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 $$T_i(u) = \operatorname{Var}\left(\langle L_i, u \rangle\right) \quad \text{for } i \in \{1,\ldots,d\}.$$ (a) Show that for all $u \in \mathscr{M}_{d,1}(\mathbb{R})$, $y \in \mathscr{M}_{1,d}(\mathbb{N})$ and $n \geqslant 0$, $$\mathbb{E}\left(\langle X_{n+1}, u \rangle^2 \mathbf{1}_{X_n = y}\right) = \mathbb{P}(X_n = y)\left(\langle y, Mu \rangle^2 + \langle y, T(u) \rangle\right).$$ (One may use without proof the fact that, for all $n \geqslant 0$, the random variables $\sum_{j=1}^{d} u_j L_{i,j}^{n,k} \mathbf{1}_{X_n = y}$ are pairwise independent when $k$ and $i$ vary.) (b) Show that for all $u \in \mathscr{M}_{d,1}(\mathbb{R})$ and $n \geqslant 0$, $$\mathbb{E}\left(\langle X_{n+1}, u \rangle^2\right) = \mathbb{E}\left(\langle X_n, Mu \rangle^2\right) + \langle x_0 M^n, T(u) \rangle.$$
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
$$T_i(u) = \operatorname{Var}\left(\langle L_i, u \rangle\right) \quad \text{for } i \in \{1,\ldots,d\}.$$
(a) Show that for all $u \in \mathscr{M}_{d,1}(\mathbb{R})$, $y \in \mathscr{M}_{1,d}(\mathbb{N})$ and $n \geqslant 0$,
$$\mathbb{E}\left(\langle X_{n+1}, u \rangle^2 \mathbf{1}_{X_n = y}\right) = \mathbb{P}(X_n = y)\left(\langle y, Mu \rangle^2 + \langle y, T(u) \rangle\right).$$
(One may use without proof the fact that, for all $n \geqslant 0$, the random variables $\sum_{j=1}^{d} u_j L_{i,j}^{n,k} \mathbf{1}_{X_n = y}$ are pairwise independent when $k$ and $i$ vary.)
(b) Show that for all $u \in \mathscr{M}_{d,1}(\mathbb{R})$ and $n \geqslant 0$,
$$\mathbb{E}\left(\langle X_{n+1}, u \rangle^2\right) = \mathbb{E}\left(\langle X_n, Mu \rangle^2\right) + \langle x_0 M^n, T(u) \rangle.$$