$K= \begin{pmatrix} 0.3 & 0.3 & 0.4\\ 0.2 & 0.7 & 0.1\\ 0.2 & 0.3 & 0.5 \end{pmatrix}$ is transition matrix of Markov-Chain with state space $S=\left\{0,1,2\right\}$. Compute $f_{00}(5)$ and $\mu_0=E(T_0 | X_0=0)$ where $T_0= \inf\left\{n \geq 0: X_n=0\right\}$
This is last task from exam and there was no solution but I try it because they can ask similar question in test next week. I want know I do it correct and if not how to do it correct and good?
For first part we need calculate $f_{00}(5)$. If I understand this notation correct, we need to take matrix $K$ to power $5$ and the cell in row $0$ column $0$ is what we looking for:
$$K^5 = \begin{pmatrix} 0.22223 & 0.49488 & 0.28289\\ 0.22223 & 0.50512 & 0.27266\\ 0.22223 & 0.49488 & 0.28290 \end{pmatrix} \Rightarrow f_{00}(5) = 0.22223$$
For calculate $\mu_0$ I'm not sure what is correct to do. I think I need to know what is the power of the matrix $K$ to $\infty$. I use software for this and the software tell me that the matrix towards power $\infty$, e.g. $$K^{\infty} \approx \begin{pmatrix} 0.22222 & 0.5 & 0.27778\\ 0.22222 & 0.5 & 0.27778\\ 0.22222 & 0.5 & 0.27778 \end{pmatrix}$$
But now is my problem what cell we need to take from this matrix? Because we have $X_0=0$ it must be something from row $0$ of the matrix but I don't know which column we need to use? :s
It is a slightly odd spectacle to see you try to reconstruct the theorems relevant to your questions about Markov chains, and, naturally, to fail big time.
For instance, every set of notes on the subject explains that, to compute $t_0=E(T_0\mid X_0=0)$, where $T_0=\inf\{n\geqslant1\mid X_n=0\}$, a simple approach is to consider every quantity $t_x=E(T_0\mid X_0=x)$, and the affine system they solve together.
In your case, one can further note that the Markov chain $X$ on $\{0,1,2\}$ induces a Markov chain $Y$ on $\{u,v\}$, where $u=\{0\}$ and $v=\{1,2\}$, with transitions $$\begin{pmatrix}0.3&0.7\\ 0.2&0.8\end{pmatrix}$$ hence, for this reduced chain, the mean hitting times $s_u=E(S_u\mid Y_0=u)$ and $s_v=E(S_u\mid Y_0=v)$, where $S_u=\inf\{n\geqslant1\mid Y_n=u\}$, solve $$s_u=1+0.7s_v\qquad s_v=1+0.8s_v$$ which yields readily $$t_0=s_u=4.5$$
But it is not needed to be that observant and one can simply solve for $t_0$ the $3\times3$ system $$t_0=1+0.3t_1+0.4t_2\qquad t_1=1+0.7t_1+0.1t_2\qquad t_2=1+0.3t_1+0.5t_2$$ which is not that difficult either and yields the same result.
Finally, if one insists on using the matrix $K^\infty$, one can as well since a somewhat more involved result, also in your notes, states that, for every $x$, $$t_x=\frac1{\pi_x}$$ where $(\pi_x)$ is the stationary distribution of the chain, and another equally important result states that, for every states $(x,y)$, $$K^\infty(x,y)=\pi_y$$ Thus, in your setting, for every $x$, $$t_0=\frac1{K^\infty(0,x)}$$ which may serve as a numerical confirmation of the computations above.
And of course, both methods are actually equivalent algebraically, and proving that they are is an enlightening exercise in matrix theory...