I am currently studying Markov models. I am presented with the following definition, theorem, and proof:
Definition 3. If
$$\pi^*_{ij} := \lim_{n \to \infty} m_{ij}(n)/n$$
exists for all $i, j \in S$ and $\sum_{j \in S} \pi^*_{ij} \equiv 1$, then $\vec{\pi}^*_i = (\pi^*_{ij} : j \in S)$ is called an occupation law. In matrix form,
$$\Pi^* = [\pi^*_{ij}] = \lim_{n \to \infty} n^{-1} (\mathcal{I} + \mathcal{P} + \dots + \mathcal{P}^n). \tag{4}$$
Theorem 5. An occupation law is a stationary law.
Proof. We are assuming that the defining limits exist, so the right-hand side of (4) equals
$$\lim_{n \to \infty} (\mathcal{I}/n) + \lim_{n \to \infty} n^{-1}(\mathcal{P} + \dots + \mathcal{P}^n) = [0] + \lim_{n \to \infty} n^{-1}(\mathcal{I} + \dots + \mathcal{P}^{n - 1})\mathcal{P} = \Pi^* \mathcal{P},$$
i.e., $\Pi^* = \Pi^* \mathcal{P}$. The rows of this matrix identity have the form (3), i.e., $\vec{\pi}^*$ is stationary.
The definition for the stationary laws is as follows:
Stationary laws
Definition 2. A stationary law is any non-negative solution $\vec{\pi}$ of the balance plus mass equations, i.e.,
$$\vec{\pi} = \vec{\pi}\mathcal{P} \ \ \ \ \ \text{&} \ \ \ \ \ \sum_{j \in S} \pi_j = 1. \tag{3}$$
Note that if $\vec{\pi}\mathcal{P} = \vec{\pi}$, then $\vec{\pi} \mathcal{P}^n = \vec{\pi}$ for all $n$. This follows by applying the first result iteratively.
And a related theorem and proof is as follows:
Definition 1. If the limit
$$\pi_{ij} = \lim_{n \to \infty} p^{(n)}_{ij}$$
exists for all $i, j \in S$, and if
$$\sum_{j \in S} \pi_{ij} = 1, \ \ \ (i \in S), \tag{1}$$
then for each $i$ we say that the row vector $\vec{\pi}_i = (\pi_{ij} : j \in S)$ is a limit law (or limiting distribution).
If $\vec{\pi}_i$ is a limit law, then $\vec{\pi}_i = \vec{\pi}_i \mathcal{P}$, i.e.,
$$\pi_{ij} = \sum_{k \in S} \pi_{ik}, \ \ \ (i, j \in S) \tag{2}$$
Proof: Define the square matrix $\Pi = [\pi_{ij}]$, which exists by assumption. Definition 1 can be restated as $\Pi = \lim_{n \to \infty} \mathcal{P}^n$. But $\mathcal{P}^n = \mathcal{P}^{n - 1}\mathcal{P}$, so
$$\Pi = \lim_{n \to \infty} P^n = \lim_{n \to \infty} \mathcal{P}^{n - 1} \mathcal{P} = \Pi \mathcal{P}$$
I am having difficulty understanding this part of the first proof:
$$[0] + \lim_{n \to \infty} n^{-1}(\mathcal{I} + \dots + \mathcal{P}^{n - 1})\mathcal{P} = \Pi^* \mathcal{P}$$
Specifically, I do not understand how we get the result that the limit
$$\lim_{n \to \infty} n^{-1}(\mathcal{I} + \dots + \mathcal{P}^{n - 1})\mathcal{P}$$
results in
$$\Pi^* \mathcal{P}.$$
I would greatly appreciate it if people would please take the time to clarify this.
This is really not particularly tied to Markov chains. It is simply the following calculation that makes sense for matrices $A$ with complex entries: if $L:=\lim_{n \to \infty} \frac{\sum_{k=0}^n A^k}{n}$ exists, then
$$L=\lim_{n \to \infty} \frac{I}{n} + \lim_{n \to \infty} \frac{\sum_{k=1}^n A^k}{n} = 0 + \left ( \lim_{n \to \infty} \frac{\sum_{k=0}^{n-1} A^k}{n} \right ) A.$$
This follows by linearity of the limit operation (first step) and continuity of linear transformations (second step). Now that inner limit is again $L$, because $\frac{\sum_{k=0}^{n-1} A^k}{n} = \frac{\sum_{k=0}^{n-1} A^k}{n-1} \frac{n-1}{n}$ and $\frac{n-1}{n} \to 1$. So $L=LA$. You could also have obtained $L=AL$, but that is not particularly useful in the Markov chain context, in which you want the rows of $L$ to be invariant under $p \mapsto pA$.