BACKGROUND
I'm working on a physics problem whereby I want to check whether a vector $\vec{v}$ of $D \in \mathbb{N}$ measurements $v_d \in [-1, 1]$, $d \in \{1, \cdots, D\}$ can result from the statistical averaging of $K \in \mathbb{N}$ possible measurement outcomes $\vec{w}^{(k)}$, $k \in \{1, \cdots, K \}$ where $w_d^{(k)} \in \{-1, 1\}$. (Note that unlike the elements of $\vec{v}$, those of $\vec{w}^{(k)}$ can only be $\pm 1$).
In other words, I want to check whether there exists a probability distribution $\vec{p}$ of dimension $K$ such that the following three conditions are satisfied, namely
$ \begin{equation} v_d = \sum\limits_{k=1}^{K} p_k w^{(k)}_d, \end{equation} $ $\sum\limits_{k=1}^{k} p_k = 1$, and $p_k \in [0, 1]$.
This is therefore an overcomplete problem if $K > D$, and, if I understand correctly, boils down to finding whether $\vec{v}$ lies within the convex hull formed by $\{\vec{w}^{(k)}\}\mid_{k=1}^K$.
QUESTION
How does one resolve the problem above, both analytically and algorithmically? Also, if it turns out that $\vec{v}$ can't be explained as a statistical mixture of $\vec{w}^{(k)}$'s, then can we quantify how far it is from being so, i.e., presumably, how far it lies from the convex hull?
To check containment of convex hull we usually have to compute the convex hull and then decide whether the new point is actually within this convex hull.
While there are many algorithms to compute the convex hull, checking the containment of a point within a convex hull is usually done using linear programming solver. (also see that it is roughly equivalent here).
To actually do this in practice I recommend relying on existing software (as understanding these algorithms and implementing them efficiently takes a lot of time). For example scipy has an arbitrary dimension convex hull algorithm built in, and requires only a few lines of code.
The only point that I see that might make your problem a little bit easier is the fact that your $w^{(k)}$ are the vertices of some hyper cube. (So a simple thing to do would first be checking whether $v \in [-1,1]^D$ so you can avoid doing the more expensive computations.)