Moments of complex random variables

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 $$1/k!$$.

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 $$t\in[0,2\pi)$$; $$0 and $$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.

• Thanks, this is very clear. Jul 10 '20 at 12:40

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.

• 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 $1/k!$? Jul 5 '20 at 17:54
• 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. Jul 6 '20 at 8:04
• 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 $\mathbb{E} u^{k} \simeq \frac{1}{k! \mathbb{E}X^{k}}$. Jul 6 '20 at 15:19
• 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? Jul 7 '20 at 11:03