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I want to analyse whether two return series are different. I was told to run the following regression:

diff = return series 1 - return series 2
constant = beta * diff

Where I set the constant equal to 1. I use a t-test to evaluate this regression with HAC standard errors. In R this would look like this:

diff <- as.numeric(series1[startRow:endRow]) - as.numeric(series2[startRow:endRow])
reg <- lm (formula = diff ~ 1, na.action = na.omit)
coeftest(reg, vcov=NeweyWest(reg, lag = 1, prewhite=FALSE), df=length(diff)-1)

The results are as follows:

t test of coefficients:

               Estimate | Std. Error | t value | Pr(>|t|)
(Intercept) -8.7425e-05 | 9.3240e-05 | -0.9376 |  0.3485

However, I do not know how to interpret the results. What does the constant do? What does the beta mean? What does it mean when the results are significant or insignificant?

(Problems with my sample: return series are dependent & individual return series are non-normal)

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1 Answer 1

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It is similar to a Diebold and Mariano test. It tests whether series1 minus series2 is positive or negative, while taking into account the possibility there is autocorrelation. If you had normal i.i.d data you could just look at series1-series2 and do a Student t-test as to whether the differences are on average zero or not. This is a fancy way of doing it, more general because of the HAC methodology. In your case series2 is better than series1 [i.e. on average series2[i] > series1[i] for all i] but not significantly so.

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  • $\begingroup$ Ok thank you. Can you also learn me why I need to run this on a constant and set the constant equal to one? And what this constant of one implies for my beta? $\endgroup$
    – user15050
    Aug 12, 2015 at 5:54
  • $\begingroup$ It is called "regression on a constant" because the model being estimated is series1[i]-series2[i] = Constant +epsilon[i] where Constant is to be estimated, with the assumption epsilon[i] have mean 0 but not necessarily homoscedastic and uncorrelated. The syntax diff ~ 1 is just how you say this in the R language. $\endgroup$
    – nbbo2
    Aug 12, 2015 at 6:44
  • $\begingroup$ Ah ok, thank you! $\endgroup$
    – user15050
    Aug 12, 2015 at 7:31

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