CDF of a convergent positive series

Question

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.

Answer 1

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).