Timeline for Tightest bounds on sample variance given sample size, mean, minimum, and maximum
Current License: CC BY-SA 4.0
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| May 7, 2018 at 22:04 | comment | added | Steve Kass | Thanks for your comments. I removed “And more obviously...” and replaced it with “For example, suppose $m=-1$, $M=10$, and $n=10$. Then $\nu=-\frac{m+M}{n-2}=-\frac98<m$.” (It seems to me that for $\mu=0$ from data with minimum $-1$ and maximum $10$, there must be at least $11$ data points. Am I still missing something?) As for the discrete case, (I will think more about it) can some better bound be given if, say, the data values are known to be integers? | |
| May 7, 2018 at 22:01 | history | edited | Steve Kass | CC BY-SA 4.0 |
added 22 characters in body
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| May 7, 2018 at 12:41 | comment | added | equaeghe | “If the data are discrete, the bounds with that additional assumption are generally tighter”: I've now clarified that I consider the real-valued case. Also, I am quite sure that no general expression can be given in the discrete case, as you are faced with a non-linear integer programming problem to determine which sets of samples are feasible. (Of course, if you make additional assumptions, this picture may change.) | |
| May 7, 2018 at 12:20 | comment | added | equaeghe | “If n is too small, what you call $\nu$ will be outside the range $[m,M]$”: I don't think so; did you take into account that $\mu=0$ is assumed? | |
| May 7, 2018 at 12:17 | comment | added | equaeghe | “And more obviously, it must be the case that $m\leq\mu\leq M$”: I mention this as $m\leq0\leq M$ in my derivation, where $0=\mu$. | |
| May 7, 2018 at 11:52 | comment | added | equaeghe | “I think you left out n near the top of your question”: Thanks, clarified. | |
| May 4, 2018 at 20:57 | comment | added | whuber♦ | We have several threads that establish the mathematical equivalent of $$\sigma^2 \le (\mu-m)(M-\mu)$$ for arbitrary distributions supported on $[M,m]$ and show that it is tight: the maximum is attained for a two-point distribution giving probability $(M-\mu)/(M-m)$ to $m$ and probability $(\mu-m)/(M-m)$ to $M.$ In light of this, the bounds given in the question look plausible (but I haven't checked the derivation). For your case the situation is more complicated, because $m,\mu,M$ alone determine a relatively small finite number of possible values of $\sigma^2.$ | |
| May 4, 2018 at 20:14 | review | First posts | |||
| May 4, 2018 at 20:18 | |||||
| May 4, 2018 at 20:10 | history | answered | Steve Kass | CC BY-SA 4.0 |