Markov chain covariance calculation

Question

Suppose $Y_t=1$ with probability $\pi_s$ and $0$ with probability $1-\pi_s$. Moreover, we know that $Y_t$ is a Markov chain with transition probabilities $Pr(Y_t=1|Y_{t-1}=1)=(p^3+(1-p)^3)/p_s$ and $Pr(Y_t=0|Y_{t-1}=0)=1/2$. How do I calculate $\operatorname{Cov}[Y_t,Y_{t-1}]$?

Answer 1

Assume $P(Y_t)=P(Y_{t-1})$ that is the state space distribution are in a steady state.

Covariance is calculated as

$$Cov[Y_t,Y_{t-1}]=E[Y_t,Y_{t-1}]-E[Y_t]E[Y_{t-1}]$$

Where $E[Y_t]=1P(Y_t=1)+0(P(Y_t=0)=\pi_s$ and $E[Y_{t-1}]=\pi_s$ by a similar argument and the steady state assumption.

For $E[Y_tY_{t-1}]$ we need the joint distribution $(Y_t,Y_{t-1})$.

$$P(Y_t=0,Y_{t-1}=0)=P(Y_t=0|Y_{t-1}=0)P(Y_{t-1}=0)=\frac{1-\pi_s}{2}$$ $$P(Y_t=1,Y_{t-1}=0)=P(Y_t=1|Y_{t-1}=0)P(Y_{t-1}=0)=\frac{1-\pi_s}{2}$$
$$P(Y_t=1,Y_{t-1}=1)=P(Y_t=1|Y_{t-1}=1)P(Y_{t-1}=1)=\frac{(p^3+(1-p)^3)\pi_s}{p}$$ $$P(Y_t=0,Y_{t-1}=1)=P(Y_t=0|Y_{t-1}=1)P(Y_{t-1}=1)=(1-\frac{(p^3+(1-p)^3)}{p})\pi_s$$

then $$E[Y_tY_{t-1}]=\frac{(p^3+(1-p)^3)\pi_s}{p}$$

and covariance is

$$Cov(Y_t,Y_{t-1})=\frac{(p^3+(1-p)^3)\pi_s}{p}-\pi_s^2$$

Note this is all dependent on the steady state assumption.