Infinite sum of Gamma random variables with same shape parameter but different rate parameter

Question

Let $$Z_i\sim\chi_{(1)}^2\sim\Gamma(\frac{1}{2},\frac{1}{2})$$ be i.i.d. chi-square random variables. We define: $$W_n =\Bigl[\sum_{i = 1}^{n-1}\frac{Z_i}{2^{i}}\Bigr] + \frac{Z_n}{2^{n-1}} \mbox{ \forall n\geq 2}$$ For example: $$W_2 = \frac{1}{2}Z_1+\frac{1}{2}Z_2$$, $$W_3 = \frac{1}{2}Z_1+\frac{1}{4}Z_2+\frac{1}{4}Z_3$$ and so on.

I would like to determine the distribution of $$W_{\infty}$$ defined as: $$W_{\infty} = \lim_{n\to\infty}W_n = \sum_{k = 1}^{\infty} \frac{Z_k}{2^{k}} \mbox{ where Z_k\sim \Gamma\Bigl(\frac{1}{2},\frac{1}{2}\Bigr) are i.i.d.}$$ More specifically I have interest in computing the probability $$P(W_{\infty} > 1$$).

References

Note that this problem has references, in particular in this paper by Mathai:

https://www.ism.ac.jp/editsec/aism/pdf/034_3_0591.pdf

Reference [4]:

In my case, following its notations $$\alpha = \frac{1}{2}$$, $$X_i\sim\Gamma(\frac{1}{2},\frac{1}{2})$$ are i.i.d. and $$n\to\infty$$. The main problem is that I didn't manage to find Prabhu's work about this fact, and in general I didn't manage to find any other references to this problem.

My attempt

I was trying to approach this problem using the Levy-criteria for the convergence in distribution.. in particular I was trying to calculate (since they are indipendent random variables) the infinite product of their characteristic functions, which is the characteristic function of $$W_{\infty}$$.. but I wasn't able to determine a closed form for it. In particular if $$H_k \sim \Gamma(1/2,2^k)$$ then $$\phi_{H_k}(t) = \Bigl(1-\frac{t}{2^k}\Bigr)^{-\frac{1}{2}}$$ and so: $$\phi_{W_{\infty}}(t) = \prod_{k = 0}^{\infty}\frac{1}{\sqrt{1-\frac{it}{2^k}}}$$

Code and simulation

I'm sure that $$W_n$$ converges in distribution by simulating it using R, in particular the following code:

it = 10000
n = 300
mat = matrix(rep(0,n*it),n,it)
w = vector()
for(i in 1:n) {
mat[i,] = rchisq(it,1)
}
for(i in 1:n) {
mat[i,] = mat[i,]/2^i
}
for(j in 1:it) {
w[j] = sum(mat[,j])
}
hist(w,freq = F,breaks = seq(0,10,0.1))


produces the following output: $W_{300}$" />

which is the distribution of $$W_{300}$$. I don't really know if this is a well known distribution, and I don't really know if its density function has a closed form. I don't know how to proceed. Thank you in advance for your help!

New approximation of $$P(W_{\infty} > 1)$$

Using the software R and the library "CompQuadForm":

I manage to approximate the value of $$P(W_{\infty} > 1)$$ with the code:

library(CompQuadForm)
n = 10000
q = 1
lambda = vector()
for(i in 1:n) {
lambda[i] = 2^(-i)
}
acc1 = 10^(-15)
approx = davies(q, lambda, h = rep(1, length(lambda)), delta = rep(0,length(lambda)), sigma = 0, lim = 500000, acc = acc1)$Qq sprintf("%.30f",approx)  which gives: $$P(W_{\infty} > 1) \approx 0.371741079532780016592141691945$$ which I think are the real first 30 digits. (I'm sure for the first 15 since I set an accuracy of $$10^{-15}$$). Of course I would prefer an "analytic" form of this number.. • That's actually the mgf, not the cf. It's expressible in terms of the$q$-Pochhammer symbol, and at least one of these results should allow you to prove$\phi_{W_\infty}(it)$is a cf. Obtaining a pdf and$P(W_\infty>1)$are the hard parts. – J.G. Nov 25, 2023 at 12:54 • Oh.. my bad. Just forgotten the (fundamental)$i$in the expression. Thank you! Yes.. I have seen by using the software Mathematica that for a fixed$n$the cf of$W_n$involves the$q$-Pochhammer symbol.. although simply taking the limit of this expression as$n\$ approaches infinity seems to crash the software. I'm sure that this problem has been already studied as I pointed out in the question.. It's just a matter to understand how! Or where it has been proved. Nov 25, 2023 at 15:17
• See math.stackexchange.com/questions/4033607/… for a similar question. Nov 25, 2023 at 15:21
• @Aruralreader I'm checking it right now.. maybe it could help since the exponential distribution is a particular case of a gamma. Thank you. Nov 25, 2023 at 15:23
• @LucaOnnis, you’re correct. Good luck! Nov 25, 2023 at 16:15