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
$$ \tilde \sigma = y\frac{1}{1-1/n}\sqrt{\pi/2}$$
Is it unbiased?
 A: The proposed estimator is not unbiased, at least if we indeed know the true mean, $\mu$, and if we are dealing with a normal sample as the title says, where the distribution is symmetric and unimodal and the mean equals the median. Informally, knowing the true mean, makes the mean absolute deviation equal in value to the probability limit of the same expression with $\bar X$ instead of $\mu$. We have
$$ y=\frac{1}{n} \sum_{i=1}^n |X_i - \mu| = \frac{1}{n}\left[\sum_{X_i\geq \mu} (X_i - \mu)+\sum_{X_j< \mu} (\mu - X_j)\right]$$
Denote $m_1$ the count for the first sum, and $m_2$ the count for the second sum (both are random variables). Then, using also Wald's equation
$$E(y) = \frac{1}{n}\Big[E(m_1)E(X\mid X\geq \mu) - E(m_1)\mu + E(m_2)\mu - E(m_2)E(X\mid X\leq \mu)\Big]$$
Since we have the true mean, which is equal to the median, we get $E(m_1)=E(m_2) = n/2$, so the two middle terms cancel, while substituitng for the expected values of the counts, taking common factors and simplifying we arrive at
$$E(y) = \frac{1}{2}\Big[E(X\mid X\geq \mu)  - E(X\mid X\leq \mu)\Big]$$
For the truncated normal distribution, these expected values are 
$$E(X\mid X\geq \mu) = \mu + \sigma \frac{\phi(0)}{1-\Phi(0)} = \mu +\sigma\sqrt{2/\pi}$$
$$E(X\mid X\leq \mu) = \mu - \sigma \frac{\phi(0)}{\Phi(0)} = \mu -\sigma\sqrt{2/\pi}$$
So
$$E(y)=\frac{1}{2}\Big[\mu +\sigma\sqrt{2/\pi}  - \mu +\sigma\sqrt{2/\pi}\Big] = \sigma\sqrt{2/\pi}$$
So the correction factor in $\tilde \sigma$ should be $\sqrt{\pi/2}$ only, for it to be unbiased.
I note that since $X_i - \bar X = (1-1/n)X_i - (1/n)\sum_{j\neq i}X_j$ one suspects that we should examine the case where we do not know $\mu$ and we use the sample mean instead, which I may find the time to do later.
