1
$\begingroup$

I know that the exponentially weighted moving average (EWMA) volatility estimator drapes a decaying weight function over historical returns in order to weight the past according to the decay of their serial correlation functions (ACF) during the estimation of volatility,

but if I instead only apply the same weighting scheme to a univariate asset return series, without going the step further of estimating volatility, does this new weighted return series have less or no autocorrelation compared to the unweighted returns?

$\endgroup$
  • $\begingroup$ Have you considered accepting the answer? Let me know if something is still unclear. $\endgroup$ – Richard Hardy Dec 13 '20 at 18:12
3
$\begingroup$

EWMA (and other sort of moving averages) introduces positive autocorrelation into otherwise uncorrelated returns. The fitted values of EWMA are linear combinations of past returns, and the constituent elements of these combinations overlap. Therefore, positive autocorrelation arises.

If you have autocorrelated returns to begin with, they would in all likelihood be only moderately autocorrelated, and so similar implications of EWMA could be expected.

Does there exist an autocorrelation structure where EWMA would actually reduce rather than increase autocorrelation? I think it does, and it also depends on what lags we are looking at. E.g. if the autocorrelation of returns at lag 1 is negative, EWMA might reduce the autocorrelation towards zero or even make it positive via the mechanism described in the first paragraph.

$\endgroup$
  • $\begingroup$ so if returns are autocorrelated and EWMA doesn't reduce the autocorrelation, what's the point of ewma? it just captures what is already present? $\endgroup$ – develarist Sep 2 '20 at 12:28
  • $\begingroup$ @develarist: ewma is just an estimation procedure that trades off how much one weights the distant past with the closer past. So, depending on the value of $\lambda$, one is giving either more or less weight to the distant past of the thing one is trying to estimate. One may not want to give the same weight to a data point that was a year ago as you do to a point that was a week ago. The value of $\lambda$ determines the weighting scheme. Also, note that the first paragraph you quoted is not correct. The EWMA is not a function of the serial correlation of the series it being used on. $\endgroup$ – mark leeds Sep 2 '20 at 12:36
  • $\begingroup$ yeah i know that it weights based on recency, which addresses long memory, but ewma does not fundamentally alter the characteristics or properties of returns. it addresses issues without altering statistical properties, it only alters values of observations. the statistical properties of the weighted returns are equivalent, identical to the original returns $\endgroup$ – develarist Sep 2 '20 at 12:42
  • 1
    $\begingroup$ @develarist, I think the statement about altering statistical properties is incorrect. EWMA introduces autocorrelation, smooths out peaks etc. etc. The fitted values of EWMA will have quite different statistical properties (both conditional and unconditional) than returns do. EWMA is mainly a forecasting technique AFAIK; the fitted values are of less interest. It can be interpreted as a latent state of the observed variable, though, as EWMA is a state-space type of model. $\endgroup$ – Richard Hardy Sep 2 '20 at 13:09
  • 1
    $\begingroup$ @develarist: I just want to add that EWMA must alter the statistical properties of the original series because it's a non-linear transformation of the original series. But I also don't think the statistical properties of the resulting EWMA sequence are terribly important. The EWMA is used as an estimator so it's the most recent value of the EWMA that matters. Note though, if you're interested in the process, the resulting EWMA can be viewed as an ARIMA(0,1,1) applied to the original series where the MA coefficient is a function of $\lambda$. If I remember correctly, $\theta = 1 - \lambda$. $\endgroup$ – mark leeds Sep 3 '20 at 3:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.