# why is standard deviation a biased estimator [duplicate]

In this post: Why is sample standard deviation a biased estimator of $\sigma$?, I am having difficulty understanding some of the steps. We have :

(a) $$s^2=\frac{1}{n-1}\sum_{i=1}^{\infty}(x_i-x\bar)^2$$

and further down,

(b) $$E(s)= \sqrt\frac{\sigma^2}{n-1} E(\sqrt\frac{s^2(n-1)}{\sigma^2})$$

I understand the 2nd and 3rd lines, but how do we get from (a) to (b)? (My knowledge of maths is basic)

• (a) to (b) is clear now; I am trying to work this through step-by-step (sorry I am bit slow). If (b) is line 4, then on line 5, I gather than alpha is n-2, and beta=2. (i) On line 5, I am not clear how we get X^(n/2)-1, since on line 4, we have the following x terms: x^(1/2).x^(n/2).x^(-1/2), which should give x^(n/2). (ii) I am also not clear how we get from line 7 to 8 (i understand the integrand=1, but am not sure how the sigma^2 gets substituted by 2 in the 1st term) Commented Oct 30, 2020 at 7:43
• This is Jensen's inequality in reverse: since$$x\longmapsto\sqrt{x}$$is a concave function$$\mathbb E[\sqrt{S}]=\mathbb E[\sqrt{S^2}]<\mathbb E[S^2]^{1/2}$$ Commented Oct 30, 2020 at 11:07
• I think you need to do two things (a) edit your question to show the steps you are following and where you are stuck understanding them (b) add the self-study tag and read its wiki entry. Commented Oct 30, 2020 at 16:20

$$E[s]=E\left[\sqrt{s^2}\right]$$
Now introduce some fractions that are equal to 1 and don't make a difference $$E\left[\sqrt{s^2}\right]=E\left[\sqrt\frac{n-1}{n-1}\times\sqrt\frac{\sigma^2}{\sigma^2}\times\sqrt{s^2}\right]$$ Now, the $$\sqrt{\sigma^2}$$ in the numerator and the $$\sqrt{(n-1)}$$ in the denominator are constants and can come out of the expectation $$E\left[\sqrt\frac{n-1}{n-1}\times\sqrt\frac{\sigma^2}{\sigma^2} \times\sqrt{s^2}\right]=\sqrt{\frac{\sigma^2}{n-1}}E\left[\sqrt\frac{n-1}{1}\times\sqrt\frac{1}{\sigma^2}\times\sqrt{s^2}\right]=\sqrt{\frac{\sigma^2}{n-1}}E\left[\sqrt{\frac{n-1}{\sigma^2}\times s^2}\right]$$ which was to be shown.