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)?



von Mises Distribution property
Yes, this is true.


Indeed, the first [circular moment](https://en.wikipedia.org/wiki/Von_Mises_distribution#Moments) 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](https://en.wikipedia.org/wiki/Von_Mises_distribution#Definition). $\quad\Box$


