Let $Z\_n=\sum\_{k=1}^n a\_k X\_k$ with $(a\_k)$ a strictly decreasing sequence of positive real numbers that tend to zero. The random variables $X\_k$ are independent and satisfy $P(X\_k=1) =p\_k, P(X\_k=-1)=1-p\_k$. Here $p\_k=\frac{1}{2}$. The normalized series is defined as $Z^\*\_n=(Z\_n-\mbox{E}[Z\_n])/\sqrt{\mbox{Var}[Z\_n]}$. Note that all odd moments, including $\mbox{E}[Z\_n]$, are zero. In particular, $\mbox{Var}[Z\_n]=\mbox{E}[Z\_n^2]$.


**My questions are**:


* Is my computation of the moments correct (see below)?
* What is the limiting distribution of $Z^\*\_n$ if $a\_k=1/k^s$ with $s=1/2$? Or
if $a\_k=1/2^k$? Or if $a\_k=1/3^k$?


**Progress made so far:**


The characteristic function is $\phi(t)=\prod\_{k=1}^n \cos(a\_k t)$. In particular, if $a\_k=1/2^k$, then $\phi(t)=(\sin t)/t$ when $n=\infty$. If $a\_k=1/3^k$, I expect the resulting distribution to have an awkward support domain, and related to Cantor sets. This is because $(a\_k)$ converges too fast to zero. But if $a\_k=1/k^s$, with $1/2 < s < 1$, the resulting distribution, for $Z\_n$, behaves like any smooth continuous distribution with support domain equal to the set of real numbers. If $s\leq 1/2$, the variance is infinite and we need to use $Z^\*\_n$ rather than $Z\_n$. However, I am wondering what the limiting distribution of $Z\_n^\*$ is. Is it Gaussian? Is the Central Limit Theorem applicable? I think so if $s<1/2$, but what if $s=1/2$ (the critical point)? I would not be surprised if this is connected to the behavior of the Riemann Zeta function (its roots) in the critical strip, especially if you allow $s$ to be a complex number.


Turning to the moment generating function, we have
$$\mu(t)\equiv \mbox{E}[\exp(t Z\_n)] = \prod\_{k=1}^n \cosh(a\_k t).$$


Thus, taking the logarithm and differentiating we get:
$$\mu'(t)=\mu(t)g(t), \mbox{ with } g(t)=\sum\_{k=1}^n a\_k\tanh(a\_k t).$$


We can iteratively differentiate to get $\mu''(t)=\mu'(t)g(t)+\mu(t)g'(t)$ and so forth, and recursively compute the moments $\mbox{E}[Z^m]=\mu^{(m)}(0)$ for $m=1, 2$ and so on. I used Mathematica to compute $m$-th derivative of $\tanh(a\_k t)$ at $t=0$, and this is what I eventually ended up with:


$$\mbox{E}[Z\_n^4]=3\Big(\sum\_{k=1}^n a\_k^2\Big)^2 - 2 \Big(\sum\_{k=1}^n a\_k^4\Big),\\
\mbox{E}[Z\_n^6]=15\Big(\sum\_{k=1}^n a\_k^2\Big)^3 - 30 \Big(\sum\_{k=1}^n a\_k^2\Big)\Big(\sum\_{k=1}^n a\_k^4\Big)+16\Big(\sum\_{k=1}^n a\_k^6\Big).
$$
Of course $\mbox{E}[Z\_n^2]=\sum\_{k=1}^n a\_k^2$ is well-known, but I have never seen the 4-th and 6-th moments published anywhere. It would we nice if my computations could be double-checked, as it may help statisticians using this type of distribution for model fitting, by choosing $(a\_k)$'s that provide a good fit with some empirical moments.


Assuming my computations are correct and using the notation $Z=Z\_\infty$, for the random harmonic series $(a\_k=1/k)$ we would have:


$$\mbox{E}[Z^2]=\frac{\pi^2}{6}, \mbox{ } \mbox{ } \mbox{ } \mbox{E}[Z^4]=\frac{11\pi^4}{180}, \mbox{ } \mbox{ } \mbox{ } \mbox{E}[Z^6]=\frac{233\pi^6}{7560}.$$


The first identity is well-known.



Generalized random harmonic series
By the [Berry--Esseen inequality](https://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem#Non-identically_distributed_summands), the limit distribution of $Z\_n^\*$ is the standard normal distribution if $a\_k=1/k^{1/2}$.




---


If $a\_k=1/2^k$, then the limit distribution of $Z\_n^\*$ is uniform on the interval $[-\sqrt3,\sqrt3]$. This follows (i) because $(X\_k+1)/2$ may be viewed as the $k$th binary digit of a random number uniformly distributed on the interval $[0,1]$ and (ii) $Var\,Z\_n\to1/3$ as $n\to\infty$.


For the more general case with $a\_k=t^k$ for $t\in(0,1)$, see e.g. [this survey](https://arxiv.org/abs/1608.04210).




---


Using the formula (cf. e.g. [formula 1.411.6 for $\tanh=(\ln\circ\cosh)'$](http://fisica.ciens.ucv.ve/%7Esvincenz/TISPISGIMR.pdf))
$$\ln\cosh x=\sum\_{j=1}^\infty \frac{2^{2 j-1} \left(4^j-1\right) B\_{2 j}}{j (2 j)!}\,x^{2j}$$
for all $x\in(-\pi/2,\pi/2)$, where the $B\_{2 j}$'s are the Bernoulli numbers, we get all the [cumulants](https://en.wikipedia.org/wiki/Cumulant#Definition) and thence the [moments](https://en.wikipedia.org/wiki/Cumulant#Cumulants_and_moments) of $Z\_n$, in terms of [Bell polynomials](https://en.wikipedia.org/wiki/Bell_polynomials#Bell_polynomials).


