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Given a sample X of size n from a normal distribution $N(\mu,\sigma)$ one can estimate $\sigma$ by $\hat{\sigma}$ from X. Then we know that:

$p(\sigma|\hat{\sigma},n)\propto\hat{\sigma}\sqrt{\frac{n-1}{\chi^2(\nu=n-1)}}$

Now I am looking for the likelihood function: $p(\hat{\sigma}|\sigma,n)$.

Is there an analytic solution for this or is it necessary to derive the likelihood function from case-by-case simulations?

Can someone help me with this relation? Or better, point to a reference where this relation has been used before?

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  • $\begingroup$ Because it is unnecessary to consider $\sigma$ to be a random variable, your first formula is irrelevant. To get your bearings, take a look at the $\chi$ distribution. $\endgroup$ – whuber Jun 11 '14 at 15:39
  • $\begingroup$ @whuber WolframMathworld seems to describe the answer. The distribution of the sample standard deviation is indeed closely related to the chi-distribution. $\endgroup$ – Sam Jun 12 '14 at 9:18
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WolframMathworld does provide the likelihood function $p(\hat{\sigma}|\sigma,n)$.

I confirmed the validity of this answer by simple Monte Carlo simulation.

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