# CDF of a convergent positive series

Let $$Y_0, Y_1, \ldots$$ be an i.i.d. random sequence such that $$\mathbb{P}(Y_k = 0) \;=\; 1 - \mathbb{P}(Y_k = 1) \;=\; p \qquad \text{for each k\ge 0}.$$ I am interested in the following random sequences $$U_k = \prod_{i=0}^{k-1}\gamma^{Y_i} \theta^{1 - Y_i},$$ $$Q_n = \sum_{k=1}^n U_k,$$ where $$0<\theta<1<\gamma$$.

I have proved that $$Q_\infty = \sum_{k=1}^\infty U_k$$ converges a.s. when $$p> \log \gamma/\log (\theta^{-1}\gamma)$$. That means under this case, $$Q_\infty$$ is a random variable and it makes sense to consider its CDF (cumulative distribution function).

Let us use $$F_n$$ and $$F_\infty$$ to denote the CDFs of $$Q_n$$ and $$Q_\infty$$, respectively. I also proved that actually $$F_n \downarrow F_\infty$$ when $$n\to\infty$$. Now I am curious about what is exactly $$F_\infty$$. Can we have a beautiful explicit formula for it? Or can we have a tight lower bound of it?

I also use MATLAB to draw the pictures of $$F_n$$ as follows (I set $$\gamma=2$$, $$\theta=0.5$$, and $$p=0.51$$), but I can only have $$n$$ around 20 since the time of computing this $$F_n$$ is $$\mathcal{O}(2^n)$$.

Any comments are welcome.

Suppose that $$\gamma>\theta>0$$. Noting that $$X_i=\sum_{i=0}^{k-1} Y_i\sim\text{Binomial}(k,1-p)$$ is the sum of $$k$$ independent and identically distributed $$Y_{i}\sim\text{Bernoulli}(1-p)$$ random variables, we have \begin{align} P(U_k\leq u)&=P\left(\gamma^{X_k} \theta^{k-X_k}\leq u\right),\\ &=P\left(X_k\leq \frac{k\ln\theta-\ln u}{\ln\theta-\ln\gamma}\right).\\ \end{align} By the de Moivre–Laplace theorem, as $$k\rightarrow\infty$$, we have $$X_k\sim\mathcal{N}(n(1-p),np(1-p))$$ so that \begin{align} F_{U_{\infty}}(u)&=F_{X_{\infty}}\left(\frac{k\ln\theta-\ln u}{\ln\theta-\ln\gamma}\right),\text{ and }\\ p_{U_{\infty}}(u)&=F_{U_{\infty}}'(u),\\ &=\frac{1}{u\left(\ln\gamma-\ln\theta\right)}F_{X_{\infty}}'\left(\frac{k\ln\theta-\ln u}{\ln\theta-\ln\gamma}\right),\\ &\propto \frac{1}{u}\exp\left(-\frac{1}{2}\left(\frac{\frac{k\ln\theta-\ln u}{\ln\theta-\ln\gamma}-n(1-p)}{p(1-p)}\right)^2\right),\\ &\propto \frac{1}{u}\exp\left(-\frac{1}{2}\left(\frac{\left(\ln u-\left(n\ln\gamma-np\ln\gamma+np\ln\theta-n\ln\theta+k\ln\theta\right)\right)^2}{(np(1-p)(\ln\gamma-\ln\theta))^2}\right)^2\right),\\ \end{align} and $$U_{\infty}\sim\text{Lognormal}(n\ln\gamma-np\ln\gamma+np\ln\theta-n\ln\theta+k\ln\theta,np(1-p)(\ln\gamma-\ln\theta))$$. Now, the PDF (or CDF) for the sum of $$n$$ independent and identically distributed lognormal random variables is a famously hard problem but, as a rule of thumb, it is approximately lognormal. See here for the $$n=2$$ case.
As $$n\rightarrow\infty$$, we may incite CLT (since we have a defined mean and variance for $$U_{\infty}$$). I'm aware that you are interested in summing $$U_k$$ and not $$U_{\infty}$$ but your plot of the CDF tends towards something approx. lognormal and so I hope my argument will work anyway (if only in an asymptotic sense).
• Thank you for your comment! But I found a problem in your idea, that is, all your "$n$" should be changed to "$k$", and we can have $U_k \sim \operatorname{Lognormal}(k(\ln \gamma - p \ln (\theta^{-1}\gamma)), kp(1-p)(\ln \theta^{-1}\gamma)^2)$ as $k\to\infty$, which implies $U_k\to 0$. Although it is not the answer I want, thank you again! Commented Aug 20, 2023 at 12:07