Can someone please tell me how the acf and pacf look like for a unit root process?

  • 1
    $\begingroup$ Why? PACF and ACF are useless for unit root process. There will be slow decay on ACF and cut off on PACF $\endgroup$
    – Aksakal
    Apr 20, 2018 at 20:36
  • $\begingroup$ Why not create one with, say, 10,000 observations, and plot the ACF and PACF? $\endgroup$
    – jbowman
    Apr 20, 2018 at 22:19
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    $\begingroup$ Slow decay of the ACF is at least a hint that differencing might needed. $\endgroup$ Apr 21, 2018 at 0:01
  • $\begingroup$ Thanks for your replies.I am trying hard to understand these concepts-Why is ACF and PACF useless for unit root process? What are the number of lags(both for AR and MA) for such a process? $\endgroup$
    – Bun
    Apr 21, 2018 at 8:47

1 Answer 1


Write the RW as (assuming a starting value $Y_0=0$)

$$Y_t = \sum_{s=1}^{t} \epsilon_s$$ and so $$ \gamma_{tj} = E\left(\sum_{s=1}^t \epsilon_s \sum_{s=1}^{t-j}\epsilon_s\right) = (t-j)\sigma^2$$ Hence, the autocorrelations (see the comments above for the pacf) are $$\rho_{jt} = \frac{\gamma_{jt}}{\sqrt{\gamma_{0t}}\sqrt{\gamma_{0 (t-j)}}}=\frac{t-j}{\sqrt{t}\sqrt{t-j}}=\frac{\sqrt{t-j}}{\sqrt{t}}= \sqrt{1-\frac{j}{t}}$$ Clearly, the decay is very slow, and completely washes out for $t\to\infty$ -shocks do not die out.


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