Let $P$ the transition probability matrix and $\mu$ the row vector of initial distribution. $$P_\mu(X_n=j)=\sum_j\mu(i)p^n(i,j)=\mu p^n(j)$$
I don't want to make a proof of that, I want to understand it. Suppose I have a $3\times3$ transition probability matrix $$\begin{bmatrix}P_{00}&P_{01}&P_{02}\\P_{10}&P_{11}&P_{12}\\P_{20}&P_{21}&P_{22}\end{bmatrix}$$ and I have that $P(X_0=0)=p_0$, $P(X_0=1)=p_1$,$P(X_0=2)=p_2$ and I want to calulate $E[X_2]$
From what I know $$E[X_2]=0\cdot P(X_2=0)+1\cdot P(X_2=1)+2\cdot P(X_2=2)$$ but how I can calculate that probabilities?
For me it looks like $$P(X_2=0)=P(X_2=0|X_0=0)p_0+P(X_2=0|X_0=1)p_1+P(X_2=0|X_0=2)p_2$$ $$P(X_2=1)=P(X_2=1|X_2=0)p_0+P(X_2=1|X_0=1)p_1+P(X_2=1|X_0=2)p_2$$ $$P(X_2=2)=P(X_2=2|X_0=0)p_0+P(X_2=2|X_0=1)p_1+P(X_2=2|X_0=2)p_2$$ then
$$P(X_2=0)=P_{00}^2p_0+P_{10}^2p_1+P_{20}^2p_2$$ $$P(X_2=1)=P_{01}^2p_0+P_{11}^2p_1+P_{21}^2p_2$$ $$P(X_2=2)=P_{02}^2p_0+P_{12}^2p_1+P_{22}^2p_2$$
but that's not make sense for me, because $$P(X_2=0|X_0=0)p_0+P(X_2=0|X_0=1)p_1+P(X_2=0|X_0=2)p_2$$ $$(*)P(X_2=0\cap X_0=0)+P(X_2=0\cap X_0=1)+P(X_2=0\cap X_0=2)\neq^? P(X_2=0)$$
perhaps this notation is confusing me, grateful for any explanation.
EDIT: Can someone explain to me in detail why $(*)$ is true?