# Markov Chain: pmf at future time steps?

I have the following markov chain with the state-transition probability matrix: $$W = \begin{bmatrix} 0.7 & 0.3 & 0\\ 0.75 & 0.05 & 0.2\\ 1 & 0 & 0 \end{bmatrix}$$

I know the chain is irreducible.

Starting from a discrete uniform distribution, obtain the state pmf at the second and third time steps

I know the following:

Let the pmf of $$\mathbb{X_0}$$ be $$\lambda_0$$ then: $$\lambda_n = \lambda_{n-1}W = \lambda_0W^n$$ where $$W^n$$ represents an $$n$$-step transition matrix.

The pmf of a discrete uniform distribution is $$\frac{1}{n}$$ where $$n$$ is the number of values in this case.

Question:

So, is $$\lambda_0 = \frac{1}{3}$$ since there are three states? and I simply plug $$\lambda_0$$ into the above equation, compute powers of $$W$$ and I'm done? Or am I missing anything?

You are basically correct, except that discrete uniform means you end up in all states with the same probability, so $$\lambda_0 = \frac13 \begin{bmatrix} 1 & 1 & 1 \end{bmatrix}$$ and therefore $$\lambda_1 = \lambda_0 W = \frac13 \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 0.7 & 0.3 & 0\\ 0.75 & 0.05 & 0.2\\ 1 & 0 & 0 \end{bmatrix}$$