Is $\exp(-2\sin^2t)$ a characteristic function?

Question

Is $\exp(-2\sin^2t)$ the characteristic function of some random variable?

Answer 1

I'll consider a generalization. First, observe that $$\phi(t):=\exp(-2 z\sin^2 t)=e^{-z}\exp(z \cos 2t)$$ is a $\pi$-periodic function and so has a Fourier series. (The question above is the case of $z=1$.) To be explicit, recall the Jacobi-Anger expansion of $e^{z \cos t}$ (see equation 10.35.3 in the DLMF) to write $$\phi(t)=\sum\limits_{k=-\infty}^\infty I_k(z)e^{-z}\,e^{2i k t}.$$ Since the complex exponential $e^{in t}$ is the Fourier transform of $\delta(x-n)$, we conclude that $\phi(t)$ is indeed the CF of a discrete random variable taking values at even integers. At this point, working out the particular probabilities (and checking normalization) is a simple exercise which I'll leave to the reader.

Answer 2

An approach which requires zero knowledge about modified Bessel functions $I_\nu$ and is valid for every nonnegative real number $u$ is to expand the function $$\phi:t\mapsto\exp(-4u\sin^2t),$$ using the identities $2-4\sin^2t=2\cos(2t)=\mathrm e^{2it}+\mathrm e^{-2it}$, as $$\phi(t)=\mathrm e^{-2u}\exp(u\mathrm e^{2it})\exp(u\mathrm e^{-2it})=\mathrm e^{-2u}\sum_k\frac{u^k}{k!}\mathrm e^{2ikt}\sum_\ell\frac{u^\ell}{\ell!}\mathrm e^{-2i\ell t},$$ that is, $$\phi(t)=\sum_n\left(\mathrm e^{-2u}\sum_{k-\ell=n}\frac{u^{k+\ell}}{k!\ell!}\right)\mathrm e^{2in t}.$$ Every parenthesis is nonnegative (this is where one uses that $u\geqslant0$) hence $\phi$ is indeed the characteristic function of an integer valued random variable $X$, with a symmetric distribution such that, for every $n\geqslant0$, $$P(X=2n)=P(X=-2n)=\mathrm e^{-2u}\sum_{k-\ell=n}\frac{u^{k+\ell}}{k!\ell!}=\mathrm e^{-2u}u^n\sum_{\ell=0}^\infty \frac{u^{2\ell}}{\ell!(\ell+n)!}. $$