Let $v \in \mathbb{R}^n$. I would like to find probability distribution $p \in \mathbb{R}^n$ such that $\mathbb{E}_p[v] = 0$. Meaning that, find $p \in \mathbb{R}^n$ such that $p \geq 0, \sum_ip_i=1$ and $$ p^Tv = 0 $$
- Is there any closed form solution for $p$?
- If not, is there any algorithm that can find such $p$?
Note: To make sure that there exists a solution for the problem, I'm assuming that $v$ has both positive and negative coordinates.
Consider the related problem of finding all non-negative solutions $p$ to $p^T v = 1$ where $v$ is a vector with positive components. Clearly $(1/v_1,0,0,...,0)$, $(0,1/v_2,0,...,0)$, ..., $(0,0,...,0,1/v_n)$ are solutions. The set of all non-negative solutions $p$ is the convex hull of these fundamental solutions.
To solve your problem, let's rearrange the coordinates of $v$ to express it as $v = (v_+,v_0,v_-)$. We can then take any solutions to $p_+^T v_+ = 1$ and $p_-^T (-v_-) = 1$, along with any non-negative vector $p_0$ and form $p = (p_+,p_0,p_-)$. Normalizing this to sum to $1$ will give you the required probability distribution. Indeed, this procedure describes the space of all possible probability vectors orthogonal to $v$ (well, almost: we also have to include the case where $p_+$ and $p_-$ are zero, but $p_0$ is non-zero).