von Mises Distribution property

Question

The von Mises distribution has the highest entropy for a given first circular moment. It seems that the converse is true: for a fixed entropy, the von Mises distribution has the highest first circular moment (in magnitude). This seems like something that should already be proven somewhere, but I haven't been able to find it, nor have my efforts at proving it met with success so far. Does anyone know where a proof of this is (or failing that, how to prove it)?

Answer 1

Yes, this is true.

Indeed, the first circular moment for a probability density $f$ on the interval $[0,2\pi]$ is $$m_1(f):=\int_0^{2\pi}e^{ix}f(x)\,dx.$$ So, $$\begin{aligned}&|m_1(f)| \\ &=\max_{u\in[0,2\pi]}\Big(\cos u\,\int_0^{2\pi}\cos x\,f(x)\,dx +\sin u\,\int_0^{2\pi}\sin x\,f(x)\,dx\Big) \\ &=\max_{u\in[0,2\pi]}\int_0^{2\pi}\cos(x-u)\,f(x)\,dx. \end{aligned}$$

So, a density $f$ maximizing $|m_1(f)|$ given the entropy $H:=-\int_0^{2\pi}f\ln f$ is maximizing $$I_u(f):=\int_0^{2\pi}\cos(x-u)\,f(x)\,dx$$ for some $u\in[0,2\pi]$, again given the entropy. Using Lagrange multipliers, we see that for such a maximizer $f$ we have $$\rho\cos(x-u)=\lambda\ln f(x)+\mu$$ for some real $\rho,\lambda,\mu$ with $\rho^2+\lambda^2+\mu^2>0$ and almost all (a.a.) $x\in[0,2\pi]$. So, for some real $a$ and $b$ and a.a. $x\in[0,2\pi]$ we have $\ln f(x)=a\cos(x-u)+b$ and hence $f(x)=ce^{a\cos(x-u)}$ for $c:=e^b$. Thus, $f$ is a von Mises density. $\quad\Box$