Unbiased estimator of standard deviation of a normal distribution, using gamma function [duplicate]

Question

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• Why is sample standard deviation a biased estimator of $\sigma$? 3 answers

According to the Wikipedia article, the following estimator of the standard deviation $$s=\sqrt{\frac{1}{n-1}\sum_{k=1}^n(x_i-\bar{x})^2}$$

for a normal variable, verifies $E[s]=C_4(n) \sigma$, where

$$ C_4(n) = \sqrt{\frac{2}{n-1}}\cdot\frac{\Gamma(\frac{n}{2})}{\Gamma(\frac{n-1}{2})}=1 - \frac{1}{4n} - \frac{7}{32n^2} - \frac{19}{128n^3} + O(n^{-4}) $$

Therefore we have an unbiased estimator for the standard deviation of a normal variable,

$$\hat{\sigma}_\text{unbiased} = \frac{1}{C_4(n)} \sqrt{\frac{1}{n-1}\sum_{k=1}^n(x_i-\bar{x})^2}$$

How could you prove this?

Answer 1

How could I prove it? I'd do it by expanding the two gamma functions as series.

I'd probably just start with the Stirling series, taken to a few terms for the numerator and denominator, simplify as far as possible, then expand the inverse of the denominator out in another series expansion and collect the low-order products.

That said, there's probably an easier way to do it.

Answer 2

The random variable found by multiplying (n-1) times the sample variance and dividing by the population variance has a chi-squared distribution with degrees of freedom equal to (n-1). You can find the expected value of the square root of this variable by using the definition of the expected value of a function of a random variable. The integral will give you a gamma function with argument (n/2). The other gamma function comes from the pdf.