I am using an EWMA model to calculate correlations and I would to know if there are any papers available that would help me estimate an optimal decaying factor.

try using Excel/Solver to minimise the variance between your predicted values and realised values over some fitting period, ie minimise cell variance by changing cell lambda.If I remember rightly RiskMetrics always quoted .94 as a broadly useful number, you may be able to find one of their free papers.

I have a relatively easy question about the same area...In JP Morgan's RiskMetrics Technical Documentation, Table 5.7 shows the number of historical observations used by EWMA for a given tolerance level and a given lambda. For example, a decay factor of 0.85 at 1% tolerance level requires 28 daily returns, whereas a decay factor of 0.98 at 0.01% requires 456 daily returns.I think I read somewhere that the number of daily returns required given the decay factor and tolerance level is a result of statistical proof and not empirical analysis. Could I please confirm this? Is there a reference for this?Thanks very much.

K = Ln(L)/Ln(lumda) defines the minimum daily return to reach the tolerance level L given the decay factor lumda.Often lumda is between 0.9 and 1 so at 0.1% T.L., the minimum data is between 66 and 687.

I seem to remember there is a classic error in the text which goes with that section were they state something 99.9 % of the information is therefore withing that tolerance band - in fact they forget to integrate the tale, and the cut of means you are taking about 95 percent of the info.But then I think I was born a pedant....

To the original poster:Probably the best (practical) reference to EWMA is : http://www.riskmetrics.com/pdf/rrmfinal.pdf?as its at the heart of the riskmetrics sytem.As somebody on the forum said before;'yeah, 120 pages on standards deviations'.You will have to register to get access (it should be free).

QuoteOriginally posted by: EdwynK = Ln(L)/Ln(lumda) defines the minimum daily return to reach the tolerance level L given the decay factor lumda.Often lumda is between 0.9 and 1 so at 0.1% T.L., the minimum data is between 66 and 687.Thanks for the formula, but what is the theory behind ln(L)/ln(lambda)? Thanks again.

There is no theory - it is a simple argument - at what date does your weighting to a historical return fall bellow a certain fraction of your weighting to the most recent days return?With exponential weighting you could go on adding historical returns for ever, but given data constraints this is impractical so you have to cut off the series somewhere.

looking for stronger theoretical background , may be you will find usefull to look at the garch(1,1) representation of the Ewma.

- Correlation is the ratio of (1) the covariance, and (2) the product of standard deviations.- if you estimate variances correctly to start with, you are on the right path for covariance.- for the variance, on each assets individually, the lambda to be choosen is the one that minimizes the square of differences between your estimates and observed daily ATM vols. The function is convex, so you can calibrate both with visual methods and using a newton approach. You have to be extremely careful with assets like Crude or NatGas, where the implied vol is an expected average of (growing) instantantaneous. vol. Your EWMA is then likely going to give you a measure of inst. vol but not of the expected average of it...- The thing now is to calibrate lambda for the pair, which is different to values obtained for assets individually. You can use an average between the two lambdas obtained if you can't observe correlation. But if there is a way to ask for a spread option price, back it out using lambdas from the two assets for variances. Otherwise if you dont have access to prices assume that covariance is slightly more stable than variance, increase your lambda by the square root of 1.05 (0.94 gives 0.966). This gives more stability to your estimates and allows further heding with vega.Now, if you work on things like Energy commodities, you can't do this and you are on your own...hope it helpsJezza