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Consider a time series generated by the following model:

$y_t = 0.8y_{t-1} + \epsilon_t$

$E(\epsilon_t) = 0$ and $Var(\epsilon_t) = 1$. Note that for any s < t, $y_s$ is independent with $\epsilon_t$. The above time series is a strictly stationary sequence and each element is normally distributed. Hence, $E(y_t)$ and $Var(y_t)$ are both constants. In particular, $E(y_t) = E(y_{t−1})$ and $Var(y_t) = Var(y_{t−1})$. Denote $E(y_t)$ by µ and $Var(y_t)$ by $σ^2$. Find the values of µ and $σ^2$.

Here is what I have so far:

$E(y_t) = 0.8E(y_{t-1}) + E(\epsilon_t) = 0.8y_{t-1}$

$Var(y_t) = 0.64Var(y_{t-1}) + Var(\epsilon_t) = 0.64σ^2_{y_{t-1}} + 1$

I do not know how to find the specific values. I thought at first the question was asking for formula answers but the specific values are required for future questions. Also, how can $E(y_t) = E(y_{t-1})$ when $E(y_t) = 0.8E(y_{t-1})$?

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1 Answer 1

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Hint:

\begin{align} E(y_t) &= E(y_{t-1}) = \mu\\ Var(y_t) &= Var(y_{t-1}) = \sigma^2 \end{align}

Can you make the substitutions and solve for $\mu$ and $\sigma^2$?

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  • $\begingroup$ Does $μ = 0$ and $σ^2 = 0.0278$? $\endgroup$
    – Ethan Smith
    Commented Oct 18, 2018 at 0:43
  • $\begingroup$ @EthanSmith Your $\mu$ is correct. Did you misplace a decimal point? $\sigma^2 = 2.78$. $\endgroup$
    – Almacomet
    Commented Oct 18, 2018 at 2:22

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