Durbin and Watson's statistic, from their 1950 and 1951 papers, is
DW = sum over t of (e_t - e_(t-1))^2 / sum over t of e_t^2How do you interpret the Durbin-Watson statistic?
It runs from 0 to 4, sits near 2 when successive residuals are unrelated, falls below 2 under positive autocorrelation and rises above 2 under negative autocorrelation, with the approximation DW ≈ 2(1 − ρ̂) tying the statistic to the residuals' own lag-1 correlation. The test is inconclusive over a range, which is why the published tables give two bounds rather than one critical value. At n = 20 with a single explanatory variable, the Savin and White (1977) tabulation puts d_L at 1.20 and d_U at 1.41: below 1.20 reject the null of no positive autocorrelation, above 1.41 do not reject, and between the two the test declines to answer.
In an ordinary cross-sectional regression, serial correlation is a nuisance that inflates your apparent significance and is fixed with a corrected standard error. On a trend model it is something else entirely. The residuals of a correctly specified trend should be a sequence of unrelated shocks around the line, so a residual that is systematically positive for six quarters, then systematically negative for nine, then systematically positive for five is not noise with an inconvenient covariance structure. It is the model telling you that the line is the wrong shape.
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That sign pattern is exactly what the revenue series produces under a linear fit: DW = 0.4006, implied ρ̂ = 0.7997, and only three runs of sign across twenty residuals. Fit the same data in logs and DW rises to 1.8500 with nine runs. The R² also rises, from 0.9623 to 0.9931, but the R² comparison is not admissible on its own because the two regressions have different dependent variables. The Durbin-Watson comparison is admissible, because both statistics are measuring the same property of the same kind of object.
When a trend model fails the test even after the transformation, the honest next move is to stop modelling calendar time and start modelling the series' own past. That is the autoregressive model, and the reason the two sit in the same node is that the failure of one is the motivation for the other.
How do you check residuals at more than one lag?
A correctly specified autoregressive model makes a strong and testable claim: the residual autocorrelation should be indistinguishable from zero at every lag, not merely at lag 1. The standard error of a sample residual autocorrelation from a fitted model is approximately 1/√T, so a lag-k autocorrelation is significant at roughly the 5% level when |ρ̂_k|√T exceeds 2. Run that check at four or six lags before you accept an order. On the AR(1) fit in Worked Example 4, T = 27, the standard error is 0.1925, and the four t-statistics are +0.19, +0.40, −0.72 and +1.38. None clears 2, so one lag is enough. Had the lag-2 statistic come in at 2.9, the answer would have been an AR(2), and no amount of comfort with the lag-1 result would have rescued the AR(1).