Can we tell anything at all about $\frac {S_n^2}{\sigma^2}$ if we know that the distribution is normal?

Suppose we don't know anything about a distribution except that it is a Normal distribution with mean $$\mu$$ and variance $$\sigma^2$$.

Can we argue, theoretically, the sample variance, $$S_n^2$$ will be close to the population variance, and thus can we say something about the ratio $$\frac {S_n^2}{\sigma^2}$$?

If this holds no basis in theory -- can we at least argue this in practice (mostly)?

1 Answer

Normal distribution is fully defined by its mean and variance, that is, if you know $$\mu$$ and $$\sigma^2$$ you know everything about the distribution.

The data points in the sample are however drawn randomly, so the sample variance will be close to $$\sigma^2$$ in probabilistic sense. In the case of a normal distribution $$S_n^2$$ follows a $$\chi^2$$-distribution with $$n-1$$ degrees of freedom.

I suggest looking for derivations of Student t-test, which usually cover the relation between real and measured means and variances.