Hi, I can’t reconcile standard deviation T years out decreasing (relative to current annual S.D.) at 1/Sqrt(T) i.e. Annual S.D. 2 years out would be 1/Sqrt(2) (in the contect of quality control charts) with: Annual Standard deviation = daily standard deviation * Sqrt(252) Can anyone shed any light on this for me? Thanks, APP

in this formula, Annual Standard deviation = daily standard deviation * Sqrt(252) our sample data is for 1 year. 252 is trading days. we use daily std times sqrt of number of samples to get annualized std. now let’s expand the sample to 2 yrs or 504 trading days. assume daily std doesn’t change, which is quite likely, apply the same formula, you will use daily std times sqrt of 504. this overstates the annual std. so you need to divide sqrt 2. hope this helps.

What’s the point to this ? Does it mean that as time increases, then standard deviation decreases?

Boris, Thanks for the reply. I was with you all the way to ‘this overstates the annual std. so you need to divide sqrt 2’… but don’t worry. I’ve looked at this enought to just remember it. In the black-scholes section of Hull’s options/derivatives book (p284) it seems to be to do with continuous compounding. 'The continuously compounded rate of return per annum is normally distributed with mean U- @/2 and S.D: @/Root(T) U = E(Return) on stock/year @ = Vol of stock price / year I got this from talking to our options guy (who incidentally couldn’t derive this). Best way I’ve found to think about it is that you can be alot more sure about standard deviation with a big sample size vs. a small one. Bidder. Yes - according to cfai the Standard deviation does decrease as time (and hence sample data) increases. APP

hull is bible it is actually a question on converting daily std to annual. std usually presents in annual term. level 3 textbook talks about frequency of data. just remember you need to divide sqrt of # of data. if you are using 2 years of daily data, you divide sqrt 504, which equals to sqrt252 times sqrt2. so 2 yr to 1 yr is divide sqrt2. it took me a while to understand this as well, and it is hard to explain. i actually learned it by calculate spx std in excel. in this way i see the changes in numbers. i learn black-schole in excel, otherwise there is no way i can remember it.

allépourpêcher Wrote: ------------------------------------------------------- > Hi, > > I can’t reconcile standard deviation T years out > decreasing (relative to current annual S.D.) at > 1/Sqrt(T) i.e. Annual S.D. 2 years out would be > 1/Sqrt(2) (in the contect of quality control > charts) with: > > Annual Standard deviation = daily standard > deviation * Sqrt(252) > > Can anyone shed any light on this for me? > > Thanks, > > APP I assume that is the case because your hypothesis is about alpha which is the mean. Standard deviation of the mean declines with the number of observations.