The test statistic for the Durbin Watson test can range from 0-4 from what I have gathered. Now the lower limit of 0 makes sense considering the test statistic consists of two summations which are both squared and divided by each other; but what gives us our upper limit of 4? Is this limit incorrect? Also, if the upper limit to this test statistic is not 4, than what is it?


1 Answer 1


The limits are both correct.

The DW statistic is, for $e_t$ the residuals of an appropriate regression, \begin{align*} DW&=\frac{\sum_{t=2}^T(e_t-e_{t-1})^2}{\sum_{t=1}^Te_t^2}\\ &=\frac{\sum_{t=2}^T(e_t^2+e_{t-1}^2-2e_{t}e_{t-1})}{\sum_{t=1}^Te_t^2}\\ &=\frac{\sum_{t=2}^Te_t^2}{\sum_{t=1}^Te_t^2}+\frac{\sum_{t=2}^Te_{t-1}^2}{{\sum_{t=1}^Te_t^2}}-2\frac{\sum_{t=2}^Te_{t}e_{t-1}}{\sum_{t=1}^Te_t^2} \end{align*} The first two fractions are obviously between 0 and 1 (both entries are positive, and we sum more positive terms in the denominator). In fact, they will almost always be very close to 1, as the numerators only differ from the denominator by $e_1^2$ and $e_T^2$, respectively, which, for $T$ reasonably large, will be negligible.

The third one can be bound to be between -1 and 1 by the Cauchy-Schwarz inequality: \begin{align*} \left(\sum_{t=2}^Te_{t}e_{t-1}\right)^2&\leq\sum_{t=2}^Te_{t}^2\sum_{t=2}^Te_{t-1}^2\\ &\leq\sum_{t=1}^Te_{t}^2\sum_{t=1}^Te_{t}^2=\left(\sum_{t=1}^Te_{t}^2\right)^2, \end{align*} so that $$ -\sum_{t=1}^Te_{t}^2\leq\sum_{t=2}^Te_{t}e_{t-1}\leq\sum_{t=1}^Te_{t}^2 $$ Somewhat less rigorously, but more intuitively: We have that the DW statistic can approximatively be written as $$ DW\approx 2(1-\hat\rho),$$ where $\hat\rho$ is the estimated $AR(1)$ coefficient. For $\hat\rho\to\pm1$, we see that the statistic tends to the bounds.

That this is not rigorous follows from the fact that $|\hat\rho|$ can be bigger than one. For large $T$ that should not happen very often when the true $\rho$ is less than one in absolute value, as it is required to be by assumption. But it can happen:


T <- 50
u <- rep(0,T)
rho <- -0.999

for (i in 2:T) {
  u[i] <- rho*u[i-1]+rnorm(1)

regr <- lm(u~1)

# by hand:
uhat <- regr$residuals

rhohat <- lm(uhat[2:T]~uhat[1:(T-1)]-1)$coef[1]
(DWstat <- sum(diff(uhat)^2)/sum(uhat^2))
[1] 3.799259
(ApproxDWStat <- 2*(1-rhohat))
uhat[1:(T - 1)] 
  • $\begingroup$ Can you explain to me why go through the trouble of finding a statistic like this with a complicated equation when we can directly measure the correlation between the error and its lag? $\endgroup$
    – Dom Jo
    Jan 29, 2020 at 13:27
  • $\begingroup$ I guess because we do not just want to measure, but test. I.e., to test if the measured correlation is significantly different from zero, we need a suitable test statistic along with critical values, like in other testing situations. $\endgroup$ Jan 30, 2020 at 7:26
  • $\begingroup$ Is the $DW \approx 2(1 - \hat{\rho})$ due to $\rho$ being the coefficient of the assumed $AR(1)$ process? $\endgroup$ Nov 23, 2021 at 21:37
  • $\begingroup$ I'd say of the fitted AR(1) process. None of the results seem to require that the actual process actually is an AR(1) process. $\endgroup$ Nov 24, 2021 at 4:56

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