Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 37 / 1 / a

A stationary causal time series exists. The infinite moving-average representationconverges in the sense of mean-square convergence because the driving variables are independent random variables and ${\sum }_j{4}^{-j}<\infty$. Shifting the series gives $X_t=\frac{1}{2}{X}_{t-1}+{\epsilon }_t$. Its expected value is zero and its autocovariance isThus it is a weakly stationary process; since the driving variables have a normal distribution, it is also a Gaussian process and a strictly stationary process. Throughout the stationarity discussion, take ${\sigma }^2>0$; degenerate zero noise permits trivial constant solutions.

No weakly stationary process solves the model with nondegenerate noise. Iteration would giveFor a weakly stationary process with finite variance $V$, the Cauchy-Schwarz inequality instead gives $\mathrm{Var}(X_t-X_{t-m})=2V-2\gamma (m)\le 4V$. These two bounds contradict each other for large $m$. This argument does not assume that $X_{t-m}$ is independent of the intervening noise, so it excludes noncausal solutions too. The corresponding unit-root autoregressive process has persistent random walk behavior rather than stationary fluctuations.

A noncausal stationary autoregression exists. On the two-sided time axis defineThis series converges in the sense of mean-square convergence, and direct subtraction gives $X_t-2X_{t-1}={\epsilon }_t$. The expected value and autocovariance areConsequently this is a weakly stationary process and, by Gaussianity, a strictly stationary process. It is an example of noncausal stationary autoregression, since $X_t$ uses future noise. In particular, $X_{t-1}$ is correlated with ${\epsilon }_t$, so the usual causal variance recursion is inapplicable. If an additional assumption required the driving noise to be independent of past observations, this solution would be excluded; that assumption is not stated here.