pennyless

I come back to this issue because I still can't get it. Anyone can help ?

dyga

1. By lengthening the measurement interval - std deviation increases.

Usually annual std dev > weekly std dev.
But what is a way this number can be gamed?
if you calculated a weekly std deviation and converted it into a annual std deviation using
annual std dev calculated = weekly std dev * sqrt(52) (since there are 52 weeks in a year) the annual calculated std dev will be a smaller number.

but numerator likewise would not be affected. It would be more or less the same number (r weekly * 52 would be approximately equal to r-annual.)

but now since a lower annual std dev calculated is used - the sharpe ratio now would be a HIGHER number (since the denom. is lower).

2:
Returns are compounded. So numerator = (1+rweek1)*(1+rweek2)*...(1+rweek52) - 1
std dev = monthly std dev*sqrt(12). (not compounded).

return would most likely be equivalent to the annual return (or off by a very little bit).
but std deviation calculated as above would be lower - so end result Sharpe Ratio would be higher.

CP

bodhisattva

CP, thank you very much for your response !

1. What you meant are :
Acutually (usually) : Yearly SD > monthly SD > weekly SD > daily SD
if the Yearly SD is calculated as : monthly SD x 12^1/2 or weekly SD x 52^1/2 or
daily SD x 250^1/2, the calculated Yearly SD will be lower than the actual yearly SD,
i.e., the calculated yearly SD is an under-estimated one and the calculated Sharpe ratio
will be HIGHER than the actual/real one.
Am I right ?

2. In your example, I think the return in the numerator shall be : (1+r month1)*
(1+r Month 2) * ...(1+r month 12) -1 , since you use "monthly" SD in the denomintor.
What is the monthly SD here ? The SD of a specific single month's return ?
And this will lead to higher Sharpe ratio (than real one) ?
Since sqrt(12) = 3.464 and the annualized yearly return (compounded from monthly
returns) is not necessarily 3.464x greater than the annualized yearly return, therefore,
I think it is not necessarily that the calculated Sharpe ratio will be HIGHER than the
actual/real one. i.e., there is a likelyhood that the the calculated Sharpe ratio will be
LOWER than the actual/real one. But in any case, the calculated Sharpe ratio will be a
distorted one.
Am I correct ?