Given an AR(1) process: $$ X_t = \phi X_{t-1}+ \epsilon_t, \quad \epsilon\sim WN(0, \sigma^2) $$ I know that if $|\phi|<1$, then the process is stationary (weakly). Thus, the first and second moment of $X_t$, $E(X_t)$ and $E(X_t^2)$, are constant. More specifically, they don't depend on $t$.

But, how about its third moment, $E(X_t^3)$, does it depend on $t$?

  • $\begingroup$ No. The third moment is zero for all $t$ $\endgroup$ Commented Jan 11, 2022 at 3:17

1 Answer 1


It may or may not be:

If $\epsilon_t$ is independent WN, the $MA(\infty)$ representation $X_t=\sum_{j=0}^\infty\phi^j\epsilon_{t-j}$ gives, for $|\phi|<1$,
$$ E(X_t^3)=\sum_{j=0}^\infty\phi^{3j}E(\epsilon_{t-j}^3), $$ as pairs $\epsilon_i,\epsilon_j,\epsilon_k$ for which we do not have $i=j=k$ will yield terms of the form, e.g., $E(\epsilon_j^2)E(\epsilon_k)=0$.

If $E(\epsilon_{t-j}^3)$ is constant over time, and if we denote that quantity by $\gamma$, we obtain $$ E(X_t^3)=\frac{\gamma}{1-\phi^3} $$

A little illustration:

n <- 21000
k <- 2
epsilon <- rchisq(n, k)-k # a skewed mean zero distribution

phi <- 0.9
X <- arima.sim(model = list(ar=phi), n = n-1000, innov = epsilon[-(1:1000)], n.start = 1000, start.innov=epsilon[1:1000])

gamma <- k*(k+2)*(k+4) - 3*k^2*(k+2) + 3*k^3 - k^3 # 3rd moment of epsilon, see https://en.wikipedia.org/wiki/Chi-squared_distribution


  • $\begingroup$ Good answer. Thanks. I know that if $\epsilon \sim N(0,\sigma^2)$, the third moments is zero. Can you say that all white noise has its third moment constant over time? $\endgroup$
    – Fam
    Commented Jan 11, 2022 at 18:29
  • 1
    $\begingroup$ I am not sure that all definitions of WN are fully compatible, but mainly, they only assert mean zero and uncorrelatedness (see e.g. Wikipedia), so that third moments could be time-varying. $\endgroup$ Commented Jan 12, 2022 at 5:23
  • $\begingroup$ +1 But I think your derivation requires independent WN. If the WN is only uncorrelated, then it not necessarily the case that e.g. $E(\epsilon_j^2\epsilon_k)=E(\epsilon_j^2)E(\epsilon_k)=0$ $\endgroup$ Commented Oct 26, 2022 at 17:18
  • $\begingroup$ True, thanks, I edited! $\endgroup$ Commented Oct 27, 2022 at 4:15

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