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I have a Gaussian random variable, which I can use to generate a sequence of values. So, I've generated a sequence of values of arbitrary length, and each set of 50 data become a sample. Now, consider a new variable, which is the variance of the samples made as described. Which is the distribution of this variable? Is this distribution the same if I consider the standard deviation instead of the variance?

Update. I've searched on the web and I've found that surely samples variance has not a Gaussian distribution (instead, samples mean follows a Gaussian distribution). It could be a Chi-square or a Gamma distribution, but I don't know precisely which of these ones could be right.

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    $\begingroup$ What do you mean by "clusterize"? Are your data coming from a single population / Gaussian, or is it from 2 populations / a mixture of 2 Gaussians? If you assign the data to 2 classes, do you have true class labels or will this be unsupervised? What algorithm are you going to use to do so either way? Do you just want to know the distribution of the variance from a Gaussian (& whether it is the same as the dist of the SD)? Those are straightforward & answerable questions. $\endgroup$ – gung - Reinstate Monica Jan 13 '16 at 16:36
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    $\begingroup$ I would to know the distribution of the variance from a Gausisan and whether it is the same as the distribution of the standard deviation. Excuse me if I've not been so clear. $\endgroup$ – foolcool Jan 13 '16 at 18:29
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    $\begingroup$ That is a perfectly good question, @foolcool. Why don't you edit your post to make that clearer & we can re-open & answer it for you. $\endgroup$ – gung - Reinstate Monica Jan 13 '16 at 18:31
  • $\begingroup$ The new phrasing is better, but still a bit ambiguous. How are you dividing the data into 2 samples? Are you simply doing this randomly, or are you using some method (eg, all values >x go into sample 1, else sample 2)? Can we simply ignore the dividing issue & address the question of the distribution of the variance from a Gaussian & whether it is the same as the dist of the SD? $\endgroup$ – gung - Reinstate Monica Jan 13 '16 at 22:33
  • $\begingroup$ I've not sayed at all that I've divided the sequence into two samples! The sampling is done sequentially, value by value. Imagine to have an array populated by these values: start from the scratch and group them in different samples of the same size (e.g. 50, but now it's not important). How many samples? Depends on the length of the sequence. $\endgroup$ – foolcool Jan 13 '16 at 22:48
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It's not clear to me exactly what you are looking for, but I will make an attempt at an answer. If the population is normal with variance $\sigma^2,$ then the quantity $${{(n-1)}s^2 \over {\sigma^2}}$$ has a chi-squared distribution with $n-1$ degrees of freedom, where $n$ is the sample size and $s^2$ is the sample variance using $n-1$ in the denominator.

A chi-squared random variable is a special case of a gamma random variable, and a gamma random variable has the property that if you multuply it by a constant, it is still a gamma random variable, just with a different scale parameter.

Using this property, we can find that the sample variance $s^2$ has a gamma distribution with shape parameter equal to ${{(n-1)} \over {2}}$ and a scale parameter equal to ${{2 \sigma^2} \over {n-1}}$

The distribution of $s,$ being a positive power of $s^2,$ has a generalized gamma distribution. Using Wikipedia's parameterization, the values of the parameters are $$p=2, \ d=n-1, \ a= \sigma \sqrt{ {{2} \over {n-1}} } $$

Again from Wikipedia, we have the expectation of a generalized gamma as $$E[X] = { {a \Gamma \left( {{d+1} \over {p}} \right) } \over { {\Gamma \left( {{d} \over {p} } \right) } } } $$

We then have immediately that the expected value of $s$ is

$$E[s] = { {\sigma \sqrt{{{2} \over {n-1}}} \Gamma \left( {{n} \over {2}} \right) } \over { {\Gamma \left( {{n-1} \over {2} } \right) } } } $$

This shows it as a biased estimator and suggests how to modify it if you want an unbiased version.

Note to address OP request: The formula for $s^2$ is $$s^2 = {{1} \over {n-1}} \sum_{i=1}^n {\left( x_i - \bar x \right)^2}$$

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  • $\begingroup$ What does s stands for in that formula? Anyway, my question synthetically is: "Take the variabile v which is the estimated variance of a sample. This values is computed for each sample with this formula: $$ \sum \limits_{i=1}^n (x_i - \overline{x})^2 \over n-1 $$ Which is the distribution of v? And which would be the distribution of the variable that corresponde to the estimated standard deviations?" $\endgroup$ – foolcool Jan 16 '16 at 17:58
  • $\begingroup$ I would know which is the distribution of $s^2$ in your formula. I'm not interested to the distribution of the whole formula. $\endgroup$ – foolcool Jan 17 '16 at 11:06
  • $\begingroup$ Thanks a lot for your support! Anyway, I've found a PDF from Mit OpenCourseWare where they say that the sample variance of a Guassian r.v. has a Chi-Squared related distribution. You can find this assertation at page 3 of the pdf. What do you think about it? $\endgroup$ – foolcool Jan 20 '16 at 10:39
  • $\begingroup$ It is completely consistent with what I have posted. Do you see any differences or issues? Also, at your request, I have given two items they don't supply, which is the distribution of $s^2$ and the distribution of $s.$ If you like the answer, please accept it. $\endgroup$ – soakley Jan 20 '16 at 14:39
  • $\begingroup$ I dont'see issues. I would only know what do you thnik about it seeing as you have a good knowledge about stastistics theory. $\endgroup$ – foolcool Jan 20 '16 at 14:54

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