Unbiased estimator of standard deviation

Let $X$ be a random variable of distribution $N(\mu , \sigma ^2)$, where we know the value of $\mu$ and we don't know the value of $\sigma$.

My task is to choose number $d$, such that random variable $Y = d \sum_{i=1}^{n} |x_i - \mu|$ was unbiased estimator of $\sigma$.

So my attempt:

We want $E(Y) = \sigma$.

$E(Y) = d E(\sum_{i=1}^{n} |x_i - \mu|) = d E( \sum_{i=1}^{n} \sqrt{(x_i - \mu)^2})$

... And what can I do next? It's not like the estimator with the coefficient $c_4$, because we have sum of square roots and not the square root of the sum... What steps should I take? How to evaluate this expression? Please, help.

Hint: you know the density of a normal $Z\sim\mathcal{N}(0,\sigma^2)$ right? Call it $f_Z$. Then $$\mathrm{E}[|Z|]=\int_{-\infty}^{+\infty}|z| \cdot f_z(z) dz = 2 \int_{0}^{+\infty} z\cdot f_z(z)dz.$$ Now compute the integral.
• In what you say, $Z= |x_i - \mu|$, not the sum of them, right? – Anne Nov 13 '13 at 21:06
• @Anne: yes but the Z_s are independent right? I am assuming your $x_i$ is a random sample. In this case you just have to multiply $\mathrm{E}[|Z|]$ by your sample size. Your expression will be $d\times n \times \mathrm{E}[|Z|]$ – Sergio Parreiras Nov 13 '13 at 21:14
• But are you sure that $|x_i - \mu|$ is a random variable of distribution $N(0, \sigma ^2)$? I guess $x_i - \mu$ would be of that distribution, but I think that taking mudule changes something... – Anne Nov 13 '13 at 21:21
• @Anne : I never said that $|x_i-\mu|$ is $\mathcal{N}(0,\sigma^2)$. I said $Z=x_i-\mu$ was. Notice in my computations you have $|Z|$ and not Z. – Sergio Parreiras Nov 13 '13 at 21:39