Imagine a two-state Markov chain which hops between the states $\pm 1$ with probability $p<1/2$, so that the autocorrelation function after $k$ steps is


$\rho\_k = (2p-1)^k$


If I take an exponential moving average of this series with weighting parameter $\lambda$, what does the distribution of values of the new series look like?


Probably the answer is "gaussian, centered on 0" but what is the variance? Is there a known result that makes this computation trivial?



Statistics of a simple Markov chain
The answer is certainly not "Gaussian". What you describe is often called Bernoulli convolutions and, even in the independent case (in your setting, $p=1/2$), the limiting object is quite complicated and interesting since it involves some deep number theoretic properties of $\lambda$.


To begin with, let $(X\_n)\_{n\in \mathbb{Z}}$ denote the $\pm1$-valued Markov chain with probability $p$ of switching states. Let $(Y\_n)\_{n\in \mathbb{Z}}$ denote the exponential moving average of parameter $\lambda$ with $0<\lambda < 1$ you are interested in, that is,
$$
Y\_n=\sum\_{k=0}^{+\infty}\lambda(1-\lambda)^kX\_{n-k}.
$$
The Markov chain is centered and has correlation $E(X\_nX\_{n+k})=(1-2p)^k$ for every integers $n$ and $k\ge0$. (Hence you should check your formula.) From there, one sees that $Y\_n$ is centered and one can compute its variance. If I am not mistaken, one finds something like
$$
E(Y\_n^2)=\frac{1-2p(1-\lambda)/(2-\lambda)}{1+2p(1-\lambda)/\lambda}.
$$
The stationary distribution of the moving average is a different story. It is often best described as a measure-valued fixed point problem, as follows. First, $Y\_n=X\_nY\_+$ where $Y\_+$ and $X\_n$ are independent, and $Y\_+$ is distributed like $Y\_n$ conditioned on $[X\_n=+1]$. Second, $Y\_+$ is distributed like $\lambda+(1-\lambda)ZY\_+$, where $Z=\pm1$, $P(Z=+1)=1-p$, $P(Z=-1)=p$, and $Z$ independent of $Y\_+$.


This indirect description of the stationary distribution is often the most useful tool to get some information on it.


As regards your original "Gaussian" hint, note that conditioning on $(X\_{n-k})\_{0\le k\le N-1}$ for a given $N$ yields that $Y\_n$ is in one of $2^N$ intervals of length $2(1-\lambda)^N$. If $2(1-\lambda)<1$, this simple remark shows that the distribution of $Y\_n$ is concentrated on a Cantor set of Lebesgue measure zero (hence this probability distribution does not even have a density with respect to the Lebesgue measure, and it has no atom either). The argument uses only the fact that each $X\_n=\pm1$ almost surely and not the structure of the process $(X\_n)\_n$.


Another easy case is when $\lambda=1/2$. Then, if $(X\_n)\_n$ is in fact independent ($p=1/2$), one recognises the usual binary expansion of a random number hence $Y\_n$ is uniformly distributed on $[-1,1]$, but for every other value of $p$, the distribution of $Y\_n$ is concentrated on a subset of $[-1,1]$ of Lebesgue measure zero.


For much more on the stationary distributions of moving averages like the ones which interests you, some starting points could be the paper *Sixty years of Bernoulli convolutions* by Peres, Schlag and Solomyak, and the book *Some random series of functions* by Kahane.


