it is true that those are the only functions under the differentiability assumption, can we say similar things in higher dimensions up to orthogonal rotation?

Edit 1: An useful fact that could be helpful is:

  1. It must be true that $ \lim_{x \to \infty} f(x) = \infty, \quad \lim_{x\to -\infty}f(x) = -\infty, $ or the vice-versa.

Edit 2: If $f$ is continuously differentiable then $f$ must be $x$ or $-x$. Thanks to the comments of @Jukka Kohonen and @zhoraster, as $f^\prime \neq 0$, by continuity of $f^\prime$ we can conclude that $f^\prime>0$ or $f^\prime <0$. Thus, either way $f$ is monotone and the result follows from the following equation: WLOG, assuming $f$ is increasing, we have $$ \Phi(x) = P(Z\leq x ) = P(f(Z)\leq x) = P(Z\leq f^{-1}(x)) = \Phi(f^{-1}(x)). $$ This implies $f^{-1}(x)=x$, i.e., $f(x)=x$. But the question is still open when we only have differentiability assumption on $f$.

  • $\begingroup$ If you do something like $\Phi^{-1}\chi_{\frac 1 2} (X^2)$ (square X getting a chi-squared, take cumulant transform getting uniform, take inverse cumulant transformation) then it has diffentiability problems ? It doesn't look like it should net to be x or -x $\endgroup$
    – mike
    Commented Apr 19, 2022 at 11:01
  • $\begingroup$ @mike : this function is not differentiable because at x=0 it takes the value negative infinity but takes finite value elsewhere. $\endgroup$
    – De vinci
    Commented Apr 19, 2022 at 13:22
  • 1
    $\begingroup$ If $f'(x)=0$ at some point $x$, then the density of $Z$ at $f(x+h)$ would go to infinity as $h \to 0$, which would contradict $Z$ being standard normal. Would that help? $\endgroup$ Commented Apr 19, 2022 at 16:55
  • $\begingroup$ @JukkaKohonen I am not yet sure whether this fact will help or not. But can you give your reasoning why the density of $Z$ will blow up at $f(x+h)$ as $h\to 0$? I can see this is true when $f$ is invertible and $f^{-1}$ is differentiable. Not sure how the result follows without these assumptions. $\endgroup$
    – De vinci
    Commented Apr 19, 2022 at 20:43
  • $\begingroup$ If $f'(x_0) = 0$, then for any $\epsilon>0$ there exists some $\delta>0$ such that $|f(x)- f(x_0)|<\epsilon |x-x_0|$ whenever $|x-x_0|<\delta$. Therefore, for any $\eta<\delta$, $$\mathrm{P} (|f(X) - f(x_0)|<\epsilon \eta)\ge \mathrm{P}(|X-x_0|<\eta)\sim 2\phi(x_0)\eta, \eta \to 0.$$ On the other hand, $$\mathrm{P} (|f(X) - f(x_0)|<\epsilon \eta)\sim 2\phi(f(x_0))\epsilon\eta, \eta \to 0.$$ This implies that $\phi(f(x_0))\epsilon\ge \phi(x_0)$ for any $\epsilon>0$, which is absurd. $\endgroup$
    – zhoraster
    Commented Apr 20, 2022 at 12:59


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