A series has a sample autocorrelation of 0.40 at lag 1 and essentially zero at every lag beyond. Fit an MA(1), solve for the moving average coefficient, and explain why two values fit and which one you report.

A series has a sample autocorrelation of 0.40 at lag 1 and essentially zero at every lag beyond. Fit an MA(1), solve for the moving average coefficient, and explain why two values fit and which one you report.

Approach: Write the lag-one autocorrelation of an MA(1) as a function of the coefficient and solve the resulting quadratic, then decide between the roots.

0.5. For the moving average model x_t = e_t + theta*e_{t-1} the autocovariances are (1 + theta^2)*sigma^2 at lag 0, theta*sigma^2 at lag 1 and zero beyond, so rho(1) = theta/(1 + theta^2) and the cut-off after lag 1 matches the observed pattern. Solving 0.4*theta^2 - theta + 0.4 = 0 gives theta = (1 plus or minus 0.6)/0.8, so theta = 0.5 or theta = 2.0. The two roots are reciprocals, because rho(1) is unchanged when theta is replaced by 1/theta, and the two processes have identical second-moment structure. You report 0.5 because only |theta| < 1 is invertible, meaning the process can be written as a convergent autoregression in its own past so the shocks are recoverable from observed data and forecasts use them. The maximum of rho(1) for an MA(1) is 0.5, attained at theta of 1, so a sample value above 0.5 rules the model out entirely.

Follow-up: An MA(1) fit returns theta_hat exactly equal to 1. What has usually happened to the data before that fit?

Key concepts: MA(1), autocorrelation function, invertibility, moving average.