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

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