Timeline for Why is asset volatility easier to estimate than the asset mean if it contains the mean?
Current License: CC BY-SA 4.0
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| Nov 17, 2020 at 18:36 | comment | added | Richard Hardy | @DaveHarris, is anyone arguing against that? The question is, however, what should actually prevent the comparison? The fact that the estimator is given does not imply we cannot evaluate it, and if we can do that, then why not compare its performance across targets? So far I think I get your statement but not the argumentation behind it. In any case, I have received an interesting answer here. | |
| Nov 17, 2020 at 18:33 | comment | added | Dave Harris | @RichardHardy yes, the MVUE is the MVUE is the MVUE, which may be an inappropriate estimator due to loss or utility reasons, but if you are minimizing quadratic loss, then it is non-sensical to compare the standard error of $\hat{\mu}$ with the standard error of $\hat{\sigma}^2$. GIVEN quadratic loss and unbiasedness as a constraint, it is what it is. They have different sampling distributions. Neither are "easier" to find. The estimators are as precise as they can be, given the constraints. Change the constraints and you change the precision. | |
| Nov 12, 2020 at 11:15 | comment | added | develarist | I like the approach of this answer since it is illustrated with probability distributions rather than only formulas, but it lacks dedication to addressing financial data, given that we already know that the ease of mean estimation is easier than s.d. estimation outside of finance. I think it's wrong to say the reverse is an illusion when you're not even addressing the finance topic, otherwise you're shooting down some very highly-decorated and credible finance journals | |
| Nov 12, 2020 at 7:35 | comment | added | Richard Hardy | Thank you for your edit and comments. As I understand you now, it does not make sense to compare estimation precision/accuracy for different target parameters. That would imply all the answers that are actually doing the comparison are missing the point. Is that right? | |
| Nov 12, 2020 at 7:30 | history | edited | Richard Hardy | CC BY-SA 4.0 |
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| Nov 11, 2020 at 21:34 | comment | added | Dave Harris | @RichardHardy to give a trivial example, consider one thousand observations of the sample mean from an unknown distribution with a second moment. One solution for a 95% interval would be to use the standard result using Student's t that is taught in first semester courses. An equally valid confidence interval would be to choose $[x_{25},x_{975}]$. They are both optimal intervals. There is an infinite number of them. Confidence intervals are not measures of precision. Their narrowness or width cannot be interpreted that way. | |
| Nov 11, 2020 at 21:29 | comment | added | Dave Harris | @RichardHardy what I did not address in his post was the issue of the confidence interval. There is an infinite number of possible confidence intervals. The textbook ones usually meet some preferred standard of the author, but are not unique and are only optimal under certain conditions. Addressing confidence intervals would require a full paper. | |
| Nov 11, 2020 at 20:48 | history | edited | Dave Harris | CC BY-SA 4.0 |
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| Nov 11, 2020 at 20:26 | comment | added | Dave Harris | @RichardHardy I will provide an edit. | |
| Nov 10, 2020 at 22:05 | comment | added | Richard Hardy | The answer by fesman shows that standard deviation is estimated more precisely than the mean for the normal distribution with parameter values that are relevant in models of financial returns. You show that for a particular distribution with particular parameter values, mean is estimated more precisely than variance. How do we reconcile the two answers? And regarding your answer, how exactly should it generalize to an arbitrary distribution with arbitrary parameter values? | |
| Nov 10, 2020 at 20:40 | history | answered | Dave Harris | CC BY-SA 4.0 |