Let $(X_k)$ be a sequence of independent Bernoulli random variables, such that $\Pr[X_k = 1] = p$. Then for $0\le\alpha<1$ the sum $$\sum_{k=0}^\infty \alpha^k X_k$$ is real random variable in the range $[0, 1/(1-\alpha)]$.

Does this variable follow a well-known distribution? I have tried to calculate it's characteristic function and moments, but I can't quite figure out how to approach it.

  • $\begingroup$ math.stackexchange.com/questions/2570573 seems related, but it only considers the case $p = \alpha = 1/2$ which appears easier. $\endgroup$ Commented May 8, 2018 at 19:28
  • 1
    $\begingroup$ If $\alpha < 1/2$, the sum has a singular continuous distribution, supported on a generalized Cantor set. $\endgroup$ Commented May 8, 2018 at 19:31
  • $\begingroup$ @RobertIsrael Right, it probably makes sense to restrict to the case $\alpha>1/2$. For $\alpha\to1$ it appears we get convergence to a normal (perhaps binomial) distribution. $\endgroup$ Commented May 8, 2018 at 19:44
  • $\begingroup$ What convergence? As $\alpha \to 1-$ with $p$ fixed, the sum goes to $\infty$ almost surely. $\endgroup$ Commented May 8, 2018 at 20:04
  • 1
    $\begingroup$ @RobertIsrael right, I guess you need the right normalization. Looking at it again, it is actually obvious, since the mgf (as you wrote in your answer) becomes $(1-p+p e^t)^n$ when $\alpha\to 1$ (if you limit the sum at $n$) which is exactly the mgf for a binomial distribution. $\endgroup$ Commented May 8, 2018 at 20:11

2 Answers 2


The moment generating function of a sum of independent random variables is the product of the mgf's of the summands. Thus in your case

$$ M(t) = \prod_{k=0}^\infty \mathbb E[\exp(t X_k)] = \prod_{k=0}^\infty \left(1 + p (e^{t \alpha^k}-1)\right) $$ I don't think this has a closed form in general.

  • $\begingroup$ Right, I guess what I meant to say was that I was looking to make a Laplace transformation of the mgf (or a Chernoff bound) in order to find the tail probabilities. In this case I need to solve an equation on $M'(t)$ which seems hard. $\endgroup$ Commented May 8, 2018 at 20:09
  • $\begingroup$ One option is to use the power means inequality to pull $\alpha^k$ out of the expectation. Then we get $M(t)\le (pe^t+1-p)^{1/(1-\alpha)}$. It seems we might be able to also lower bound $(pe^{t\alpha}+1-p)\ge (pe^{t}+1-p)^{f(\alpha,p)}$, that is uniformely in $t$, but I don't know how to get the best value for $f$. $\endgroup$ Commented May 13, 2018 at 13:01

I got a bit closer to an answer myself.

Consider $$\begin{align} EX^n &= E\left(\sum_k \alpha^k X_k\right)^n \\&= \sum_{k_1,\dots,k_n}\alpha^{k_1+\dots+k_n}E(X_{k_1}\cdots X_{k_n}) \\& = \sum_{P\in\text{partitions($n$)}}{n\choose P}p^{|P|}\sum_{k_1,\dots,k_{|P|}}\alpha^{P_1k_1+\dots P_{|P|}k_{|P|}}[\forall_{i,j}k_i \neq k_j] \\& \le \sum_{P\in\text{partitions($n$)}}{n\choose P}p^{|P|} \prod_{s\in P}\frac{1}{1-\alpha^s}, \end{align}$$ where partitions($n$) is the integer partitions of $n$, e.g. partitions($5$) = $\{\{5\},\{4,1\},\{3,2\},\{3,1,1\},\{2,2,1\},\{2,1,1,1\},\{1,1,1,1,1\}\}$. We let ${n\choose P} = {n\choose P_1, \dots, P_{|P|}}$ be the number of ways a particular partition can appear.

Now we know from Ramanujan that $|\text{partitions}(n)| \sim \exp(\pi\sqrt{2n/3})$. Hence, if we only want to know $EX^n$ up to exponential terms, it suffices to find the largest element of the (all positive) sum. We may guess that the largest partitions are those where all elements of $P$ are the same, hence we consider for $n=sm$:

$$\begin{align} \log\left({n\choose \underbrace{s, \dots, s}_{\text{$m$ times}}}p^m\left(\frac1{1-\alpha^s}\right)^m\right) &= \left(n\log\frac ns+o(n)\right)+m\log p+m\log\frac{1}{1-\alpha^s} \\&= n\left(\log\frac ns+o(1)+\frac1s\log\frac{p}{1-\alpha^s}\right). \end{align}$$

This is decreasing in $s$, so it suggests the bound $\log EX^n\le n\log\frac{np}{1-\alpha}+o(n)$. It may be that this upper bound is too lose to get an equivalent (up polynomial terms) lower bound and that we need to not throw away the $[\forall_{i,j}k_i\neq k_j]$ condition.

Update: Using this result of Hitczenko: $\|\sum a_i X_i\|_n\sim\sum_{i\le p}a_i + \sqrt{n}\sqrt{\sum_{i>n}a_i^2}$, we can find $\|X\|_n = (EX^n)^{1/n}$ up to a constant:

$$ \|X\|_n \sim \sum_{1\le i\le n}\alpha^{i-1} + \sqrt{n}\sqrt{\sum_{i > n}\alpha^{2i-1}} = \frac{1-\alpha^n}{1-\alpha} + \sqrt{n}\frac{\alpha^n}{\sqrt{1-\alpha^2}}. $$

This is for $p=1/2$ of course.

For $p\neq 1/2$ we might use the biased Khintchine inequalities by Wolff and Oleszkiewi to show:

$$\begin{align} \|X\|_n &\le \sqrt{\frac{q^{2-2/n}-p^{2-2/n}}{p^{1-2/n}q-q^{1-2/n}p}}\frac1{\sqrt{1-\alpha^2}} \\&\sim\begin{cases} \sqrt{\frac{p^{2/n}}{p\,(1-\alpha^2)}} & \text{if}\quad \frac1{n-1}\le\log\frac1p\\ \sqrt{\frac{(n-1)\,p\log1/p}{1-\alpha^2}} & \text{if}\quad \frac1{n-1} >\log\frac1p \end{cases} \end{align}$$

where $q=1-p$. However, this isn't necessarily tight.

Note the previous suggested bound was $\|X\|_n\le\frac{np}{1-\alpha}+o(1)$ which is mostly less than the hypercontractive bound. Presumably, the right answer is somewhere in between.

Perhaps a generalized (in the sense of Hitczenko) biased Khintchine is needed to solve this problem.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .