Distribution of variance of Gaussian variable 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.
 A: 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 $$s^2 = {{1} \over {n-1}} \sum_{i=1}^n {\left( x_i - \bar x \right)^2}$$
