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 $1/k!$.


2 Answers 2


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.

  • $\begingroup$ Thanks, this is very clear. $\endgroup$ Jul 10, 2020 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.