Consider the Markov chain $(\theta_n, \phi_n)$ on $S^1 \times S^1$ constructed in the following way. For $\xi_n$ a sequence of i.i.d. normal random variables and $\kappa > 0$ a fixed number, we set $$ \theta_{n+1} = \theta_n + \kappa \xi_n\;,\qquad \phi_{n+1} = \arg((3+2\sqrt 2)\cos \phi_n,\sin \phi_n) + \theta_n\;, $$ where $arg(v)$ denotes the angle of the vector $v$ with $(1,0)$. For every $\kappa > 0$, it has a unique invariant measure $\mu_\kappa$ and it is obvious that its first marginal $\pi_\theta^* \mu_\kappa$ is just Lebesgue measure $\lambda$ for every $\kappa$.

Of course, one has $\lim_{\kappa \to \infty} \mu_\kappa = \lambda \otimes \lambda$, so that $\lim_{\kappa \to \infty} \pi_\phi^*\mu_\kappa =\lambda$. More surprisingly, it follows from an explicit but lengthy calculation that one also has $\lim_{\kappa \to 0} \pi_\phi^*\mu_\kappa =\lambda$. This begs the question: is it true that $\pi_\phi^*\mu_\kappa =\lambda$ for every $\kappa > 0$? I've checked it numerically to rather high accuracy for several values of $\kappa$ and it seems to be the case. Given the simplicity of the statement, it feels like there should be a rather simple argument if it is true, but it eludes me at the moment.

According to computer simulations, the statement seems independent of the specific value $3+2\sqrt 2$ appearing in the definition above although i) it appears to fail when the value is replaced by a negative number and ii) the specific value $3+2\sqrt 2$ simplifies the calculation of the limit $\kappa \to 0$ somewhat.

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    $\begingroup$ Do your simulations suggest that normality of $\xi_n$ plays an essential role, or can $\xi_n$ be a more general (non-lattice) i.i.d. sequence? $\endgroup$ Jun 14 '18 at 12:54
  • $\begingroup$ @Mateusz Kwaśnicki Good point. Normality (or centering for that matter) doesn't seem to play much of a role. I just did a run replacing Gaussians with squares of Gaussians and it looks very close to uniform after 30M steps. $\endgroup$ Jun 14 '18 at 13:54
  • $\begingroup$ If you replace $3+2\sqrt{2}$ with $1$, you will have $\phi_n=\phi_0+n\theta_0+\kappa\sum_{i=1}^{n-1}(n-i)\xi_i$, so the convergence to Lebesgue should be relatively easy and robust in the distribution of $\xi_i$. $\endgroup$
    – Algernon
    Jun 14 '18 at 14:49
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    $\begingroup$ @Algernon You are right, in this case $\mu_\kappa = \lambda\otimes \lambda$ for all $\kappa$. But the interesting fact is that both marginals appear to be uniform for all positive values of that parameter although one has $\mu_\kappa \neq \lambda\otimes \lambda$ in general. $\endgroup$ Jun 14 '18 at 15:26
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    $\begingroup$ You conveniently forgot to tell us that your circle homeomorphism $\Psi$ is odd and that $z\mapsto \Psi(z)^2$ is analytic in the unit disk. These two facts establish the claim in no time by the consideration of moments on the circle, don't they? $\endgroup$
    – fedja
    Jun 14 '18 at 17:48

Since I was requested to elaborate, here goes. First, let's look at the automorphism of the unit circle induced by this mapping (written in the least revealing way). With $z=e^{it}$, as usual, we have $2\cos t=z+z^{-1}, 2i\sin t=z-z^{-1}$, so for positive $3+\sqrt 2$ (I absolutely loved this red herring) the direction is that of $z+\delta z^{-1}$ with $|\delta|<1$. Normalizing, we get $$ \Psi(z)=\frac{z+\delta z^{-1}}{|z+\delta z^{-1}|} $$ and, most importantly, $$ \Psi(z)^2=\frac{(z+\delta z^{-1})^2}{|z+\delta z^{-1}|^2}= \frac{z+\delta z^{-1}}{z^{-1}+\delta z}=\frac{z^2+\delta}{1+\delta z^{2}}\,. $$

Now the full transformation for random variables $U=e^{i\theta}$, $W=e^{i\varphi}$ is $U,W\mapsto ZU, U\Psi(W)$ where $Z=e^{i\xi_n}$ is something smeared a bit and independent of $U,W$. Since it doesn't matter for the convergence to the limiting distribution with what joint distribution to start, we'll start with independent $U,W$ uniformly distributed over the circle.

Claim 1: All four pairs $(\pm U,\pm W)$ have the same distribution at every iteration step.

Proof: It is clearly true in the beginning and, since $\Psi$ is odd, this property is preserved by the mapping.

This alone immediately kills all moments $\mathcal E [U^kW^\ell]$ with $k$ or $\ell$ odd.

Claim 2: If $k,\ell\ge 0$ with $k+\ell>0$, then $\mathcal E [U^{2k}W^{2\ell}]=0$.

Proof: It is clearly true in the beginning. Now just recall that $\Psi^2$ is even analytic, so $\Psi(z)^{2\ell}=\sum_{m\ge 0} a_{\ell,m}z^{2m}$ and, thereby, $$ \mathcal E [U_{\text{new}}^{2k}W_{\text{new}}^{2\ell}]= \mathcal E [(ZU)^{2k}(U\Psi(W))^{2\ell}]= \mathcal E [Z^{2k}]\sum_{m\ge 0}a_{\ell,m}\mathcal E [U^{2k+2\ell}W^{2m}]=0\,. $$ In particular $\mathcal E[W^{2\ell}]=0$ for $\ell>0$ (and, by conjugation, for $\ell<0$ too) and we are done.

The above recursion can also be used to find the Fourier coefficients of the limiting distribution with $k<0,\ell>0$ row by row.


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