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Why does the function VarianceMLE give a different result from Variance?

And what is it in Mathematica 11.3?

enter image description here

Please see the picture above t665he MLE is 621 and the other is 665.

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  • $\begingroup$ In Mathematica 11.3, << Statistics` produces an error message and Variance@data // N gives 665.524 $\endgroup$ – m_goldberg Jan 14 at 7:47
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    $\begingroup$ I'm going to guess that VarianceMLE is the maximum likelihood variance estimator rather than the unbiased one (assuming normally distributed data). The difference between the two is that Variance divides by N-1 (N == Length[data]) while the MLE estimator divides by N. $\endgroup$ – Sjoerd Smit Jan 14 at 9:22
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    $\begingroup$ And for future reference: please post copyable code in your question rather than a screenshot. This makes it much easier for someone else to copy your code and try things out. $\endgroup$ – Sjoerd Smit Jan 14 at 9:23
  • $\begingroup$ Which book did you see this in? It looks like a scan. $\endgroup$ – Szabolcs Jan 14 at 11:45
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I just checked my guess in my comment and I was right. VarianceMLE is the maximum likelihood variance estimator (see, e.g. here).

data = {34, 56, 28, 62, 32, 90, 20, 10, 12, 35, 63, 78, 12, 25, 68};
Variance[data]

13976/21

myVariance[lst_List] := Total[(lst - Mean[lst])^2]/(Length[lst] - 1);
myVarianceMLE[lst_List] := Total[(lst - Mean[lst])^2]/Length[lst];

myVariance[data]
myVarianceMLE[data]

13976/21

27952/45

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VarianceMLE computes a biased, maximum likelihood estimate of the population variance. Variance computes an unbiased estimate of the population variance. It can be shown that VarianceMLE underestimates the variance of the population.

Let $\{y_i : 1 \leq i \leq n\}$ be a sample of $n$ values from a population. The variance (central second moment) of the sample is $$ \sigma_y^2 = \frac{1}{n} \sum_{i=1}^n (y_i - \bar{y}) \text{,} $$ where $\bar{y} = \frac{1}{n} \sum_{i=1}^n y_i$ is the sample mean. This $\sigma_y^2$ is computed by VarianceMLE.

If $\sigma^2$ is the population variance, with some work, one can show that the expected value of $\sigma_y^2$ is $\frac{n-1}{n} \sigma^2$, so the sample variance is a biased estimator of the population variance. We can make this an unbiased estimator via $$ s^2 = \frac{n}{n-1} \sigma_y^2 = \frac{1}{n-1}\sum_{i=1}^n (y_i - \bar{y}) \text{.} $$ This $s^2$ is computed by Variance. From the documentation (in the Details):

"Variance[list] is equivalent to Total[(list-Mean[list])^2]/(Length[list]-1) for real-valued data."

The very sparse documentation for VarianceMLE indicates that it is implemented in terms of Variance:

"VarianceMLE[data_] := Variance[data] (Length[data] - 1)/Length[data]"

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