# Resampled efficient frontier length of simulation

I was provided with a VBA program from my lecturer that applies the resampled efficient frontier. We have an investment horizon $T$ (6 years) and he uses a multivariate normal distribution with parameters $\mu$ and $\Sigma$ and the code simulates $12$ years of monthly data. He does this a few hundred times, but only takes the top half of the returns matrix, $R$, (i.e. takes a $6$ year slice of it) to calculate the estimates of $\mu$ and $\Sigma$ in the REF procedure.

Does it make any difference whether we simulate $6$ years or $12$ years under MVN if we're just going to be slicing off $6$ years of simulated returns from the top of the simulated $R$ no matter what?

-
Hi quantface, I notice you have the same IP address as user3031 and Harokitty. Are these your accounts too? Oh, and all three of these accounts have practically the same email address. –  chrisaycock Oct 1 '12 at 13:33

Maybe you want to use the second part of $R$ to perform an out of sample test of the REF to show that the strength of the REF lies not in its in-sample performance (where it must be inferior to MV optimization) but rather in its out-of-sample performance but we can only guess on this part. It could also be an error in the code but I think to hold half of the data back for the out-of-sample test should be the reason.
So you're saying that if we compared $N$ simulations of the TOP HALVES of the 12 year $R$ (a (6*12)*No.Assets matrix) to $N$ simulations of a 6 year $R$ (also a (6*12)*No.Assets matrix), there would be some kind of difference? –  user2921 Oct 1 '12 at 12:23
I'm not sure that this is for a simulation study. It takes a gigantic amount of time to calculate one resampled frontier over $200$ simulations in VBA, let alone simulating multiple. –  user2921 Oct 1 '12 at 12:28
No i was just implying that if you estimate $\mu$ is formed of $N=6\times 12$ years then, all else being equal, you will in genereal end up with a wider confidence interval for $\mu$ than if it is based on a sample $N=12\times 12$ points. Taking the top half makes no difference since by definition a random sample is independent (i.e. the values in the rows of your matrix are independent) or, if you take a time series viewpoint, the increments of the brownian motion have no autocorrellation. –  vanguard2k Oct 1 '12 at 12:32