Let $y_t = 0.8y_{t-1} + \epsilon_t$ where $\mathbb{E}[\epsilon_t] = 0$ and $\mathrm{Var}[\epsilon_t] = 1$. The time series is a strictly stationary sequence and each element is normally distributed. Hence $\mathbb{E}[y_t]$ and $\mathrm{Var}[y_t]$ are both constants. In particular, $\mathbb{E}[y_t]$ = $\mathbb{E}[y_{t-1}]$ and $\mathrm{Var}[y_t]$ = $\mathrm{Var}[y_{t-1}]$. Let $\rho(T)$ be the correlation between $y_t$ and y_{t-1}. Find $\rho(1)$ and $\rho(2)$.

Here is what I did for ?(1):

?(1) = Corr(y_t-1, y_t) = Cov(y_t, y_t-1) / (?_yt * ?_yt-1)

Cov(y_t , y_t-1) = Cov(y_t-1, 0.8y_{t- 1}+?_t) = 0.8Cov(y_t-1, y_t-1) + Cov(y_t-1, ?_t)

= 0.8Var(yt-1)+ E((y_t-1 - (0.8y_t-1))(?_t - E(?_t))) = 0.8Var(y_t-1) + E((0.2y_t-1)(?_t))

= 0.8Var(y_t-1) + 0.2E(y_t-1)E(?_t) = 0.8Var(y_t-1) =>?((0.8Var(y_t-1)))² / Var(y_t-1)²)= ?0.8 = 0.8944

Did I do this right and if so, wouldn't ?(2) be the same thing? I was also confused as to how $\mathbb{E}[y_t]$ could be equal to $\mathbb{E}[y_{t-1}]$ when $\mathbb{E}[y_t]$ = 0.8$\mathbb{E}[y_{t-1}]$