Generalized random harmonic series

Question

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.

Answer 1

By the Berry--Esseen inequality, the limit distribution of $Z_n^*$ is the standard normal distribution if $a_k=1/k^{1/2}$.

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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.

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Using the formula (cf. e.g. formula 1.411.6 for $\tanh=(\ln\circ\cosh)'$) $$\ln\cosh x=\sum_{j=1}^\infty \frac{2^{2 j-1} \left(4^j-1\right) B_{2 j}}{j (2 j)!}\,x^{2j}$$ for all $x\in(-\pi/2,\pi/2)$, where the $B_{2 j}$'s are the Bernoulli numbers, we get all the cumulants and thence the moments of $Z_n$, in terms of Bell polynomials.