We use the setup of the third part. For $u \in \mathscr{M}_{d,1}(\mathbb{R})$, we denote $T(u) = (T_i(u))_{1 \leqslant i \leqslant d} \in \mathbb{R}^d$ the vector defined by $$T_i(u) = \operatorname{Var}\left(\langle L_i, u \rangle\right) \quad \text{for } i \in \{1,\ldots,d\}.$$ Show that for all $n \geqslant 0$, $$\mathbb{E}\left(\langle X_n, u \rangle^2\right) = \mathbb{E}\left(\langle X_0, M^n u \rangle^2\right) + \sum_{k=0}^{n-1} \langle x_0 M^k, T\left(M^{n-1-k} u\right) \rangle$$ (with the convention that the sum indexed by $k$ is zero if $n = 0$).
We use the setup of the third part. For $u \in \mathscr{M}_{d,1}(\mathbb{R})$, we denote $T(u) = (T_i(u))_{1 \leqslant i \leqslant d} \in \mathbb{R}^d$ the vector defined by
$$T_i(u) = \operatorname{Var}\left(\langle L_i, u \rangle\right) \quad \text{for } i \in \{1,\ldots,d\}.$$
Show that for all $n \geqslant 0$,
$$\mathbb{E}\left(\langle X_n, u \rangle^2\right) = \mathbb{E}\left(\langle X_0, M^n u \rangle^2\right) + \sum_{k=0}^{n-1} \langle x_0 M^k, T\left(M^{n-1-k} u\right) \rangle$$
(with the convention that the sum indexed by $k$ is zero if $n = 0$).