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Let $(Z_n)_{n\geq 0}$ be a sequence of i.i.d. random variables with $\mathbb{P}(Z_i=1)=\mathbb{P}(Z_i=-1)=1/2$. Define $S_n=\sum_{k=1}^nZ_k/k$. Since $(S_n)_{n\geq 0}$ is a martingale that is bounded in $L^p$ for $p>1$, it converges a.s. and in $L^p$ to some r.v. $S_\infty$.

Can we determine the distribution of $S_\infty$?

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  • $\begingroup$ Well, the even moments $<S^{2j}_\infty>$ are $\zeta(2j)$, since $<S^{2j}_\infty>=\sum_k{<Z^{2j}> \over k^{2j}}$ $\endgroup$
    – user619894
    Jan 7, 2021 at 9:05
  • $\begingroup$ @user619894 We can find the characteristic function which is $\prod_{n=1}^\infty\cos(\xi/n)$. It seems that $S_\infty$ has a density but I'm not sure it can be expressed in terms of usual functions $\endgroup$
    – charlus
    Jan 7, 2021 at 12:03
  • $\begingroup$ The moments are finite, doesn't that imply that the PDF exists? $\endgroup$
    – user619894
    Jan 7, 2021 at 14:16
  • $\begingroup$ I'm not sure I see which property you're referring to? One other way to do things is to say that the characteristic function is integrable so $S_\infty$ has a density. I would like to know how to write this density but I'm not able to find anything conclusive $\endgroup$
    – charlus
    Jan 7, 2021 at 14:25
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    $\begingroup$ Here is a AMM article about this distribution, preprint here (click on "random harmonic series"). $\endgroup$ Jan 8, 2021 at 23:48

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