August 5th, 2003, 3:31 pm
QuoteOriginally posted by: AthleteScholarVaR = [ F * exp(-0.5 * Sigma^2 * 1/252 + Sigma * (1/252)^(.5) * 2.33) - F) * (A)VaR = 2.33 * Sigma * F * (252)^(-.5) * A For small Sigma, the Sigma^2 term in the first equation can be neglected and exp(Sigma*2.33*252^-0.5) is approximately 1 + Sigma*2.33*252^-0.5. Using that substitution, the two formulae are identical.The second formula is the VaR if you assume P&L has a normal distribution. This is the usual practice. The first formula is the VaR assuming P&L follows a lognormal distribution and Sigma is the standard deviation of the underlying normal, not the lognormal. This is not a common practice. It only matters for large Sigma, it takes an 82% annual standard deviation to make a 5% difference in the VaR calculations. There are much more important errors inthe VaR calculation than this one, and I see no reason that it would give a more accurate VaR.