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Ex. series, say stock prices 103 101 102 150 101 102 100 First differenced 2 1 48 -49 1 -2 Notice you could guess a very large negative number following the very large positive in the first differenced numbers.

Is it the case that negative serial correlation is common in first differenced data and do you correct for this using MA models? (including the lagged error term?) I couldn't find any hints about this on a search so perhaps something is wrong in my logic and I need a vacation..

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    $\begingroup$ Differencing can certainly lead negative autocorrelation in several ways - (i) outliers can do it, as you suggest, as can (ii) overidfferencing, but it can also be quite legitimate in a perfectly appropriate model, such as (iii) an IMA(1) model where (in the differenced series) the MA coefficient leads to negative serial correlation at lag one. $\endgroup$
    – Glen_b
    Commented Jul 5, 2014 at 1:08

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Short answer is yes, differencing will introduce a negative autocorrelation into the differenced series in most situations. Assuming a mean centered variable to make the notation a bit simpler, the covariance between the differenced series can be represented as:

$$Cov(\Delta X_t,\Delta X_{t-1}) = E[\Delta X_t \cdot \Delta X_{t-1}]$$

Where

  • $\Delta X_t = X_t - X_{t-1}$
  • $\Delta X_{t-1} = X_{t-1} - X_{t-2}$

Breaking this down into the original variables, we then have:

\begin{align} E[X_t \cdot X_{t-1}] &= E[(X_t - X_{t-1}) \cdot (X_{t-1} - X_{t-2}) ] \\ &= E[X_tX_{t-1} - X_tX_{t-2} - X_{t-1}X_{t-1} + X_{t-1}X_{t-2}] \end{align}

The multiplications are then just variances and covariances of the levels:

$$Cov(X_t,X_{t-1}) - Cov(X_t,X_{t-2}) - Var(X_{t-1}) + Cov(X_{t-1},X_{t-2})$$

So here we can see that many different situations will result in negative autocorrelations of the differenced series - basically only in the case that the auto-correlations of the levels are really large (e.g. an integrated series) will the differences have a small negative auto-correlation.

With random data the autocorrelation of the differences will be approximately -0.5, as with random data those covariance terms among the levels will be 0, so it is just $-Var(X_{t-1})$ for the numerator, but with the differences is $Var(X_t) - Var(X_{t-1})$ in the denominator.

This is typically called over-differencing. The solution is to not over-difference the data to begin with.

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    $\begingroup$ Andy's answer is excellent here. I think this discussion helps formalize the idea of 'regressing to the mean' which crops up in a bunch of places: en.wikipedia.org/wiki/Regression_toward_the_mean $\endgroup$
    – Mark T Patterson
    Commented Jul 4, 2014 at 19:35
  • $\begingroup$ Yes I agree @MarkTPatterson - this is just a transcript of my notes for a colleague to show why this negative autocorrelation occurs in differenced panel data and why interpreting it is meaningless in substantive terms. $\endgroup$
    – Andy W
    Commented Jul 4, 2014 at 19:42
  • $\begingroup$ An equivalent question is here. $\endgroup$
    – Andy W
    Commented Mar 8, 2015 at 15:16
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    $\begingroup$ Regarding your penultimate paragraph, with integrated series there will be no tendency towards positive vs. negative autocorrelation. A pure random walk, when differenced, will produce uncorrelated errors. $\endgroup$
    – Richard Hardy
    Commented Jan 26, 2017 at 11:43
  • $\begingroup$ @AndyW "With random data the autocorrelation of the differences will be approximately the variance of the levels" - is this statement correct? With random data (uncorrelated data), autocorrelation of the differenced series should be -0.5, not the variance of the levels, right? $\endgroup$
    – shani
    Commented Nov 2, 2018 at 5:59

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