\section{Proof of theoretical results}
\label{sec:proof}

\begin{proof}[Simplified sketch of proof of \ref{lem:1}].
 Let $\mathbf{x}_\text{ID}$ (resp. $\mathbf{x}_\text{OoD}$) such that $\mathbf{x}_\text{ID} \sim \mathbb{P}_{\text{ID}}(\mathbf{x})$ (resp. $\mathbf{x}_\text{OoD} \sim \mathbb{P}_\text{OoD}(\mathbf{x}))$.
We suppose without loss of generality that for both inputs, softmax layer yields $y_\eta \in \llbracket 1,K\rrbracket$ as a label and same cosine similarity. 


Moreover we recall that $\mathbf{x}_\text{OoD}$ is such that $\mathbf{x}_\text{OoD}$ verify $(\mathbf{x}_{\text{OoD}}, y_\eta) \in \{(\mathbf{x},y) |\exists j \in {1,...K}, \cos(\phi_j) \geq 1- \varepsilon \}$.  Because MSP correctly classified $\mathbf{x}_\text{ID}$ as ID and $\mathbf{x}_\text{OoD}$ as OoD, it yields  \begin{equation}
    \mathbb{P}(y_\eta|\mathbf{x}_\text{OoD}) \propto \exp(\|\mathbf{w}_\eta\|\|\mathbf{z}_\text{OoD}\|\cos(\phi_\eta)) \leq \mathbb{P}(y_\eta|\mathbf{x}_\text{ID}) \propto \exp(\|\mathbf{w}_\eta\|\|\mathbf{z}_\text{ID}\|\cos(\phi_\eta)).
\end{equation}
Using that $x \mapsto \log(x)$ is increasing, it yields $\|\mathbf{z}_{\text{OoD}}\| \leq \|\mathbf{z}_{\text{ID}} \|$. \\
For a general proof, it should utilize the Neural Collapse (NC) property, which implies that all classification vectors have the same norm:  $\|\mathbf{w}_1\| = \dots = \|\mathbf{w}_K\|$. and should account for its angular property, which arises from the way  $\mathbf{x}_\text{OoD}$ is selected.\end{proof}

\begin{lemma}
Let \(V : \Omega \to \mathbb{R}^d\) be a random vector distributed uniformly on the spherical cap
\[
C_\Phi(\boldsymbol{\varphi}) \;=\; \bigl\{ x \in S^{d-1} : x^\top \boldsymbol{\varphi} \ge \cos \Phi \bigr\},
\]
where \(\boldsymbol{\varphi} \in S^{d-1}\) is a fixed unit vector and \(\Phi \in [0,\pi]\) is a given angle. Then, the first moment of \(V\) is given by
\[
\mathbb{E}[V] \;=\; c\,\boldsymbol{\varphi},
\]
\end{lemma}

\begin{proof}[Proof of \ref{thm:theorem43}]
 
 
 For all $ i \in \llbracket 1,M\rrbracket$ we set $\mathbf{z}^{(i)} = \|\mathbf{z}\|\boldsymbol{\alpha}^{(i)}$.  Recall that $\Var(Z_M) = \text{Tr}\left(\frac{1}{M-1} \sum_{i=1}^M (\mathbf{z}^{(i)} - \boldsymbol{\mu})(\mathbf{z}^{(i)} - \boldsymbol{\mu})^T\right)$. 
 Then : 
 \begin{equation}
          \Var(Z_M) = \frac{1}{M-1}\sum_{i=1}^M\|\mathbf{z}^{(i)} - \boldsymbol{\mu}\|^2.
 \end{equation}
Expanding the sum knowing that $ \forall i ,\|\mathbf{z}_i\| = \|\mathbf{z}\|$ and $\boldsymbol{\mu} = \frac{1}{M}\sum_{i=1}^M z_i = \frac{\|\mathbf{z}\|}{M}\sum_{i=1}^M  \boldsymbol{\alpha}^{(i)}$ we have: 
\begin{equation} \label{eq:sumu}
    \sum_{i=1}^M \|\mathbf{z}^{(i)} - \boldsymbol{\mu}\|^2 = \sum_{i=1}^M \left( \|\mathbf{z}\|^2 - 2\|\mathbf{z}\|\langle \boldsymbol{\alpha}^{(i)}, \boldsymbol{\mu} \rangle + \|\boldsymbol{\mu}\|^2 \right) = M\|\mathbf{z}\| - 2\frac{\|\mathbf{z}\|^2}{M} \langle \sum_{i=1}^M \boldsymbol{\alpha}^{(i)}, \sum_{j=1}^M \boldsymbol{\alpha}^{(j)} \rangle + M\|\boldsymbol{\mu}\|^2.
\end{equation}
Expanding $\|\boldsymbol{\mu}\|^2 = \frac{\|\mathbf{z}\|^2}{M^2} \|\sum_{i=1}^M \boldsymbol{\alpha}^{(i)}\|^2 =  \frac{\|\mathbf{z}\|^2}{M^2} \left( \sum_{i=1}^M \|\boldsymbol{\alpha}^{(i)}\|^2 + \sum_{i \neq j} \langle\boldsymbol{\alpha}^{(i)},\boldsymbol{\alpha}^{(j)}\rangle \right)$, same for $\langle \sum_{i=1}^M \boldsymbol{\alpha}^{(i)}, \sum_{j=1}^M \boldsymbol{\alpha}^{(j)} \rangle$, we have :

\begin{equation}
\label{eq:bigsum}
\frac{1}{M-1}\sum_{i=1}^M \|\mathbf{z}^{(i)} - \boldsymbol{\mu}\|^2 = \frac{M}{M-1}\|\mathbf{z}\|^2 - \frac{\|\mathbf{z}\|^2}{M(M-1)}\left( \sum_{i=1}^M \|\boldsymbol{\alpha}^{(i)}\|^2 + \sum_{i \neq j} \langle\boldsymbol{\alpha}^{(i)},\boldsymbol{\alpha}^{(j)}\rangle \right).
\end{equation}

As $\boldsymbol{\alpha}^{(i)},... \boldsymbol{\alpha}^{(M)}$ are i.i.d., uniform on the spherical cap, we have $\mathbb{E}[\boldsymbol{\alpha}^{(i)}] = c\boldsymbol{\varphi} $, taking expectation, we have :
\begin{equation}
\mathbb{E}\left[\sum_{i=1}^M \|\boldsymbol{\alpha}^{(i)}\|^2 + \sum_{i \neq j} \langle\boldsymbol{\alpha}^{(i)},\boldsymbol{\alpha}^{(j)}\rangle  \right] = M + M(M-1)c^2.
\end{equation}
Therefore we have the following \textbf{expectation equality}: 
\begin{equation}
    \mathbb{E}[\Var(Z_M)] = \frac{M\|\mathbf{z}\|^2}{M-1} - \frac{\|\mathbf{z}\|^2(1+(M-1)c^2)}{M-1}  = \|\mathbf{z}\|^2 \left( \frac{M-1}{M-1}- \frac{(M-1)c^2}{M-1} \right) = (1-c^2)\|\mathbf{z}\|^2.
\end{equation}

Taking back Eq. \eqref{eq:sumu} and observing that by the strong law of large number\addEmi{s}, \begin{equation}
    \|\boldsymbol{\mu}\|^2 = \frac{\|\mathbf{z}\|^2}{M^2} \|\sum_{i=1}^M \boldsymbol{\alpha}^{(i)}\|^2 \to_{a.s} c^2\|\mathbf{z}\|^2 ,\quad M \to + \infty.
\end{equation}
Therefore, we have the \textbf{following asymptotic convergence}: \begin{equation}
    \Var(Z_M) \to_{a.s} (1-c^2)\|\mathbf{z}\|^2, \quad M \to + \infty.
\end{equation}

\end{proof}