Markov Chain $3$-step state $X_3 \mid X_0=1$ conditional probabilities, mean and variance. Consider the Markov chain with one-step transition probability matrix:
$$P=
\begin{bmatrix}
1/3 &amp;1/3  &amp;1/3 \\ 
3/4 &amp;1/8  &amp;1/8\\ 
1/4 &amp;1/2  &amp;1/4
\end{bmatrix}$$ and initial probabilities $p_{0} = p_{1} = p_{2} = 1/3.$


Compute the three-step transition probability matrix.
Find $ P (X_{0} = 1, X_{3} = 0).$
Find $P (X_{3} = 0).$
Find $P (X_{0}= 1\mid X_{3}= 0)$.
Find $E[X_3 \mid X_0 = 1]$ and $Var(X_3 \mid X_0 = 1).$


I need help to solve to solve part 5.
In part 1. you have calculated the matrix $$P^{(3)}=P^3=(p^{(3)}_{ij})_{i,j \in \Bbb X}$$ where $\Bbb X=\{0,1,2\}$. So, for $k\in \Bbb X$ you have that $$P(X_3=k\mid X_0=1)=p^{(3)}_{1k}$$ (these will be the entries of the first line of the matrix $P^{3}$). These values completely determine the pmf of the random variable $X_3 \mid X_0=1$ and the computation of $E[X_3 \mid X_0=1]$ and $Var(X_3 \mid X_0=1)$ are then standard. 



Just to compare your result: $$P^3=P=
\begin{bmatrix}
193/432 &amp; 529/1728 &amp; 427/1728\\ 
\dots &amp; \dots &amp;\dots\\ 
\dots &amp;\dots  &amp; \dots
\end{bmatrix}$$ Hence $$E[X_3\mid X_0=1]=0\cdot\frac{193}{432}+1\cdot\frac{529}{1728}+2\frac{427}{1728}$$ and $$Var(X_3\mid X_0=1)=E[X_3^2\mid X_0=1]+\left(E[X_3\mid X_0=1]\right)^2$$ with $$E[X_3^2\mid X_0=1]=0^2\cdot\frac{193}{432}+1^2\cdot\frac{529}{1728}+2^2\frac{427}{1728}$$Can you finish the calculations?
For a) 

we just calculate $P^{3}$

For b) 

$ P[X_{0}=1,X_{3}=0]=P[X_{3}=0|X_{0}=1]*P[X_{0}=1]=P^{3}_{01}*P_{1}=$
$P[X_{0}=1|X_{3}=0]*P[X_{3}=0] $

For c)

we multiple [ 1/3 1/3 1/3 ] by $P^{3}$ the first elements gives us $P[X_{3}=0]$

For d) 

since we know $ P[X_{0}=1,X_{3}=0]=P[X_{3}=0|X_{0}=1]*P[X_{0}=1]=P^{3}_{01}*P_{1}=$
$P[X_{0}=1|X_{3}=0]*P[X_{3}=0] $ we can just dive $P[X_{0}=1,X_{3}=0]$ by $P[X_{3}=0]$

I don't know how to solve part e) I need help