17 - Joint Distribution of a Markov Chain

Joint Distribution of a Markov Chain

Let $\lambda$ be the initial distribution (distribution of $X_0$ then

Distribution of $X_1$ is $\lambda P$
Generally distribution of $X_n$ is

$$\lambda P^n$$

Proof

Note that we treat $\lambda$ as a row vector

$$\begin{array}{rl}\mathbb{P}(X_n=j)& =\sum _i\mathbb{P}(X_0=i)\mathbb{P}(X_n=j\mid X_0=i)\\ & =\sum _i\mathbb{P}(X_0=i)\mathbb{P}(X_n=j\mid X_0=i)\\ & =\sum _i\mathbb{P}(X_0=i)p_{ij}^{(n)}\\ & =\sum _i{\lambda }_i(P^n{)}_{ij}\\ & =(\lambda P^n{)}_j\end{array}$$

Hence result follows