Expectation of Median of Absolute Random Variables Let $X_1, X_2,..., X_n$ from $N(0,\sigma^2)$.
What I want to get is not $E(median(|X|))$ , but $E(median(|X_1|,|X_2|,...,|X_n|))$

Reason why I need it is because I'm studying LOWESS, and in using bisquare wieght function, weight of residuals within the 6m(m is a median) will be 1, elsewhere its weight would be 0. And the reason why we took 6m is because, m=$\frac{2}{3}\sigma$, 6m=4$\sigma$. As $P(|Z|\ge4\sigma)$ is almost zero, weights are corresponding. 
So I need a conclusion that $E(median(|X|))=\frac{2}{3}\sigma$, which I can't derive. 
=============Edit================
There is a huge error in denoting $E(median(|X|))$. As I meant it by the expectation of the median of $|X_1|,|X_2|,...,|X_n|$, It should have been $E(median(|X_1|,|X_2|,...,|X_n|))$. Sorry for the confusion.
And I will also edit the title from 'Expectation of Absolute median' to 'Expectation of Median of Absolute Random Variables' (Please recommend other titles) as soon as we have the conclusion. Thank you so much for the interest. 
 A: The OP's question is somewhat confused.
But, I think I have been able to figure out what he is asking.
We are given a parent random variable $Z \sim N(0,\sigma^2)$. The pdf of $X =|Z|$ is a half-Normal with pdf say $f(x)$:

which appears thus, as parameter $\sigma$ changes:

The cdf $P(X<x)$ is:

where I am using the mathStatica Prob function to automate.
The population median is the value of $x$ such that  $P(X<x) = \frac12$, which yields:

The OP asks to show that something is the same as $\frac23 \sigma$. The median is NOT equal to $\frac23 \sigma$, but $\frac23 \sigma$ is a rather good simple approximation of the correct solution which we have just derived. To see this, the following diagram compares:


*

*the EXACT median, plotted as a function of $\sigma$ (blue curve), and the

*APPROXIMATE median $\frac23 \sigma$, plotted as a function of $\sigma$ (orange curve)



It's a nice fit  - but it is not the population median. Also note that it is a constant (a function of parameter $\sigma$) ... so there is no such meaningful thing as $E[median[X]]$ in this context.
