show that the sample median of normal distribution is median unbiased

Question

$X_i \sim N(\mu,\sigma^2)$, $i = 1,2,\ldots,n$. Show that the sample median is unbiased in median for $\mu$.

I have obtained the pdf of sample median for $n=2m+1$ as:

$$f(x_{(m+1)}) = \frac {(2m+1)!} {m!m!} \Phi\left( \frac{x-\mu}{\sigma} \right) \left(1-\Phi\left( \frac{x-\mu}{\sigma}\right)\right)^m \cdot \phi\left(\frac{x-\mu}{\sigma} \right)$$

How to show that the median of this distribution of sample median is $\mu$?

Please look at my edits. One need not write (x-$\mu$)/$\sigma$; one can write $(x-\mu)/\sigma$. It is absurd to alternate continually in and out of MathJax the way you did. — Michael Hardy, May 20 '14 at 17:18

score 1 · Answer 1 · answered May 20 '14 at 17:58

Forget the exact form of the distributions and use only their symmetry.

Recall that if $X$ is normal $(\mu,\sigma^2)$ then $X=\mu+Y$ where $Y$ is normal $(0,\sigma^2)$, in particular the distributions of $Y$ and $-Y$ coincide.

The sample median $M$ for $(X_i)_{1\leqslant i\leqslant2m+1}$ is $M=\mu+K$ where $K$ is the sample median for $(Y_i)_{1\leqslant i\leqslant2m+1}$ and $Y_i=X_i-\mu$ for every $i$.

Now, $(-Y_i)_{1\leqslant i\leqslant2m+1}$ is distributed like $(Y_i)_{1\leqslant i\leqslant2m+1}$ and has median $-K$ hence the distributions of $K$ and $-K$ coincide.

In particular $E(K)=0$ hence $E(M)=\mu$. Furthermore, the distribution of $M$ is symmetric with respect to $\mu$ hence the median of this distribution is $\mu$.

show that the sample median of normal distribution is median unbiased

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