Timeline for Minimum-variance unbiased estimator to estimate quantiles when the errors are normal distributed
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 1, 2023 at 13:40 | comment | added | Sextus Empiricus | Would the Lehmann Scheffé theorem also work for functions of multiple parameters? | |
| Mar 13, 2019 at 10:45 | history | edited | Sextus Empiricus |
edited tags
|
|
| Mar 12, 2019 at 21:19 | comment | added | whuber♦ | The near-but-non-duplicate is a good reference, though, because it concerns a closely related problem: estimation of a quantile is a form of tolerance limit. (That's why I asked the question about what you mean by "better," because the different kinds of tolerance limit differ about that.) | |
| Mar 12, 2019 at 21:15 | comment | added | Sextus Empiricus | The duplicate I was thinking of was stats.stackexchange.com/questions/382124/… but it is not the same. | |
| Mar 12, 2019 at 21:08 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
added 133 characters in body; edited title
|
| Mar 12, 2019 at 21:00 | history | tweeted | twitter.com/StackStats/status/1105574316119388160 | ||
| Mar 12, 2019 at 20:54 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
added 118 characters in body
|
| Mar 12, 2019 at 20:49 | comment | added | Sextus Empiricus | @whuber, I was thinking of the minimum variance unbiased estimator. | |
| Mar 12, 2019 at 20:13 | comment | added | whuber♦ | Re the edit: the question is now in a form I expected to see when it first appeared, so the time is ripe to post the comment I originally had in mind: exactly in what sense do you mean "better"? Obviously alternative procedures will be biased, so what loss function (or other quantitative objective) would you have in mind? | |
| Mar 12, 2019 at 20:08 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
added 179 characters in body
|
| Mar 12, 2019 at 20:05 | comment | added | Sextus Empiricus | I see now that I had mistaken $\frac{1}{n}$ as the correction for $\sigma^2$, as well as $\sigma$. | |
| Mar 12, 2019 at 19:59 | comment | added | whuber♦ | Your impression is correct: expectation is linear. What do you suppose the expectation of the sample standard deviation might be? (I haven't checked my formula, but certainly the expectation is not equal to $\sigma.$) | |
| Mar 12, 2019 at 19:56 | comment | added | Sextus Empiricus | aha, I see, I was under the impression that E[m + ks] = E[m] + k E[s], I will look into that. But, in any case, when I made that assumption erroneously, it is not the idea of my question. | |
| Mar 12, 2019 at 19:52 | comment | added | whuber♦ | I took expectations! (See the first comment.) Because you're requiring the estimator be unbiased, this is the very first thing to check. The point is that the requirement to be unbiased uniquely determines $k.$ That's doesn't seem to leave much to ask about. | |
| Mar 12, 2019 at 19:49 | comment | added | Sextus Empiricus | To be honest, I do not really understand where the expression $\sqrt{2/(n-1)}\Gamma(n/2)/\Gamma((n-1)/2)$ comes from. What I am asking is whether quantiles of a population can be estimated efficiently using the mean and variance of a sample taken from that population, when the population has a normal distribution. I know that this is the case when the quantile to be estimated is the median (which is a simple case because it is basically estimating the distribution parameter $\mu$), I wonder whether this is also true when the quantile that is to be estimated is not the median. | |
| Mar 12, 2019 at 19:24 | comment | added | whuber♦ | Would I be correct, then, in interpreting your question as asking whether it is the case that $$\sqrt{2/(n-1)}\Gamma(n/2)/\Gamma((n-1)/2)=1$$ for any (or even one) integral value of $n$? | |
| Mar 12, 2019 at 18:45 | comment | added | Sextus Empiricus | $k$ is a fixed parameter. It is set according to the quantile of the normal distribution that one wishes to estimate. E.g. if one wishes to estimate the 95th quantile of a population that is normal distributed, then one is indirectly estimating the value of $\mu + 1.645 \sigma$. I wonder whether $\bar{x} + 1.645 s$ is an efficient unbiased estimator of $\mu + 1.645 \sigma$ (and in general the same question for other values of $k$). | |
| Mar 12, 2019 at 18:03 | comment | added | whuber♦ | Could you please state what you think $k$ should be more specifically? After all, the unbiasedness criterion immediately implies $E[m+ks]=\mu+k\sigma\sqrt{2/(n-1)}\Gamma(n/2)/\Gamma((n-1)/2),$ uniquely determining the value of $k.$ | |
| Mar 12, 2019 at 17:21 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
added 24 characters in body
|
| Mar 12, 2019 at 17:19 | comment | added | Sextus Empiricus | I vaguely remember it being a duplicate. I believe that a question about the efficiency of the estimator $\bar{x} + k s$ to estimate $\mu + k \sigma$ has occurred before. But I could not find it easily, so I thought that this question, when it is a duplicate after all, may at least be a good pointer for searchers of the same question. | |
| Mar 12, 2019 at 17:15 | history | asked | Sextus Empiricus | CC BY-SA 4.0 |