# Is the third moment of an AR(1) dependent on $t$?

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 $$t$$?

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

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

gamma
mean(epsilon^3)

gamma/(1-phi^3)
mean(X^3)

• 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?
– Fam
Commented Jan 11, 2022 at 18:29
• 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. Commented Jan 12, 2022 at 5:23
• +1 But I think your derivation requires independent WN. If the WN is only uncorrelated, then it not necessarily the case that e.g. $E(\epsilon_j^2\epsilon_k)=E(\epsilon_j^2)E(\epsilon_k)=0$ Commented Oct 26, 2022 at 17:18
• True, thanks, I edited! Commented Oct 27, 2022 at 4:15