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Let $\tilde{X}_0$ be some random variable on $\mathbb{R}^n$, with a strictly positive p.d.f..

Define: $$X_0:=(\operatorname{var}{\tilde{X}_0})^{-\frac{1}{2}}(\tilde{X}_0-\mathbb{E}\tilde{X}_0),$$ where we take the unique positive definite matrix square root.

Further, for all $k\in\mathbb{N}$ define: $$\tilde{X}_{k+1}:=\Phi^{-1}_n(F_{X_k}(X_k)),$$ and: $$X_{k+1}:=(\operatorname{var}{\tilde{X}_k})^{-\frac{1}{2}}\tilde{X}_k,$$ where for all $z\in\mathbb{R}^n$: $$\Phi_n(z)=[\Phi(z_1),\dots,\Phi(z_n)],$$ and:$$F_{X_k}(z)=[F_{X_{k,1}}(z_1),\dots,F_{X_{k,n}}(z_n)],$$ where $\Phi$ is the CDF of a standard Normal distribution, and for $i\in\{1,\dots,n\}$, $F_{X_{k,i}}$ is the CDF of the $i^\mathrm{th}$ component of $F_{X_k}$.

It is easy to see that for all $k\in\mathbb{N}$, $X_k$ is mean zero and has covariance given by the identity matrix, and that $\tilde{X}_{k+1}$ has standard Normal marginals.

I would like to prove that there is some $X$ such that $X_k$ converges in distribution to $X$ as $k\rightarrow\infty$, and (ideally) such that $\tilde{X}_k$ also converges in distribution $X$ as $k\rightarrow\infty$.

The Banach fixed point theorem cannot be applicable as the mapping has more than one fixed point. E.g. with $n=2$, both the bivariate standard Normal distribution and the distribution with p.d.f. $(x,y)\mapsto \frac{1}{2}(1-\Phi(\max(|x|,|y|)))$ are fixed points.

Can you prove the convergence to a fixed point?

cfp
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    (1) What definition of the square root of a matrix do you use? (2) I don't see how your claim at "it is easy to see" could be generally true. When the underlying distribution is discrete it's not possible for the new variable's marginals to be Normal, because they will be discrete, too. – whuber Feb 24 '18 at 22:08
  • Fair point. I've added the restriction that $\tilde{X}_0$ must have full support, and I've clarified the definition of a matrix square root. – cfp Feb 25 '18 at 10:12
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    You will need to modify "full support" to (at a minimum) *continuous distribution* for this to have any hope of being true. – whuber Feb 25 '18 at 15:10
  • Yep. Sorry, in my mind the set-up was clear, but I clearly didn't write it down correctly. I'm not assuming something stronger, the existence of a positive p.d.f.. – cfp Feb 25 '18 at 16:01
  • Made a further more substantial edit to say that the Banach fixed point theorem is not an option, as there are certainly multiple fixed points. – cfp Feb 25 '18 at 17:36

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