# Non-affine smooth transformation of Gaussian is Gaussian

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 $$f$$.

• 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
– mike
Apr 19, 2022 at 11:01
• @mike : this function is not differentiable because at x=0 it takes the value negative infinity but takes finite value elsewhere. Apr 19, 2022 at 13:22
• 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? Apr 19, 2022 at 16:55
• @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. Apr 19, 2022 at 20:43
• 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. Apr 20, 2022 at 12:59