HISTORICAL STANDARD DEVIATION

[RenInvs]

Hi
In the part of the allen book:

"Compare, contrast and calculate parametric and non-parametric approaches for estimating conditional volatility, including: HISTORICAL STANDARD DEVIATION"

The part of :
For example, assume the previous four daily returns for a stock are 6% (n-1), 5% (m-2), 4% (n-3) and 3% (n-4). What is a current volatility estimate, applying the moving average, given that our short trailing window is only four days (m=14)? If we square each return, the series is 0.0036, 0.0025, 0.0016 and 0.0009. If we sum this series of squared returns, we get 0.0086. Divide by 4 (since m=4) and we get 0.00215. That’s the moving average variance, such that the moving average volatility is about 4.64%.


I don't understand why 6% (n-1), 5% (m-2), 4% (n-3) and 3% (n-4) have in parentheses (n-1), m-2, n-3 and n-4, what does it mean?

Thanks a lot!
Renzo

 

[ShaktiRathore]

Hi,
I think these terms are used to denote the number of the return e.g. n-1th return is 6%,n-2th return is 5% and so on. So that total of four returns n-1,n-2,n-3,n-4th returns are used to calculate moving average. There is typo of using m instead of n in m-2 si its n-2 instead of m-2.
n-1 to n-4 are the returns on days n-1 to n-4 used to calculate moving avg variance on nth day. If n=5, n-1,n-2,n-3,n-4th returns are 4,3,2,1st returns used to calculate avg variance on day 5.
Thanks

 

[RenInvs]

Hi,
I think these terms are used to denote the number of the return e.g. n-1th return is 6%,n-2th return is 5% and so on. So that total of four returns n-1,n-2,n-3,n-4th returns are used to calculate moving average. There is typo of using m instead of n in m-2 si its n-2 instead of m-2.
n-1 to n-4 are the returns on days n-1 to n-4 used to calculate moving avg variance on nth day. If n=5, n-1,n-2,n-3,n-4th returns are 4,3,2,1st returns used to calculate avg variance on day 5.
Thanks

Thanks a lot! It was a typo! When you checked after work it wasn't clear!