# Assessing discreteness of the random variable by its characteristic function

It is easy to spot a discrete integer valued random variable by looking at its characteristic function, as that is periodic with period $2 \pi$, i.e. for binomial distribution it is $\phi(t) = (1-p+p \, \mathrm{e}^{i t})^n$.

I recently came across Khintchine's result that $\phi(t) = \frac{\zeta(s + i t)}{\zeta(s)}$ is a characteristic function of a random variable for $s > 1$. After some fudging, I determined that it corresponds to $x_k = -\log(k)$, where $k$ follows Zipf distribution with parameter $s-1$. Indeed:

$$\mathbb{E}( \mathrm{e}^{ -i t \log(k)} ) = \mathbb{E}( k^{-i t} ) = \sum_{k \ge 1} k^{-i t} \frac{k^{-s}}{\zeta(s)} = \frac{\zeta(s+i t)}{\zeta(s)}$$

This characteristic function, thus, also corresponds to a discrete random variable.

This brings up a question: Can one easily spot a discrete random variable from it's characteristic function ? Or is inverting the characteristic function the only way ? How does one go about doing the inversion ? Ordinary inverse Fourier transform would produce distributions, right ?

Thank you.

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I'm not sure how to distinguish purely discrete random variables, but the formula $$\lim_{T\to\infty}{1\over 2T}\int^T_{-T}|\phi(t)|^2\,dt=\sum\mu(\{x\})^2$$ tells us whether or not the distribution $\mu$ has a discrete part.
The characteristic function is the Fourier transform of the probability density (with positive exponent, instead of the more usual convention of negative; but that's convention). This also holds if the variable is discrete, by using a distribution (sum of Dirac deltas) inside the Fourier integral, which in turn can be equivalently expressed as a Fourier sum. If the variable takes values in the integers, then we have the DTFT, and the transform $X(\omega)$ is a periodic function ($2 \pi$). To recover the original density, is nothing more nor less than computing the inverse transform. If you apply it directy to the full $X(w)$ we would get the sum of deltas; equivalently, we can integrate over a single period to get the values of the probability function. All this is explained in the linked page.