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Suppose $X_n$ are i.i.d. random variables on $\mathbb{R}$ with compact support, and define the Markov chain $Y_n=X_n +\frac{1}{Y_{n-1}}$ on $\Omega=\mathbb{R}\cup \{\infty\}$. Does the chain $Y_n$ have a finite stationary distribution $\pi$, i.e., is there a distribution $\pi$ s.t. if $P(X_n \in A)=\pi(A)$, then $P(X_{n+1}\in A)=\pi(A)$?

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  • $\begingroup$ Do you mean to say "if $P(Y_n \in A)=\pi(A)$, then $P(Y_{n+1}\in A)=\pi(A)$"? Also, are we supposed to find the distribution for $Y_0$? Are we also supposed to find the distribution for $X_n$ as well? $\endgroup$ – Hans Mar 4 '18 at 20:54
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In this case, you can give a somewhat explicit description of the stationary distribution. It’s the distribution of $$ X=Z_1+\frac 1{ Z_2+\frac 1{ Z_3+\frac 1{ Z_4+\ddots}}}, $$ Where the $Z_i$ are iid with the same distribution as $X$.

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  • $\begingroup$ Thanks! But how do you show that the right hand actually converges to a random variable? (If the Z_i are non-negative then it is known). $\endgroup$ – joeyg Mar 7 '18 at 10:47
  • $\begingroup$ I agree. This is the right question. I think you should be able to show that for almost every realization of the $Z$’s, this random variable becomes exponentially less sensitive to $Z_n$ as N\to\infty$. That is: there exists $\theta<1$ (depending on the distribution of $X$) and $K$ depending on the realization of the $Z$s such that the derivative of the continued fraction expression with respect to $Z_n$ is at most $K\theta^n$. I have not written down a proof of this. $\endgroup$ – Anthony Quas Mar 7 '18 at 15:15

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