# Brockwell/Davis seem to say more persistence implies better predictability---do I have a counterexample?

Brockwell/Davis, Introduction to Time Series and Forecasting, p. 40, write (notation slightly adapted; please refer to screenshot below)

The best linear predictor $$l(Y_{T})=aY_{T}+b$$ for a stationary time series $$Y_{T+h}$$ based on $$Y_{T}$$ minimizes $$E[Y_{T+h}-aY_T-b]^2$$ and is given by, for $$\mu$$, $$\gamma_h$$ and $$\rho_h$$ the mean and $$h$$-order autocovariance/autocorrelation, $$l(Y_T)=\mu+\rho_h(Y_{T}-\mu),$$ so that the optimal $$b=\mu(1-\rho_h)$$ and $$a=\rho_h$$. The MSE is $$E[Y_{T+h}-l(Y_{T})]^2=\gamma_0(1-\rho_h^2).$$ I can follow these steps up until here.

BD then go on to write:

This calculation shows that [...] prediction of $$Y_{T+h}$$ in terms of $$Y_T$$ is more accurate as $$|\rho_h|$$ becomes closer to 1, and in the limit as $$\rho_h → \pm1$$ [...] the corresponding mean squared error approaches 0.

I am not so sure what to make of this quote: consider for example a zero-mean AR(1) process with coefficient $$\rho$$, $$Y_t=\rho Y_{t-1}+\epsilon_t$$, and error variance $$\sigma^2$$, so that $$\gamma_h=\sigma^2\rho^h/(1-\rho^2)$$ and $$\rho_h=\frac{\gamma_h}{\gamma_0}=\frac{\sigma^2\rho^h/(1-\rho^2)}{\sigma^2/(1-\rho^2)}=\rho^h.$$

Then $$MSE_{h,\rho}=\sigma^2\frac{1-\rho^{2h}}{1-\rho^2}$$ In this AR(1) case, $$\rho_h\to1$$ iff $$\rho\to1$$. By l'Hopital's rule this tends to $$\lim_{\rho\to1}MSE_{h,\rho}=h\sigma^2,$$ which is also the MSE of an $$h$$-step ahead forecast of a random walk $$Y_t=Y_{t-1}+\epsilon_t$$, $$E(Y_{T+h}-Y_T)^2=E\left(\sum_{j=T+1}^{T+h}\epsilon_j\right)^2=h\sigma^2.$$

This is intuitive to me as the AR(1) approaches a random walk as $$\rho\to1$$, and the AR(1) $$h$$-step ahead forecasts $$\rho^hY_T$$ also approach those of a RW, $$Y_T$$.

In particular, as the variance $$\gamma_0$$ of the process is also affected by changes in the persistence $$\rho$$ of the process, this result suggests that the MSE does not converge to zero and also that it would not be the case that persistent processes are more predictable: An evaluation of

h <- 1:20
rho <- seq(.9, .99, by=.01)
MSE.AR <- (1-outer(rho, 2*h, "^"))/(1-rho^2)
MSE.RW <- h

matplot(h, t(MSE.AR), type="l", lwd=1, col=seq_len(length(rho)), lty=seq_len(length(rho)), ylim=c(0,max(h)))
lines(h, MSE.RW, lwd=5, lty=1, col="black")
legend("topleft", legend=c(rho, "RW"), col=c(seq_len(length(rho)), "black"), lty=c(seq_len(length(rho)),1), lwd=c(rep(1, length(rho)),5))


reveals that the MSE according to this result increases in the persistence of the process:

I am reluctant to claim, though, that BD make a mistake. Is anything wrong with my example?

Original discussion in BD (note in particular that their $$\sigma^2$$ is my $$\gamma_0$$):

Note: This was to be part of an answer here, but actually gave rise to a question of mine, as posted here.

UPDATE:

In personal communication with Richard Davis, he states (my summary) that for a fixed $$h$$ and a stationary process (so no AR(1) approaching a random walk), if $$\rho_h \to 1$$, then you can perfectly predict $$X_h$$ from $$X_0$$. This is related to the relationship between any two random variables $$X$$ & $$Y$$. If their correlation is 1, then they are linearly related and one can be perfectly predicted from the other.

Hence, in the context of my example, it seems we would need $$\sigma^2\to0$$ to get such perfect correlation, in which case both $$\gamma_0$$ and $$(1-\rho_h^2)$$ in $$\gamma_0(1-\rho_h^2)$$ would go to zero.

• There are predictability and forecastability tags (which I have proposed to merge), perhaps you want to add one. Commented May 5 at 9:55
• Thanks, I have done so! Commented May 5 at 10:09
• I think the crucial statement to investigate is: $\rho_h\to 1$ iff $\rho\to 1$. Commented May 5 at 11:12
• Christoph: That's interesting and my Brockwell and Davis book is in storage. What are $a$ and $b$ in B&D's example ? Is $b$ equal to $(1-\rho_h) \mu$ and $a = \rho_h$. Thanks and sorry for confusion. Commented May 6 at 10:15
• Thank you for getting to the bottom of this! Commented May 13 at 8:39

As things stand, I cannot see how the prediction MSE for any process that regularly has an error term $$\epsilon_t$$ with variance $$\sigma^2$$ coming in, independently of what happened before, can ever be smaller than $$\sigma^2$$.
• Indeed, the MSE is $\sigma^2$ also according to the expressions I found for the AR(1) at $h=1$, and increases from there on. Thanks for the idea about the email, I will consider that after having waited here for a little more! Commented May 7 at 11:04