# Variance Ratio Test in R

I would like to conduct a variance ratio test for a financial time series in order to examine whether I can apply the square root rule for the variance with the software R. I used the Automatic Variance Ratio Test vrtest::Auto.Vr and got a statistic of -0.01. Now I am wondering, is that the z-score, which is distributed standard normal under the Null hypothesis, that this ratio is 1 (or equivalent that there is no autocorrelation)? It is not specified in the describtion, I just found this:

Usage:
Auto.VR(y)
Arguments:
y financial return time series
Value:
stat Automatic variance ratio test statistic


TL;DR: the test statistic's distribution is $N(0,1)$

A bit more information about the Automatic Variance Ratio Test:

$H_0$: ${\Delta}r_t$ is serially uncorrelated (where ${\Delta}r_t=r_t-r_{t-1}$)

$H_1$: ${\Delta}r_t$ is serially correlated

The test statistic is $VR=\sqrt{T/l}[\hat{VR}(l)-1]/\sqrt{2} \quad {\xrightarrow{d}} \quad N(0,1)$

The $d$ over the arrow is important, i.e. the $VR$ converges in distribution to standard normal (hence $d$ does not imply the convergence in mean square or convergence in probability) as $T$, $l$ and $T/l$ approach infinity. The $l$ is the lag truncation point. The paper has detail on formulae for both $VR$ and $l$.

Final note: the test is two sided.

• Isn't convergence in distribution the weakest among the three forms of converncege that you mention, so that it cannot imply the other two (while the other two imply the former)? See Wikipedia. – Richard Hardy Mar 5 '17 at 12:05
• Absolutely! It goes m.s. -> p -> d. That is a typo, now fixed. – rbm Mar 5 '17 at 12:13