Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm hoping that someone could help better explain why $\sigma$ (equation 2.19) must be multiplied by $\frac{4n}{4n + 1}$. Obviously all the math is there. Perhaps someone can make this easier to understand.

Reference is here:


share|improve this question
up vote 5 down vote accepted

Before answering our question, I would like to remind you of the following definition:

Unbiased Estimator: $\hat{\theta}$ is an unbiased estimator of $\theta$ if $\mathbb{E}[\hat{\theta}]=\theta$.

Let's assume having a sample $\{x_1, x_2,..., x_n\}$, and you want an Unbiased Estimator of it's variance $\sigma^2$, then the following estimator satisfies this property :

$\hat{\sigma^2} = \frac{1}{n-1}\sum_{i=1}^{n}(x_i - \hat{\bar{x}})^2$, where $\hat{\bar{x}}=\frac{1}{n}\sum_{i=1}^{n}x_i$ (which is also an unbiased estimator of the mean $\bar{x}$).

But if you want an unbiased estimator of the standard deviation $\sigma$, then you cannot just take the square root of $\hat{\sigma^2}$, because due to the Jensen's Inequality applied to the case of a concave function $x \rightarrow \phi(x)$ (which the case of $x \rightarrow \sqrt x$) taking the square root of the previous estimator leads to an biased estimator of the standard deviation.

But fortunately, an application of Cochran's Theorem shows that:

if we denote $s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i - \hat{\bar{x}})^2}$ then $\sqrt{n-1}\frac{s}{\sigma}$ has a $\chi_{n-1}$ distribution. Which leads to :

$\mathbb{E}[s]= (\sqrt{\frac{2}{n-1}}\frac{\Gamma(\frac{n}{2})}{\Gamma(\frac{n-1}{2})}) \sigma = (1 + \frac{1}{4n} + O(\frac{1}{n^2})) \sigma = (\frac{4n+1}{4n} + O(n^{-2}))\sigma\approx\frac{4n+1}{4n}\sigma$

Hence if you want to get an unbiased estimator of the standard deviation (to a certain approximation) you need to use:

$s' = \frac{4n}{4n+1} s$ and thus $$\mathbb{E}[s']= \frac{4n}{4n+1} \mathbb{E}[s]=(\frac{4n}{4n+1})(\frac{4n+1}{4n})\sigma = \sigma$$

I hope this will help!

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.