The Exact Standard Deviation of the Sample Median

In a previous post, I gave the well-known approximation to the standard deviation of the sample median

$$\sigma \approx {1 \over 2\sqrt{n}\,f(x_m)}$$

where $f(x)$ is the probability density function and $x_m$ is the median (see Laplace and Kenney and Keeping).  Here are some examples.

Distribution Median Approx StD of Median
Standard Gaussain mean 0 std 1 0 $\sqrt{\pi \over{2 n}}$
Uniform 0 to 1 1/2 $1\over{2\sqrt{n}}$
Logistic with mean 0 and shape $\beta$ 0 ${2\beta}\over{\sqrt{n}}$
Student T with mean 0 and $\nu$ deg free 0 $\frac{\sqrt{\nu }\  B\left(\frac{\nu }{2},\frac{1}{2}\right)}{2 \sqrt{n}}$

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Computing the exact standard deviation of the sample median is more difficult. You first need to find the probability density function of the sample median which is

$$f_m(x) = g(c(x)) f(x)$$

where

$$g(x) = \frac{(1-x)^{\frac{n-1}{2}}
x^{\frac{n-1}{2}}}{B\left(\frac{n+1}{2},\frac{n+1}{2}\right)},$$
$B$ is the beta function, $c(x)$ is the cumulative distribution function of the sample distribution, and $f(x)$ is the probability density function of the sample distribution.

Now the expected value of the sample median is

$$\mu_m = \int x f_m(x) dx$$

and the standard deviation of the sample median is

$$\sigma_m = \sqrt{\int (x-\mu_m)^2 f_m(x)\ dx}. $$

 

Generally speaking, these integrals are hard, but they are fairly simple for the uniform distribution.  If the sample distribution is uniform between 0 and 1, then

$f(x) = 1,$

$c(x) = x,$

$g(x) = \frac{(1-x)^{\frac{n-1}{2}}
x^{\frac{n-1}{2}}}{B\left(\frac{n+1}{2},\frac{n+1}{2}\right)},$

$f_m(x) = g(x),$

$\mu_m = \int x g(x) dx \ =\ 1/2,$ and

$$\sigma_m = \sqrt{\int (x-\mu_m)^2 f_m(x)\ dx}\  = \ {1\over{2\sqrt{n+2}}} $$

which is close to the approximation given in the table.

 

(Technical note:  The formulas above only apply for odd values of $n$ and continuous sample probability distributions.)

If you want the standard deviation for the sample median of a particular distribution and a $n$, then you can use numerical integration to get the answer. If you like, I could compute it for you.  Just leave a comment indicating the distribution and $n$.