Follow on question to this, answered negatively by Thomas Lumley. We reprint it here for convenience.

In this SE question, it is stated that there is a central limit theorem for the sample median, namely

$$ \sqrt{n}(Y_n - m) \xrightarrow{d} N(0, [2f(m)]^{-2}), $$

as $n\to\infty$ where

  • $Y_n$ is the sample median from $n$ iid samples,
  • $m$ is the population median,
  • $f$ is the PDF (assumed to exist) of the distribution we're sampling from.

If I'm not mistaken, this result holds even if the original distribution doesn't have finite variance (e.g. the Cauchy distribution).

Is it necessarily true that the variances converge? I.e that

$$ nE[(Y_n-m)^2] \to [2f(m)]^{-2}? $$

As answered in the linked question, the answer is no. Let $F$ be the CDF of the distribution. First, note that the PDF of the sample median is

$$ f_{Y_{2k+1}}(x) = (2k+1) \binom{2k}{k} F(x)^k (1-F(x))^k f(x). $$

Thomas Lumley's counterexample is a distribution whose CDF $F$ looks like the following.

$$ F(x) \overset{x\to -\infty}{\sim} \frac{1}{\log |x|} \\ 1-F(x) \overset{x\to \infty}{\sim} \frac{1}{\log |x|}. $$

In this case, the second moment of $Y_{2k+1}$ looks like the following.

$$ \begin{align*} E[Y_{2k+1}^2] &= \int_{-\infty}^\infty x^2f_{Y_{2k+1}}(x) \mathrm{d}x \\ &= C\int_{-\infty}^\infty x^2 F(x)^k (1-F(x))^k f(x) \mathrm{d}x \\ &= C\int_{-\infty}^\infty x^2 F(x)^k (1-F(x))^k f(x) \mathrm{d}x. \end{align*} $$

this will be infinite for all $k$, from which the negative result follows.

My follow up question is: if we add the extra constraint that the CDF $F$ satisfies

$$ F(x) \overset{x\to-\infty}{=} o(|x|^{-\epsilon}) \\ 1-F(x) \overset{x\to\infty}{=} o(|x|^{-\epsilon}) \\ $$

for some $\epsilon>0$ (that is, the CDF approaches infinity polynomialy), is it now true that the variances converge?

As a particular case, is true for a Cauchy distribution?

  • $\begingroup$ I will answer trying not to repeat what has already been said in 'part 1', to which you have linked. $\endgroup$ – BruceET Apr 17 at 20:55

This is sometimes called the 'CLT for medians'. Similar, limiting distributions hold for quantiles other than the median (but not for the maximum or minimum).

As you say, the condition for the CLT for means, is that the population have finite variance.

By contrast, the key condition that assures convergence of sample medians to normal is that the value of the population density at the median $\eta$ be positive. A standard Cauchy distribution (also called Student's t distribution with 1 degree of freedom) has $f(\eta) = 1/\pi > 0,$ so that the current theorem holds for standard Cauchy data.

In particular, we might expect samples of size $n=100$ from $\mathsf{T}(\nu=1)$ to have medians $H \stackrel{aprx}{\sim}\mathsf{Norm}(0, \frac{1}{20\sqrt{\pi}}).$ This is illustrated by the following simulation in R:

h = replicate(10^5, median(rt(100,1)))
mean(h);  sd(h)
[1] 0.0001220243   # aprx 0
[1] 0.1588548      # sptc 2.57078
[1] 0.1570796      # uses limiting SE

hist(h, prob=T, br=30, col="skyblue2")
curve(dnorm(x, 0, 1/(20*dt(0,1))), lwd=2, col="orange", add=T)

enter image description here

  • 1
    $\begingroup$ This does not answer whether the variances converge. $\endgroup$ – extremeaxe5 May 3 at 20:13
  • $\begingroup$ This Q&A format works best for self-contained rather than serial questions. See previous comment. As in the link, variances converge to 'asymptotic variance' shown for normal distribution. And Yes, even for Caucny, which has non-zero pdf at its median, if you pay attention to the conditions. $\endgroup$ – BruceET May 3 at 21:17
  • $\begingroup$ Where does it say variances converge to the asymptotic variance? As stated in the question, this is not implied by convergence in distribution to a normal distribution. Also, as stated in the question variances do not necessarily converge to the asymptotic variance, as shown by the counterexample. Also, this question is self contained. I could remove the link, but doesn't it make more sense to keep it for context? $\endgroup$ – extremeaxe5 May 4 at 0:15

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