Let $y_t = 0.8y_{t-1} + \epsilon_t$ where $\mathbb{E}[y_t]$ and $\mathrm{Var}[y_t]$ are constants (or $\mathbb{E}[y_t] = \mathbb{E}[y_{t-1}]$ and $\mathrm{Var}[y_t] = \mathrm{Var}[y_{t-1}]$) and $\mathbb{E}[\epsilon_t] = 0$ and $\mathrm{Var}[\epsilon_t] = 1$. Let $\rho(T)$ be the correlation between $y_t$ and $y_{t-T}$. Find $\rho(1)$ and $\rho(2)$.
Here is what I did for $\rho(1)$:
\begin{align} \rho(1) &= Corr(y_{t-1}, y_{t})\\ &= \dfrac{Cov(y_{t}, y_{t-1})}{(σ_{y_{t}}σ_{y_{t-1}})}\\ Cov(y_{t} , y_{t-1}) &= Cov(y_{t- 1}, 0.8y_{t- 1}+ε_{t})\\ &= 0.8Cov(y_{t-1}, y_{t-1}) + Cov(y_{t-1}, \epsilon_{t})\\ &= 0.8Var(yt-1)+ E((y_{t-1} - (0.8y_{t-1}))(\epsilon_{t} - E(\epsilon_{t})))\\ &= 0.8Var(y_{t-1}) + E((0.2y_{t-1})(\epsilon_{t})) \\ &= 0.8Var(y_{t-1}) + 0.2E(y_{t-1})E(\epsilon_{t})\\ &= 0.8Var(y_{t-1})\\ \implies \dfrac{\sqrt{(0.8Var(y_{t-1})))^2}}{Var(y_{t-1})^2)}&= \sqrt{0.8} = 0.8944 \end{align} Did I do this right and if so, wouldn't ρ(2) be the same thing?