$X$ and $Y$ are independent random variables of Bernoulli with parameter $\frac{2}{3}$.
$Z=X+Y$
$\{X_n\}_{n \in \mathbb{N}}$ with values in {0,1,2} having $Z$ such as initial law and the transition matrix :
$$Q=\begin{bmatrix}0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 &\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0\end{bmatrix}$$
a. Calulate $E[ZX]$ and $Var[Z-2Y]$
b. Calculate the discrete density of the variables $X_0$, $X_1$ , $X_2$
c. Calculate $E[X_2]$ and $Var[X_1]$
d. Calculate the invariant probability measure
$ X \sim B(1, \frac{2}{3}) $ and $ Y \sim B(1, \frac{2}{3}) $
$$E[X^2+XY]=E[X^2]+E[X] E[Y]=E[X^2]+\frac{4}{9}$$ $$E[X^2]=Var[X]+E[X]^2=\frac{2}{3} \ \frac{1}{3}+ \frac{4}{9}= \frac{2}{3}$$ $$E[X^2+XY]=\frac{10}{9}=E[ZX]$$
$$Var[X-Y]=Var[X]+Var[Y]=\frac{2}{9} + \frac{2}{9}=\frac{4}{9} $$
How can I make the next steps?
Examining all the possibly cases of $Z$ values:
$$P(Z=0)=P(X=0, Y=0)=\frac{1}{9}$$
$$P(Z=1)=P(X=1, Y=0)+P(X=0, Y=1)=\frac{1}{3} \ \frac{2}{3}+\frac{1}{3} \ \frac{2}{3}= \frac{4}{9}$$
$$P(Z=2)=P(X=1, Y=1)=\frac{4}{9}$$
So, I have calculate the probability measure: $\mu=(\frac{1}{9}, \frac{4}{9}, \frac{4}{9})$ of $X_0$
$X_1$:
$$(\frac{1}{9}, \frac{4}{9}, \frac{4}{9}) \bullet (0, \frac{1}{2}, \frac{1}{2})=\frac{4}{9} $$
$$(\frac{1}{9}, \frac{4}{9}, \frac{4}{9}) \bullet (\frac{1}{2}, 0 , \frac{1}{2})=\frac{5}{18} $$
$$(\frac{1}{9}, \frac{4}{9}, \frac{4}{9}) \bullet (\frac{1}{2}, \frac{1}{2}, 0 )=\frac{5}{18} $$
$X_2$:
$$Q^2=\begin{bmatrix}\frac{1}{2} & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{2} &\frac{1}{4} \\ \frac{1}{4} & \frac{1}{4} & \frac{1}{2}\end{bmatrix}$$
$$(\frac{1}{9}, \frac{4}{9}, \frac{4}{9}) \bullet (\frac{1}{2}, \frac{1}{4}, \frac{1}{4})=\frac{10}{36} $$
$$(\frac{1}{9}, \frac{4}{9}, \frac{4}{9}) \bullet (\frac{1}{4}, \frac{1}{2} , \frac{1}{4})=\frac{13}{36} $$
$$(\frac{1}{9}, \frac{4}{9}, \frac{4}{9}) \bullet (\frac{1}{4}, \frac{1}{4}, \frac{1}{2} )=\frac{13}{36} $$
$$E[X_1]=0 \ \frac{4}{9}+ 1 \ \frac{5}{18}+ 2 \ \frac{5}{18}=\frac{15}{18}$$
$$E[X_1^2]=0 \ \frac{4}{9}+ 1 \ \frac{5}{18}+ 4 \ \frac{5}{18}=\frac{25}{18}$$
$$Var[X_1]=\frac{25}{18}-\frac{225}{324}=\frac{225}{324}$$
$$E[X_2]=0 \ \frac{10}{36}+ 1 \ \frac{13}{36}+ 2 \ \frac{13}{36}=\frac{13}{18}$$
d. The invariant probability measure: $\rho \ Q=\rho$
\begin{cases} \frac{1}{2} \ y + \frac{1}{2} \ z=x \\ \frac{1}{2} \ x + \frac{1}{2} \ z=y \\ \frac{1}{2} \ x + \frac{1}{2} \ y=z \\ x+y+z=1 \end{cases}
So, $\rho=(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$
Is it correct?
Thanks!