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My question is that how much information we can get form integer moments of a complex random variable?

Let $\mathcal{Z}$ be a complex value random variable, and assume that we can compute $$\int \mathcal{Z}^k d\mu,$$ For $k \in \mathbb{N}$ and $\mu$ be a measure.

I also am looking for an example of a complex random variable, that its moments positively approaches zero with

$$ 0< \int \mathcal{Z}^k d\mu \asymp \frac{1}{k!}.$$

If moments satisfy the above, can we occlude that $|\mathcal{Z}|$ is bounded almost surely?

One simple example: let $$ \mathcal{Z}= \sum_{m \in A} e^{imt} + \tfrac{1}{2},$$

where $A \subset \mathbb{Z}$ and $t$ is uniformly distributed in $(0, 2\pi].$ Then k-th moment of $\mathcal{Z}$ is 2^{-k}. Therefore moments decay but they fall short of decaying like $1/k!$.

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This is to rewrite the excellent example by Mateusz Wasilewski in a more conventional form.

Let $Z:=XU$, where $X$ and $U$ are independent random variables (r.v.'s); $P(X>0)=1$; $X$ is unbounded; $EX^k<\infty$ for all natural $k$; $U=e^{iT}$; $T$ is a r.v. with values in the interval $[0,2\pi)$ and pdf $p$ given by the formula $$p(t)=\frac1{2\pi}\,\Big(1+2\sum_{n=1}^\infty a_n\cos nt\Big)$$ for $t\in[0,2\pi)$; $$0<a_n\sim\frac1{n!\,EX^n};$$ and $2\sum_{n=1}^\infty a_n<1$ (so that $p>0$).

Then for all natural $k$ $$EU^k=Ee^{ikT}=\int_0^{2\pi}e^{ikt}p(t)\,dt=a_k$$ and hence $$EZ^k=EX^k\,EU^k=EX^k\,a_k\sim\frac1{k!}$$ and $EZ^k>0$, whereas the r.v. $|Z|=X$ is unbounded.

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  • $\begingroup$ Thanks, this is very clear. $\endgroup$ Jul 10, 2020 at 12:40
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Let $u$ be uniformly distributed on the unit circle and let $X$ be a positive random variable with all moments finite. If $u$ and $X$ are independent then $Z:=uX$ has all moment equal to zero, because $\mathbb{E} u^{k} = 0$ for any $k\neq 0$. If you want an example with strictly positive moments then just add to $Z$ and independent positive random variable with moments decreasing very fast. You can easily cook up in this way examples that are not bounded.

EDIT: Here comes an actual example. Our random variable $Z$ will be of the form $Z=uX$, where $u$ is some distribution on the unit circle and $X$ is positive; we assume that $u$ and $X$ are independent. Then the moments are $\mathbb{E}Z^k = \mathbb{E} u^{k} \cdot \mathbb{E} X^{k}$. We want $X$ to be unbounded, so the moments of $X$ will grow to infinity at some rate, but it is not so important. We will use the oscillatory nature of $u$ to produce examples with an arbitrarily fast decay of $\mathbb{E} u^{k}$. The distribution of $u$, being supported in the unit circle, can be represented by the Fourier series $u \sim \sum_{n\in\mathbb{Z}} a_n e^{int}$. Note that $a_n= \mathbb{E} u^{-n}$. We have $a_0=1$ and we want the coefficients $(a_n)$ to decay very fast, while ensuring that the Fourier series above represents a probability measure. The moments are supposed to be real, so $a_n = a_{-n} = \overline{a_n}$. We can therefore rewrite the Fourier series as $1 + 2\sum_{n\geqslant 1} a_n \cos(nt)$. If we take $a_n$ small enough so that $|2\sum_{n\geqslant 1} a_n \cos(nt)|\leqslant 1$, then we have a probability measure. Now just pick $(a_n)$ in such a way that $\mathbb{E} u^{k} \simeq \frac{1}{k! \mathbb{E}X^{k}}$ and you get a counterexample.

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  • $\begingroup$ Thanks, so basically, to put more in analysis form, an exponential sum like the one I mentioned plus an independent function which has moments like $1/k!$ would do the job. I am wondering if that is the only possibility, meaning that if we pull that oscillatory part out of $\mathcal{Z}$ we always end up with positive RV with same moments as $\mathcal{Z}.$ Could you give me an example of a RV with moments like $1/k!$? $\endgroup$ Jul 5, 2020 at 17:54
  • $\begingroup$ I actually realised that it's not possible to have such a fast decay of moments, even for not necessarily positive random variables. Indeed, if $X$ is a non-zero random variable with all moments finite then for some $\varepsilon>0$ the probability of the event $\{X^2>\varepsilon\}$ is positive and we have a moment bound $\mathbb{E} X^{2k} \geqslant C \varepsilon^{2k}$ which decays more slowly than the factorial. So I still don't know if there are examples satisfying your original assumptions. $\endgroup$ Jul 6, 2020 at 8:04
  • $\begingroup$ On the edit: You no longer assume that $t \sim U(0, 2\pi)?$ cause that way all the moments of $u$ are zero. Also, could you please explain a bit more on how to choose $a_n$ such that $\mathbb{E} u^{k} \simeq \frac{1}{k! \mathbb{E}X^{k}}$. $\endgroup$ Jul 6, 2020 at 15:19
  • $\begingroup$ Yes, I said that now $u$ is some distribution on the unit circle, not the uniform one. It is basically a small perturbation of the uniform distribution so that the moments are very small but positive. Could you ask more specifically, which part of the construction is still unclear? $\endgroup$ Jul 7, 2020 at 11:03

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