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? – Glen_b Sep 16 '17 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 – kjetil b halvorsen Sep 16 '17 at 14:42
• Related question seeking moments of sample median with Normal parent: stats.stackexchange.com/questions/303160 – wolfies Sep 16 '17 at 16:25
• Can you please qualify the question: standard Normal or general $N(\mu, \sigma^2)$ parent? – wolfies Sep 16 '17 at 16:27
• Glen_b : Yes. It is about the asymptotic relative efficiency. The question compare the variance of sample mean and sample median, and these samples are normally distributed. I can derive the variance of sample mean easily, but it is hard to derive the variance of sample median, and my lecture note does not clarify how to derive it. – Si Hyun Kim Sep 17 '17 at 3:26

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), ... )
}