One of the problems in Brockwell and Davis book about time series is to show that

1) if \begin{equation} x_t = a + b t \end{equation} then the sample autocorrelation ($\hat{\rho}(h)$) converges to 1 as the sample size tends to infinity for $h \geq 1$.

2) if \begin{equation} x_t = a \text{cos}(\omega t) \end{equation} then the sample autocorrelation ($\hat{\rho}(h)$) converges to $\text{cos}(\omega h)$ as the sample size tends to infinity, where $a \neq 0$ and $\omega \in [-\pi, \pi)$.

I can prove these two results, but what is the message this question is trying to give?


1 Answer 1


Referring to the first case, since x is a simple trend it increases linearly as a function of b. That is: $x_t=x_{t-1}+b$. So it is a perfectly predictable process where the first difference is deterministic and equal to b. It is a unit root process and as such it has perfect memory about all its past values because each $x_t$ depends on $x_{t-1}$ which on its turn depends on $x_{t-2}$, etc.. so it is just a sum of deterministic terms over time. So it has a perfect memory of all its past values. So logically the autocorr must be one at any lag.

Hint: try to write the autocorr function for a AR process and set the AR coefficuent to 1 (if you drop the white noise to substitute it with an intercept b and set the AR coeff ti 1 then you get exactly this non-stationary deterministic trend, so you get the same process!). For an AR process the autocorr at k lags is $\phi ^{k}$ so if $\phi =1$ then the autocorr will be always 1 regardless k.

  • $\begingroup$ How could you derive the theoretical autocorrelation function for $X_t = a + bt$, because $E(X_t) = a + bt, \forall t$, so $cov(X_{t+h}, X_t) = 0$? $\endgroup$
    – shani
    Commented Aug 19, 2019 at 12:34
  • $\begingroup$ @shani rewrite the process x as a function of x_t-1 by subtracting x_t - x_t-1 : you find b. So x_t=x_t-1+b $\endgroup$
    – Fr1
    Commented Aug 19, 2019 at 12:37
  • $\begingroup$ How it is helpful to derive the autocorrelation function of $X_t$, I think the derivation of $\phi^k$ assumes $|\phi |< 1$ $\endgroup$
    – shani
    Commented Aug 19, 2019 at 12:40
  • $\begingroup$ @shani ok then you substitute \phi = 1. Very logic, very intuitive.. $\endgroup$
    – Fr1
    Commented Aug 19, 2019 at 12:51

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