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First, let me apologize if the question is not formulated appropriately. I'm a bit rusty in math lately.

Let $X$ be a continuous random variable in $\mathbb{R}^d$, and $N\sim\mathcal{N}(0, \mathbb{1}_d)$ be the standard multidimensional normal distribution. Then, I want to know if:

$\exists H: R^d \rightarrow R^d$ such that:

  • $X = H(N)$
  • $H$ is an homeomorphism.
  • $H$ and $H^{-1}$ are differentiable.

In other words, I want to know if the existence of a continuous, invertible and differentiable mapping between any continuous random variable and the standard normal is guaranteed. I don't need an explicit formal proof, just some pointers of why this is true (if it is) or a counterexample if not.

Álvaro Parafita
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1 Answers1

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Take $d=1, X=N^{2}$. If $X=H(N)$ then $x^{2}=H(x)$ almost everywhere w.r.t. Lebesgue measure. By continuity this implies $H(x)=x^{2}$ for all $x$ so $H$ is not one-to-one.

More generally, if $K$ is any continuous function then $K(N)=H(N)$ implies $K \equiv N$. So if you start with a function $K$ which does not satisfy your requirements you get a counterexample.

Kavi Rama Murthy
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