SlowGuess/ACL_2020_OCR huggingface dataset:
The contrapositive of the above statement implies that if  $\nexists v$ , where  $\forall x_{i} \in p$  we have  $\langle v, x_{i} - p \rangle \leq 0$ , then  $p$  is on the convex hull. In fact, we also note that if  $p$  was a vertex, the inequality would be strict, which implies that one can find a test point such that the probability  $\mathbb{P}(p) \to 1$ .
The more interesting case is if the point  $p$  is on the convex hull, but not a vertex. In this case we define the set  $\Omega(p, h) = \{x_i \in P \mid \langle h, p - x_i \rangle = 0\}$ . This corresponds to the set of points lying directly on the hyperplane perpendicular to  $h$ , running through  $p$ . This set is nonempty. Then we see that:
$$
\begin{array}{l} \mathbb {P} (p) = \frac {\exp (\langle h , p \rangle)}{\sum_ {j} \exp (\langle h , x _ {j} \rangle)} \\ \leq \frac {\exp (\langle h , p \rangle)}{\sum_ {j \in \Omega (p , h)} \exp (\langle h , x _ {j} \rangle)} \\ = \frac {1}{| \Omega (p , h) |} \\ \end{array}
$$
B 3D Numerical Illustration
Figure 4: Numerical Illustration in 3D The top panels show six words in a flattened cross-section of 3D space. Points  $A$ ,  $B$ ,  $C$ ,  $D$  and  $E$  are embedded at  $(0,0,0)$ ,  $(1,0,0)$ ,  $(0,1,0)$ ,  $(1,1,0)$  and  $(0.5,0.5,1)$  respectively. In the top-left panel,  $F$  is embedded outside of the convex hull at  $(0.65,0.35,1.5)$ , and in the top-right panel  $F$  is embedded inside of the convex hull at  $(0.65,0.35,0.5)$ . Subsequent panels show cross sections of the probability of  $F$  for test points in the plains  $z = \{0.0,2.0,4.0,6.0\}$ , numerically illustrating the stolen probability effect in 3D.