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# Standard Deviation without mean in EWMA/GARCH

#### Makki

##### New Member
Hi David,

one quick question regarding the calculation of StDev for the EWMA model. While I was watching your youtube video:
I noticed that your formula to calculate the simple StDev only sums squared returns. Why is the mean not subtracted from the daily returns before squaring them?

Kind regards

Markus

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi Markus,

Only because the period is daily; when it's only a daily period, Hull says we can round to zero on an assumption that it is nearly zero anyway; e.g., +10% per annum = only 0.004. Carol Alexander goes further and suggests it is preferred to exclude the sample mean and assume zero (!), emphasis mine:
"This mean deviation form [i.e., subtracting the mean before squaring] of the estimator may be useful for estimating variance using monthly or even weekly data over a period for which average returns are significantly different from zero. However, with daily data the mean return is usually very small. Moreover, the errors induced by other assumptions are huge relative to the error induced by assuming the mean is zero, as we shall see below. Hence, we normally use the form (II.3.7) [i.e., excluding the mean: using only squared returns]" -- Carol Alexander, MRA Vol II on volatility

What i have noticed is, consider this odd outcome, what if the returns are:
-10%, -10%, -10%, -10%, ....

If we include the mean, the (population) standard deviation is zero (I resent this outcome ); but if we exclude, it is 10%. I actually prefer the 10% outcome. This is an exaggerated, ridiculous example but just illustrates the premise (I think) of Carol Alexander's preference: it may be that the historical sample returns a mean that is not best with respect to a current estimate; or technically, I think she is suggesting, if we want an estimate of unconditional volatility, due to sampling variation, it may be better to make the error of assuming E[daily return] = 0, than to trust the historical sample. Thanks,

#### ShaktiRathore

##### Well-Known Member
Subscriber
As illustrated above,
I would share an example as why would zero mean is more feasible,
if there are 5 observations with .05%,.04%,-.03%,-.01%,.02%
mean =.05%+.04%-.03%-.01%+.02%/5=.07/5=.014% which is approx. equal to zero.
if we assume mean as it is , variance=((.05-.014)^2+(.04-.014)^2-(-.03-.014)^2-(-.01-.014)^2+(.02-.014)^2)/4=0.001296+0.000676+0.001936+0.000576+0.000036=0.00452=> sqrt(0.00452)=6.72% which is well above if we assume mean as zero.
If we take mean as zero, variance=.05%^2+.04%^2-.03%^2-.01%^2+.02%^2/4=.0025+.0016+.0009+.0001+.0004/4=.0055/4=.001575=> s.d.=sqrt.(001575)=3.96% which doe not reflect the real volatility if we take the actual mean of the sample i.e. its underestimating the actual volatility.
lets look at a serious case, But if there is a positive serial correlation present , returns might look like .05%,.06%,.07%,.08%,.09%
mean=.05%+.06%+.07%+.08%+.09%/5=.07% which is significant
if we assume mean as it is, variance=.02^2+.01^2+0^2+-.01^2+-.02^2/4=.0004+.0001+0+.0001+.0004/4=.0010/4=.00025=> s.d. =sqrt(.00025)=1.58%
If we take mean as zero , variance=.05^2+.06^2+.07^2+.08^2+.09^2=.0025+.0036+.0049+.0064+.0081=0.0255=> s.d. =sqrt(0.0255)=15.98% which is almost 10 times as measured by assuming actual mean again which does not reflect the real volatility if we take the actual mean of the sample i.e. its overestimating the actual volatility.
So it might be possible that if we are taking 0 as the actual mean that the deviation we derive might actually deviate from the actual deviation several times if there is some sort of serial correlation present as cited above.
If observations are random,
.05%,-.05%,.05%,-.05% => mean =0 which is a special case if there is perfect negative serial correlation present. In this case our assumption of mean zero in finding volatility is not violated. So this assumption of zero mean is well suited for returns that are negatively correlated but as the correlation tends to turn positive the assumption seems to be violated.

thanks