# Is this an unbiased estimator for standard deviation of normal distribution?

Suppose we have $n$ samples, with mean $\mu$.

Calculate the average absolute distance from $\mu$, i.e., $$y = \frac{1}{n} \sum_{i=1}^n |X_i - \mu| \>.$$

Then, take as an estimate of the standard deviation $$\tilde \sigma = \frac{1}{\sqrt{2\pi (1-1/n)}} y \>.$$

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So, presumably $\mu$ is known a priori? –  cardinal Jul 30 '12 at 13:55
Hint: Can you find the expectation $\mathbb E |X - \mu|$? It is a scaled version of $\sigma$. Find the scale factor. –  cardinal Jul 30 '12 at 13:57
I think I proved it was unbiased estimator. E|X-u| = standard deviation * sqrt(2/pi) –  steviekm3 Jul 30 '12 at 14:04
I was wondering though why we don't see this formula that often for estimating standard deviation? –  steviekm3 Jul 30 '12 at 14:04
Good start, stevie. You have the scale factor correct, but not the conclusion regarding the unbiasedness. Your follow-up question is a good one, with a fairly deep answer. You might consider adding it to the question to make it more prominent. :) –  cardinal Jul 30 '12 at 14:32