TITLE: Limit distribution of infinite sum of Bernoulli random variables
QUESTION [6 upvotes]: I know that the finite sum of Bernoulli i.i.d. random variables is a binomial distribution, but what is the distribution of $$\lim_{n \to \infty}\sum_{k=1}^{n} \frac{x_k}{2^k}$$ where $x_k$ is a Bernoulli random variable with parameter $\frac12$?

REPLY [9 votes]: An approach using generating functions is possible.
Define $Y_k = X_k/2^k$, $k = 1, 2, \ldots$.  Then $$M_{Y_k}(t) = \operatorname{E}[e^{tY_k}] = e^0 \Pr[Y_k = 0] + e^{t/2^k} \Pr[Y_k = 2^{-k}] = \frac{e^{t/2^k} + 1}{2}.$$  Then defining $$S_n = \sum_{k=1}^n Y_k,$$ we find that the MGF for $S_n$ is $$\begin{align*} M_{S_n}(t) &= \operatorname{E}\left[\prod_{k=1}^n e^{tY_k}\right] = \prod_{k=1}^n M_{Y_k}(t) \\ &= 2^{-n} \prod_{k=1}^n (e^{t/2^k} + 1) \\ &= \frac{2^{-n}}{e^{t/2^n} - 1} (e^{t/2^n} - 1)(e^{t/2^n} + 1) \prod_{k=1}^{n-1} (e^{t/2^k} + 1) \\ &= \frac{2^{-n}}{e^{t/2^n} - 1} (e^{t/2^{n-1}} - 1) \prod_{k=1}^{n-1} (e^{t/2^k} + 1) \\ &= \frac{e^t - 1}{2^n(e^{t/2^n} - 1)} .\end{align*}$$  Taking the limit as $n \to \infty$, we easily get $$M_{S_\infty}(t) = \frac{e^t-1}{t},$$ which is the MGF of a $\operatorname{Uniform}(0,1)$ distribution.