Existence of probability density on $S^1$ with first circular moment on $S^1$

Question

I am searching for a function $f$ on the unit circle $S^1$ with the following properties.

1. $f \geq 0$ almost everywhere.
2. $\int_{S^1} f = 1$.
3. $\int_{S^1} e^{i \theta} f (\theta) d \theta = 1$.

In other words, I would like a probability density $f$ on the unit circle whose first circular moment also lies on the unit circle. Using a little Fourier series, the conditions 2 and 3 above are equivalent to saying $f$ has the form

$$ f(\theta) = \frac{1}{2 \pi} \left( 1 + 2 \cos{\theta} \right) + \sum_{n \in \mathbb{Z} \backslash \{ \pm 1 , 0 \}} a_n e^{in\theta} \tag{*} $$

for some coefficients $a_n \in \mathbb{C}$. The question then becomes whether it is possible to choose the $a_n$ so that $f$ is non-negative. My intuition tells me, if such a function exists, then it would be rather pathological.

Question: Is some able to demonstrate the existence/non-existence of such a function? If such a function exists and is "nice" (e.g. smooth), is a closed-form possible?

Context: I've been studying some directional statistics and I've noticed that many of the wrapped distributions have the property that their circular moments do not lie on the unit circle. For example, the wrapped normal distribution has circular moments $\langle e^{in\theta} \rangle = e^{in\mu - n^2 \sigma^2 / 2}$, where $\mu$ and $\sigma$ are the mean and standard deviation of the unwrapped distribution. Clearly, this lies on $S^1$ only if $\sigma = 0$, in which case the wrapped normal becomes a Dirac comb distribution. Part of me feels like this behavior is reasonable: depending on where the distribution is concentrated, it will "pull" the unit vector $e^{in\theta}$ toward the concentration center (in this case $\mu$), but not "all the way". For a uniform distribution, e.g. $\sigma \rightarrow + \infty$, the circular moments all lie at the origin (i.e., they all get "pulled" equally in all directions). I started to wonder if this is always the case for probability densities on the unit circle. Or could I, at the very least, make the first moment lie on the unit circle?

Answer 1

The point 1 is an extreme point of $S^1$, so there is no such $f$. Indeed, if $\mu$ is a probability measure on $S^1$ (where $S^1$ is identified with $[-\pi,\pi)$) such that $$\int_{-\pi}^{\pi} e^{i\theta} \, d\mu(\theta)= 1 \tag{1}\,,$$ then $\mu$ must be the Dirac measure at 1.

Proof $\;$ Taking real part of (1) and rearranging gives $$\int_{-\pi}^\pi (1-\cos \theta)\, d\mu(\theta)= 0\tag{2}\,.$$ The integrand in (2) is nonnegative, and at least $(1-\cos \epsilon)$ for $\theta \in [-\pi,\pi) \setminus [-\epsilon,\epsilon]$. Therefore $\mu([-\epsilon,\epsilon])=1$ for every $\epsilon>0$, which forces $\mu(\{1\})=1$.

https://en.wikipedia.org/wiki/Extreme_point