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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?

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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

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It doesnt make any difference. Of course with 12 years under MVN your estimates will be closer to the "true" values of the population.

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.

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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
Very nice. I like the throwback to independent and stationary increments! – user2921 Oct 1 '12 at 12:33
My guess was that maybe the approach was like the one often used in machine learning: You take your dataset, split it in two halves, train the algorithm on the first half (estimating your REF and MVF for this "training set") and then testing it on the second half to see how they perform out of sample ("test set"). – vanguard2k Oct 1 '12 at 12:34

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