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