Wikipedia:Reference desk/Archives/Mathematics/2023 August 29

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August 29

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Variance measure from 0 to 1 (or o% to 100%)

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Is there a standard way to measure variance such that it is always 0 to 1? Example of usage: I have to sets. One has values from 10 to 100. The other has values from 1000 to 1000000. I want to compare the variance of the two sets head to head. I want 0 (or 0%) to mean that there is absolutely no variance (every value in the set is the same). I want the variance measurement to max out at 1 (or 100%) for each set. Then, I could say that one set was something like 42% variance and the other was 85%, practically doubled. Using variance (or standard deviation) doesn't cap out at 1. I know I could work out the maximum possible variance based on the ranges of the values and divide by that, but it seems that this is something that was already done by someone else with a more elegant solution. 12.116.29.106 (talk) 15:59, 29 August 2023 (UTC)[reply]

Does Coefficient of variation meet your requirements? --Jayron32 16:06, 29 August 2023 (UTC)[reply]
Am I reading it correct that it shouldn't exceed 1 more than 0.2% of the time? If it is rare that it exceeds 1, I feel that I could cap the value at 1 and if it is really 1.1, I will just use 1 and pretty much get the same result. I am incorrect. It is not limited 0 to 1 in any way. It is a bit standardized and in practice I am seeing values from around 10 to 80. Strike that as well. I was dividing mean by stddev. Dividing stddev by mean gives me value of 0.04ish to 0.50ish. 12.116.29.106 (talk) 16:14, 29 August 2023 (UTC)[reply]
If the support of the density of the random variable is a finite interval   the maximal value of the variance is equal to   So if the sample variance is  , the quantity   is a measure of dispersion in the range from   to   When using the unbiased sample variance   (see Variance § Unbiased sample variance) this measure can theoretically slightly exceed 1, but in practice this is unlikely to happen.  --Lambiam 20:35, 29 August 2023 (UTC)[reply]