unbiased estimator of sigma in normal distribution 
It's mathematical statistics, but i couldn't find the category of it.
Anyway, i am not sure that i did it right. I need help.
r.s is random sample,
and u.e is unbiased estimator.
 A: This doesn't work: you've proved that if $c^2S^2$ is unbiased for $\sigma^2$ then $c=\sqrt{n/(n-1)}$, but that doesn't imply $\sqrt{n/(n-1)}S$ is unbiased for $\sigma$.
Instead, show that a suitable multiple of $S$ has the $\chi_n$ distribution. The expected value of a $\chi_n$ distribution is
$$ \sqrt{2} \Gamma((n+1)/2)/\Gamma(n/2)$$
which you can use to find what $c$ has to be.
A: What you did is wrong. You found that if $c=\sqrt{\dfrac n {n-1}}$ then $c^2 S^2$ is an unbiased estimator of $\sigma^2.$ It does not follow that $cS$ is an unbiased estimator of $\sigma.$ The expected value of a square root of a random variable is not the same as the square root of the expected value of the random variable.
Here I will use the fact that
$$
\frac 1 {\sigma^2} \sum_{i=1}^n (X_i - \overline X)^2 \sim \chi^2_{n-1}.
$$
This has been proved a number of times in answers posted here.
This probability distribution is
$$
\frac 1 {\Gamma(\frac{n-1}2)} \left( \frac x 2 \right)^{(n-1)/2-1} e^{-x/2} \, \frac{dx} 2 \quad \text{for } x>0.
$$
So you need the expected value of the square root of a random variable with this distribution. That expected value is
\begin{align}
& \int_0^\infty \sqrt x \frac 1 {\Gamma(\frac{n-1}2)} \left( \frac x 2 \right)^{(n-1)/2-1} e^{-x/2} \, \frac{dx} 2 \\[10pt]
= {} & \sqrt 2 \int_0^\infty \sqrt{\frac x 2} \cdot \frac 1 {\Gamma(\frac{n-1}2)} \left( \frac x 2 \right)^{(n-1)/2-1} e^{-x/2} \, \frac{dx} 2 \\[10pt]
= {} & \sqrt 2 \cdot \frac 1 {\Gamma\left( \frac{n-1} 2 \right)} \int_0^\infty \sqrt u \cdot u^{(n-1)/2-1} e^{-u} \, du \\[8pt]
= {} & \sqrt 2 \cdot \frac 1 {\Gamma\left( \frac{n-1} 2 \right)} \int_0^\infty u^{(n/2) - 1} e^{-u} \, du = \sqrt 2 \cdot \frac {\Gamma\left( \frac n 2 \right)} {\Gamma\left( \frac{n-1} 2 \right)}.
\end{align}
Now apply some identities involving the gamma function including the fact that $\Gamma\left( \frac 1 2 \right) = \sqrt\pi.$
