In the case of lognormal distribution, when can the median be more efficiently estimated than the mean?

Question

I understand, that in the case of normal distribution, the estimation of the mean (from samples) is more efficient (i.e. of less risk), than the estimation of the median. According e.g. to this post, the median is 64% efficient as the mean.

How about a lognormal distribution with parameters $\mu$ and $\sigma$? The population mean is then

$a = e^{(\mu + \frac12 \sigma^2)}$

and the median

$m=e^\mu$ .

To be specific, let $x_1,x_2,…, x_n$ be a sequence of i.i.d. variables drawn from $X$ ~ $logN(\mu, \sigma)$. Define $a_n = \frac1n ∑x_i$, and $m_n$ to be the median of ${x_1,x_2,…x_n}$. Which varies less around its respective true value: $a_n$ around $a$ or $m_n$ around $m$? Does the answer for the question depend on the sample size?

My guess is, that the answer depends on the value of $\sigma$. To go further, my simulations show, that for $\sigma$ < 0.5, the mean, for $\sigma$ > 0.5 the median is more efficient. Is there a proof for that?

n <- 10^6 #population size
mu.v <- 10^seq(from=-1 ,to=2,by=.5) #different values of mu
sig.v <- 10^seq(from=-2 ,to=1,by=.5) #different values of sigma

ss <- 10 #sample size

ar0 <- array(data=rep(NaN, length(mu.v)*length(sig.v)*2),
             dim=c(length(mu.v),length(sig.v), 2),
             dimnames=list(paste("mu=",mu.v),paste("sig=",sig.v),c("mean","medi"))); 
for (jmu in 1:length(mu.v)){
  for (jsig in 1:length(sig.v)){
    y <- exp(rnorm(n, mu.v[jmu],sig.v[jsig])) #generate random population

    nr <- 1e2;  #num of runs
    outtab <- matrix(data=NA,nrow=nr,ncol=2,dimnames=list(c(),c("dev_mean","dev_medi")))
    for (jr in 1:nr) { #loop runs
      y1 <- sample(y, size=ss,replace=TRUE)
      outtab[jr,] <- c((mean(y1)-mean(y))^2 ,(median(y1)-median(y))^2)
    }

    ar0[jmu,jsig,1] <- (mean(outtab[,1]))^.5
    ar0[jmu,jsig,2] <- (mean(outtab[,2]))^.5
  }

write.csv(ar0[,,1],"mean.csv")
write.csv(ar0[,,2],"medi.csv")
}

Answer 1

If the lognormal distribution has parameters $\mu$ and $\sigma$, with pdf $f$, then

\begin{align} Var[a_n] & =\frac{Var[X]}{n}=(e^{2\sigma^2}-e^{\sigma^2})\frac{e^{2\mu}}{n}\\ \\ Var[m_n] &\sim\frac{1}{4nf(m)^2}=\frac{\pi \sigma^2}{2}\frac{e^{2\mu}}{n}\\ \end{align} (The variance of the mean comes from the central limit theorem; the variance of the median comes from Laplace's asymptotic formula for the sampling distribution of the median.)

For large $n$ the mean has lower variance when $Var[a_n]<Var[m_n]$, i.e. when $$2(e^{2\sigma^2}-e^{\sigma^2}) < \pi \sigma^2$$

So the mean has lower variance if $\sigma<0.546$ and the median has lower variance if $\sigma>0.546$.