# Proving that $\text{Cov}(Y_{t}, Y_{t-1})=\frac{\tau^2\beta^2_{1}}{1-\beta^2_{1}}$ for AR(1) process

$$\text{Cov}(Y_{t}, Y_{t-1})=\frac{\tau^2\beta^2_{1}}{1-\beta^2_{1}}$$ $$Y_{t}=\beta_{0}+\beta_{1}Y_{t-1}+u_{t}$$ How do I use stationarity to prove the first equation? I have come to the conclusion of $$=\beta_{1}\text{Var}(Y_{t-1})$$ but I am stuck, and it is the last step before I can show that the first equation is true. However, the solution skips the step where we convert it to the true form.

• Can you try solving for $Y_{t}$ in terms of $Y_0$? – Alex R. Dec 4 '16 at 21:15
• I don't know how to do that. However, for the problem, I only need to convert it back, as I have already done most of the work. – Mataunited17 Dec 4 '16 at 21:25
• Mataunited17: Pay attention to what AlexR is suggesting instead of pushing back. Hint:Can you figure out why $\operatorname{var}(Y_n) = \beta^2\operatorname{var}(Y_{n-1})+1$? – Dilip Sarwate Dec 4 '16 at 23:53

Sounds like you're still stuck so I'll try and prompt you further. I'll use the following notation.

\begin{align} X_{t}&=a+\phi X_{t-1}+Z_{t}\\ Z_{t}&\sim\text{WN}(0,\sigma^{2}) \end{align}

You want to know:

\begin{align} \gamma=\text{Cov}(X_{t},X_{t-1})&=E[X_{t}X_{t-1}]-E[X_{t}]E[X_{t-1}]\\ \text{Cov}(X_{t},X_{t-1})&=E[X_{t}X_{t-1}]-E[X_{t}]^{2}\,\,\, \text{ (stationarity)} \end{align}

Now, we know:

\begin{align} X_{t}&=a+\phi X_{t-1}+Z_{t}\\ X_{t}X_{t-1}&=aX_{t-1}+\phi X_{t-1}^{2}+X_{t-1}Z_{t}\\ E[X_{t}X_{t-1}]&=aE[X_{t-1}]+\phi E[X_{t-1}^{2}]+E[X_{t-1}Z_{t}]\\ E[X_{t}X_{t-1}]&=aE[X_{t-1}]+\phi E[X_{t-1}^{2}]+0 \end{align}

So we need the terms in the equation above. Let's start with $E[X_{t-1}]$:

\begin{align} X_{t-1}&=a+\phi X_{t-2}+Z_{t-1}\\ E[X_{t-1}]&=a+E[\phi X_{t-2}]+E[Z_{t-1}]\\ E[X_{t-1}]&=a+\phi E[X_{t-1}]+0\,\,\,\, \text{ (stationarity)}\\ E[X_{t-1}](1-\phi)&=a\\ E[X_{t-1}]&=\frac{a}{(1-\phi)}\\ \end{align}

Now we need $E[X_{t-1}^{2}]$:

\begin{align} X_{t-1}^{2}&=(a+\phi X_{t-2}+Z_{t-1})^{2}\\ X_{t-1}^{2}&=a^{2}+2a\phi X_{t-2}+2aZ_{t-1}+\phi^{2}X_{t-2}^{2}+2\phi X_{t-2}Z_{t-1}+Z_{t-1}^{2}\\ E[X_{t-1}^{2}]&=a^{2}+2a\phi E[X_{t-2}]+2aE[Z_{t-1}]+\phi^{2}E[X_{t-2}^{2}]+2\phi E[X_{t-2}Z_{t-1}]+E[Z_{t-1}^{2}]\\ E[X^{2}_{t-1}]&=a^{2}+2a\phi E[X_{t-1}]+0+\phi^{2}E[X^{2}_{t-1}]+0+E[Z^{2}_{t-1}]\,\,\,\, \text{ (stationarity)}\\ E[X^{2}_{t-1}](1-\phi^{2})&=a^{2}+\frac{2a^{2}\phi}{(1-\phi)}+\sigma^{2}\\ E[X^{2}_{t-1}]&=\frac{a^{2}}{(1-\phi^{2})}+\frac{2a^{2}\phi}{(1-\phi)(1-\phi^{2})}+\frac{\sigma^{2}}{(1-\phi^{2})}\\ \end{align}

So finally:

\begin{align} \text{Cov}(X_{t},X_{t-1})&=E[X_{t}X_{t-1}]-E[X_{t}]^{2}\\ &=aE[X_{t-1}]+\phi E[X_{t-1}^{2}]-E[X_{t}]^{2}\\ &=\frac{a^{2}}{(1-\phi)}+\phi\bigg(\frac{a^{2}}{(1-\phi^{2})}+\frac{2a^{2}\phi}{(1-\phi)(1-\phi^{2})}+\frac{\sigma^{2}}{(1-\phi^{2})}\bigg)-\frac{a^{2}}{(1-\phi)^{2}}\\ &=\frac{a^{2}(1-\phi)(1+\phi)+a^{2}\phi(1-\phi)+2a^{2}\phi^{2}-a^{2}(1+\phi)}{(1-\phi)^{2}(1+\phi)}+\phi\frac{\sigma^{2}}{(1-\phi^{2})}\\ &=\frac{a^{2}-a^{2}\phi^{2}+2a^{2}\phi^{2}+a^{2}\phi-a^{2}\phi^{2}-a^{2}-a^{2}\phi}{(1-\phi)^{2}(1+\phi)}+\phi\frac{\sigma^{2}}{(1-\phi^{2})}\\ &=\phi\frac{\sigma^{2}}{(1-\phi^{2})}\\ \end{align}