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(yt-1, yt) = Cov(yt, yt-1) / (?yt * ?yt-1)
Cov(yt , yt-1) = Cov(yt-1, 0.8yt- 1+?t) = 0.8Cov(yt-1, yt-1) + Cov(yt-1, ?t)
= 0.8Var(yt-1)+ E((yt-1 - (0.8yt-1))(?t - E(?t))) = 0.8Var(yt-1) + E((0.2yt-1)(?t))
= 0.8Var(yt-1) + 0.2E(yt-1)E(?t) = 0.8Var(yt-1)
=>?((0.8Var(yt-1)))2 / Var(yt-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$\mathbb{E}[y_{t-1}]$