# Probability distribution of markov chain

I have a Markov chain with state space $E = \{1,2,3,4,5\}$ and transition matrix below:

$$\begin{bmatrix} 1/2 & 0 & 1/2 & 0 & 0 \\ 1/3 & 2/3 & 0 & 0 & 0 \\ 0 & 1/4 & 1/4 & 1/4 & 1/4 \\ 0 & 0 & 0 & 3/4 & 1/4 \\ 0 & 0 & 0 & 1/5 & 4/5 \end{bmatrix}$$

Given the initial distribution $\pi = (1/2,0,0,1/2,0)$, how would I calculate $\mathbb{P}(X_2 = 4)$? Do I have to multiply $\pi$ by the transition matrix etc.?

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If "initial distribution" means the distribution of $X_1$, then the distribution of $X_2$ is $\pi M$, where $M$ is the transition matrix above.
If "initial distribution" means the distribution of $X_0$, then the distribution of $X_2$ is $\pi M^2$.
Thanks Michael! I think the question means the "initial distribution" means the distribution of $X_0$. Just one quick question, I cant seem to multiply $M^2 \pi$, because the dimensions (row and columns) are not correct, but I can multiply $\pi M^2$, and I get the correct answer $(59/160)$. So the distribution $M^2 \pi$ or $\pi M^2$? – Henry Apr 10 '12 at 23:41
It should be $\pi M^2$. This is a "right stochastic matrix" (rows sum up to 1), which means it's on the "right" of a matrix multiplication. It is my understanding that transition matrices are by convention situated on the right. I'm not sure why this is so; maybe someone else can enlighten us. – Yang Apr 11 '12 at 0:01