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I have estimated two different time series (A and B) using daily data, and the average returns of each of them are insignificant. However, once I construct a third time series using the difference between A and B (A-B). The average returns of A-B are significant. How to interpret these results? Can I say that?

"As both the A and B strategies do not generate positive returns; therefore, difference between A and B at least indicate that the the effect of A tends to be stronger than the effect of B.

Thank you very much for your time, really appreciated.

Kind Regards,

Mac

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1 Answer 1

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Without the data, it's hard to say, but one potential answer could be that these TSs are positively correlated.

Let's assume that you have two TSs A and B

\begin{align} y_{a,t} = \mu_A + \epsilon_t,\\ y_{b,t} = \mu_B + \nu_t,\\ \end{align}

and you test $H_{A,0}: \mu_A = 0$ and $H_{B,0}: \mu_B = 0$. You can't reject them, maybe your TS are very volatile. Then you construct the third time series $y_{c,t} = y_{a,t} - y_{b,t}$. Assume that your errors are perfectly correlated $(cor(\epsilon_t, \nu_t) = 1$). This implies that $y_{c,t} = \mu_A - \mu_B$ and hence always significant unless $\mu_A = \mu_B$. Intutition is that by combining two TSs you can gretly reduce the variablility and hence have a new significant TS. I attach the code the show the idea.

library(mvtnorm)
set.seed(1)
data <- rmvnorm(n = 1e2, mean = c(-2,-4), sigma = matrix(c(6, 5.9, 5.9, 6), nr = 2))
x <- data[,1]
y <- data[,2]

cor(x,y)

plot(density(x))
plot(density(y))


z <- x - y
plot(density(z))
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  • $\begingroup$ Thank you very much Koval, that makes sense. $\endgroup$
    – Mac8444
    Commented Aug 1, 2020 at 22:07
  • $\begingroup$ @Mac8444, you are very welcome. $\endgroup$ Commented Aug 2, 2020 at 6:41

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