# variance of sample median of normal distribution

Does anyone tell me how to find the variance of sample median when random samples are normally distributed?? I know that I need to use the order statistics, but actually I don't understand how to apply it.

• 1. Is this for coursework? Can you give more context about how the issue arises? 2. What have you learned about order statistics? See How to Ask, in relation to search and research, and see help center in relation to homework-style questions. 3. Are you just after a formula? Commented Sep 16, 2017 at 8:18
• stats.stackexchange.com/questions/41557/… and search this site for "median varianve" (a lot of posts) See also jstor.org/stable/2286545?seq=1#page_scan_tab_contents Commented Sep 16, 2017 at 14:42
• Related question seeking moments of sample median with Normal parent: stats.stackexchange.com/questions/303160 Commented Sep 16, 2017 at 16:25
• Can you please qualify the question: standard Normal or general $N(\mu, \sigma^2)$ parent? Commented Sep 16, 2017 at 16:27
• You can find thorough explanations from various viewpoints at stats.stackexchange.com/questions/45124.
– whuber
Commented Sep 17, 2017 at 15:39

I will make some illustrations with R. First, using results discussed at Central limit theorem for sample medians the asymptotic distribution of sample median of $N$ iid observations from a normal distribution is itself a normal distribution centered at the theoretical median $\mu$ and with variance $\frac{2\pi \sigma^2}{4 N}$ where $\sigma^2$ is the variance of the parent distribution. The sample median is the central order statistic, and in the case $N=2s+1$ is odd, it is the order statistic of order $s+1$, which have a distribution we can calculate exactly rather easily (theory given in the cited post and elsewhere at this site). I will use some R function I detail at the end of this post to compare the exact and approximate distributions of the sample median. We use the standard normal for parent distribution.

First with sample size $N=11$:

The approximation is already quite good! Let us look at the case $N=101$:

Here the approximation is virtually exact. Some R functions for distributions of order statistics:

porder  <-  function(x, N=1, r=1, basename="norm", ..., log.p=FALSE) {
pfun  <-  get(paste("p", basename, sep=""))
stopifnot(r  <=  N)
prob  <-  pfun(x, ...)
retval  <-  pbeta(prob, r, N-r+1, log.p=log.p)
retval
}

dorder <-  function(x, N=1, r=1, basename="norm", ..., log=FALSE)  {
pfun  <-  get(paste("p", basename, sep=""))
dfun  <-  get(paste("d", basename, sep=""))
stopifnot(r <= N)
logdens  <-  -lbeta(r, N-r+1)+(r-1)*pfun(x, ..., log.p=TRUE) +(N-r)*pfun(x, ..., lower.tail=FALSE, log.p=TRUE) + dfun(x, ..., log=TRUE)
retval  <-  if (log) logdens else exp(logdens)
retval
}

qorder  <-  function(p, N=1, r=1, basename="norm",  ..., log.p=FALSE) {
pfun  <-  get(paste("p", basename, sep=""))
qfun  <-  get(paste("q", basename, sep=""))
stopifnot( r <= N)
retval  <-  qfun(qbeta( p, r, N-r+1, log.p=log.p), ..., log.p=FALSE)
retval
}

rorder  <-  function(n, N=1, r=1, basename="norm", ...)  {
qfun  <-  get(paste("q", basename, sep=""))
qfun( rbeta(n, r, N-r+1), ... )
}