Distribution of EMA of iid $\chi_1^2$ I am wondering if it is possible to get an analytical expression for the  distribution of a random variable defined as
$$
r = \sum_{k=0}^\infty a_i b^i
$$
where $a_i$ are iid chi-square variable with one degree of freedom ($a_i \sim \chi_1^2$) and $b\in (0,1)$.
It's a long shot to get an analytical expression, but I thought maybe someone has some idea.
 A: The characteristic function of $a_i$ is $(1-2it)^{-1/2}$, so that the characteristic function of $r$ is $\prod_{k=0}^\infty (1-2itb^k)^{-1/2}=\sqrt{(2it;b)_{\infty}}$, where the inside is the $q$ analogue of the Pochhammer symbol. As far as I can tell, this is continuous at $t=0$, so by Levy's continuity theorem, it's a characteristic function. Unfortunately, a google search doesn't seem to hint at any well-known corresponding distribution. 
A: I don't think there is a closed form answer but there seems to be a good approximation that I will derive here. The cumulant generating function can be obtained by the sum of cumulant generating functions (CGF) of $\chi_1^2$ and is
$$
\begin{eqnarray}
C_r(t) &=& -\frac{1}{2} \sum_{n=0}^\infty \log(1-2b^n t) \nonumber \\
&=& \frac{1}{2} \sum_{n=0}^\infty \sum_{k=1}^\infty \frac{1}{k} (2t)^k b^{nk} \nonumber \\
&=&\frac{1}{2}  \sum_{k=1}^\infty \frac{1}{k} (2t)^k \frac{1}{1-b^k} \nonumber \\
&=&\frac{1}{1-b}  \sum_{k=1}^\infty \frac{t^k}{k} \left[\frac{2^{k-1}}{1+b + b^2 + \dots +  b^{k-1}} \right]
\end{eqnarray}
$$
The reason to write it like this is to compare it with the CGF of a $\gamma(\alpha,\beta)$ distribution
$$
C_\Gamma(t) = \alpha \sum_{k=1}^\infty \frac{t^k}{k} \frac{1}{\beta^k}
$$
Equating the first two cumulants we get
$$
\alpha = \frac{1+b}{2(1-b)}\\
\beta= \frac{1+b}{2}
$$
With this we get
$$
C_\Gamma(t) = \frac{1}{1-b}  \sum_{k=1}^\infty \frac{t^k}{k} \left[\frac{2^{k-1}}{(1+b)^{k-1}} \right]
$$
We clearly see that in the limit $b \to 0$ we get back the $\chi_1^2$ distributions CDF. The first two cumulants are the same for both from the match and are
$$
c_1 = \frac{1}{1-b} \\
c_2 = \frac{2}{1-b^2}
$$
The higher normalized-cumulants (normalized by variance) for the $\Gamma$ distribution are
$$
s_{k;\Gamma} =2^{k/2-1}(k-1)! \left( \frac{1-b}{1+b}\right)^{k/2-1} \qquad k>2
$$
and for the variable $r$ are
$$
s_{k;r} =s_{k;\Gamma} \frac{(1+b)^{k-1}}{(1+ b + b^2 + \dots + b^{k-1})} \qquad k>2
$$
Now note that while the ratio of the true cumulant to the cumulant of the $\Gamma$ distribution diverges exponentially in $k$ for $b \to 1$, the actual cumulant goes to zero because of the powers of the variance (or more profoundly because of the central limit theorem). We have already established that for small $b$ the match is good. Thus we see that the above approximation is good for both small and large $b$.
