Throughout part III, $N$ is a fixed non-zero natural integer and $(X_n)_{n \in \mathbb{N}}$ is a homogeneous Markov chain on $\llbracket 0, N \rrbracket$ with transition matrix $Q$ where $q_{i,j} = P(X_{n+1} = j \mid X_n = i) > 0$. We define $a_{i,j}(t) = q_{i,j} \mathrm{e}^{jt}$, $A(t) = \left(a_{i,j}(t)\right)_{0 \leqslant i,j \leqslant N}$, $z_j(t) = P(X_1 = j)\mathrm{e}^{jt}$, $Z(t) = \begin{pmatrix} z_0(t) \\ \vdots \\ z_N(t) \end{pmatrix}$, and $Y^{(n)}(t) = (A(t))^{n-1} Z(t)$ so that $E\left(\mathrm{e}^{tS_n}\right) = \sum_{j=0}^N Y_j^{(n)}(t)$. Let $\gamma(t)$ be the dominant eigenvalue of $A(t)$. Show that $\lim_{n \rightarrow +\infty} \frac{\ln\left(E\left(\mathrm{e}^{tS_n}\right)\right)}{n} = \lambda(t)$ where $\lambda(t) = \ln(\gamma(t))$.
Throughout part III, $N$ is a fixed non-zero natural integer and $(X_n)_{n \in \mathbb{N}}$ is a homogeneous Markov chain on $\llbracket 0, N \rrbracket$ with transition matrix $Q$ where $q_{i,j} = P(X_{n+1} = j \mid X_n = i) > 0$. We define $a_{i,j}(t) = q_{i,j} \mathrm{e}^{jt}$, $A(t) = \left(a_{i,j}(t)\right)_{0 \leqslant i,j \leqslant N}$, $z_j(t) = P(X_1 = j)\mathrm{e}^{jt}$, $Z(t) = \begin{pmatrix} z_0(t) \\ \vdots \\ z_N(t) \end{pmatrix}$, and $Y^{(n)}(t) = (A(t))^{n-1} Z(t)$ so that $E\left(\mathrm{e}^{tS_n}\right) = \sum_{j=0}^N Y_j^{(n)}(t)$. Let $\gamma(t)$ be the dominant eigenvalue of $A(t)$. Show that $\lim_{n \rightarrow +\infty} \frac{\ln\left(E\left(\mathrm{e}^{tS_n}\right)\right)}{n} = \lambda(t)$ where $\lambda(t) = \ln(\gamma(t))$.