Can one tell based on the moments of the random variable if it is continuos or not

Suppose we are given moments of a random variable $X$. Can we determine based on this if the random variable is continuous or not?

We also assume that the moments of $X$ completely determine the distribution of $X$.

In other words, do moments of continuous random variable behave fundamentally differently than moments of say discrete random variable?

Thanks, looking forward to your ideas.

Edit: It seems like there was some confusion with the questions. Let me demonstrate with an example what I have in mind.

Suppose, we are given moments of some random variable $X$ \begin{align} E[ X^n]=\frac{1}{1+n}, \end{align} for $n \ge 0$.

Can we determine if the distribution of $X$ is continuous or not?

In this example, I took $X$ to be continuous uniform on $(0,1)$.

Some Thoughts: Since we know the moments we can reconstruct the characteristic function of $X$ (I think this can be done, right? If not let as assume this) \begin{align} \phi_X(t) =\sum_{n=0}^\infty \frac{i^n E[X^n]}{n!} t^n \end{align}

We also know that $X$ has a pdf iff $\phi_X(t) \in L_1$.

So it seems it is enough to show that \begin{align} \int_{-\infty}^\infty \left| \sum_{n=0}^\infty \frac{i^n E[X^n]}{n!} t^n \right| dt \end{align} is finite or not. However, I don't think the above approach would work, as we can not switch the integration and summation.

Solution 1:

I doubt that there are some feasible universal conditions for two reasons:

1. If the moment problem is indeterminate, then there can be both discrete and continuous random variables with same moments. For example, it is known that there is an infinite family of discrete random variables having the same moments as the log-normal distribution (see e.g. Stoyanov Counterexamples in Probability).

2. One can approximate a continuous distribution with discrete ones and vice versa. So the moments of discrete distribution can be quite close to those of continuous distribution.

Of course, it is possible to formulate infinitely many sufficient conditions for a distribution to be discrete. Example: Let $\mu_n = \mathsf{E}[X^n]$. If $\mu_8 - 10\mu_6 + 33\mu_4 - 40\mu_2 + 16=0$, then $X$ is discrete (moreover, $X\in\{\pm1, \pm2\}$ a.s.).