$\newcommand{\E}{\mathbb{E}}$I was wondering if anyone could help me show that $\E(Y_t) = 0$? I have had a go myself but have got stuck. Could someone please point me in the right direction?

Consider the stationary AR(4) model for $Y_t$ $$ Y_t = \phi_1Y_{t-1}+\phi_2Y_{t-2}+\phi_3Y_{t-3}+\phi_4Y_{t-4}+e_t $$ Show that the unconditional mean is zero – so $\E(Y_t) = 0$.

My calculations thus far: \begin{align} \E(Y_t) &= \E(\phi_1Y_{t-1}+\phi_2Y_{t-2}+\phi_3Y_{t-3}+\phi_4Y_{t-4}+e_t) \\[5pt] \E(Y_t) &= \E(\phi_1Y_{t-1})+ \E(\phi_2Y_{t-2})+ \E(\phi_3Y_{t-3})+ \E(\phi_4Y_{t-4})+\E(e_t) \\[5pt] \E(Y_t) &= \E(\phi_1Y_{t-1})+ \E(\phi_2Y_{t-2})+ \E(\phi_3Y_{t-3})+ \E(\phi_4Y_{t-4})+\E(e_t) \\[5pt] \E(Y_t) &= \phi_1\E(Y_{t-1})+ \phi_2\E(Y_{t-2})+ \phi_3\E(Y_{t-3})+ \phi_4\E(Y_{t-4})+\E(e_t) \end{align} I know that $\E(e_t) = 0$ but I don’t know how to treat $\phi \E(Y_{t-j})$ & then am unsure as to what the next step is to derive $\E(Y_t) = 0$.

  • 1
    $\begingroup$ Please add the [self-study] tag & read its wiki. $\endgroup$ Commented Oct 9, 2015 at 3:48
  • $\begingroup$ Let $\mu$ be the mean, which exists by the assumption of stationarity. Then $\mu = \phi_1 \mu + \phi_2 \mu + \phi_3 \mu + \phi_4 \mu$ implies $\mu(1-\phi_1-\phi_2-\phi_3-\phi_4) = 0$ so either the mean is zero or $1 = \phi_1 + \phi_2 + \phi_3 + \phi_4$. Perhaps that helps you, but I am not sure how to show that the second condition isn't possible. $\endgroup$
    – user369210
    Commented Jan 9, 2017 at 0:09
  • $\begingroup$ This wasn't stated explicitly but I assume there are some initial conditions about $Y_0, Y_1, Y_2, Y_3$. Should be a pretty simple argument after that based on what you've already derived. $\endgroup$
    – gammer
    Commented Jan 9, 2017 at 1:04

1 Answer 1


Recall that by stationary $E(y_t)=E(y_{t-1})=E(y_{t-2})=...$. You can try proving it by contradiction, plug something in for $E(y_t)$ that is not zero and see what happens.


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