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So this is my first post in mathoverflow. I posted this problem in Mathstack, an I've also put a bounty on it, but did not get any response. If anyone can at least point out a reference on this problem, I will be grateful.

Let $X_1,X_2,\dots$ be independent random variables such that $P(X_i=1)=p=1- P(X_i=\epsilon_i)$, for some $0<\epsilon_i<1$, and let

$$Y=X_1+\frac{X_1}{X_2+\frac{X_2}{X_3+\frac{X_3}{X_4+\dots}}} \, .$$

  1. What is the distribution of $Y$?
  2. What is the characteristic function of $Y$?

We can think of less complicated version:

Let $X_1,X_2,\dots$ be i.i.d. random variables such that $P(X_1=1)=p=1-P(X_1=\epsilon)$ for some $0<\epsilon <1$

$$Y=X_1+\frac{X_1}{X_2+\frac{X_2}{X_3+\frac{X_3}{X_4+\dots}}}$$

Here we can write: $Y=Y_1$, $Y_1=X_1(1+\frac{1}{Y_2})$, $Y_2=X_2(1+\frac{1}{Y_3})$ and so on. It can be easily seen that $Y_i$'s are identical but not independent. So, I think that first we should answer questions 1 and 2 for this simpler case.

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  • $\begingroup$ @CarloBeenakker should I delete this post or I delete mathstack post? $\endgroup$
    – MAN-MADE
    Commented Jul 7, 2017 at 9:51
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    $\begingroup$ it's OK, but referencing both posts is good practice to avoid duplication of efforts. $\endgroup$ Commented Jul 7, 2017 at 9:52
  • $\begingroup$ I added the link of mathstack post in this post. $\endgroup$
    – MAN-MADE
    Commented Jul 7, 2017 at 10:04

1 Answer 1

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A general approach to problems of this type, worked out for a slightly different continued fraction, $$Y_n=X_n+1/X_{n-1},$$ is described in Random Continued Fractions: A Markov Chain Approach (2004).

A closed-form answer follows if the $X_n$'s have a Gamma distribution, $$P(X)\propto X^{\lambda-1}e^{-aX},\;\; X> 0,$$ when the $n\rightarrow\infty$ limit of $Y_n$ tends to the distribution $$P(Y)\propto Y^{\lambda-1}\exp[-a(Y+1/Y)],\;\; Y>0.$$ This result goes back to A characterization of the generalized inverse Gaussian distribution by continued fractions (1983). A Bernoulli distribution for the $X_n$'s gives a more complicated answer for $P(Y)$.

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