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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

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

Preliminary observations:

  • This cannot be deduced directly from convergence in distribution---if we take $Y_n$ and move $\frac1n$ of its probability to $n$, then convergence in distribution will still hold but the variances won't converge.
  • The first few terms in the limit above might not exist. For the Cauchy distribution, for $n=1,2,3$ (and maybe more) the sample median doesn't have finite variance.
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  • $\begingroup$ Let $X$ be the random variable and apply the CLT to $X^2.$ $\endgroup$
    – whuber
    Apr 11 at 17:47
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For simplicity, assume $m=0$ and odd $n=2r+1$. The pdf of $Y_n$ is

$$f_n(y)={n\choose r} f(y) F(y)^r(1-F(y))^r$$ and we are interested in whether $\int y^2f_n(y)\,dy$ exists.

We know $F(y)\to 0$ as $y\to-\infty$ and $1-F(y)\to 0$ as $y\to\infty$. If $F$ approaches 0/1 at some polynomial rate, then some power $r$ will approach faster than $y^{-2}$, cancelling the $y^2$ so the variances will at least exist for all large enough $n$. This includes the Cauchy case.

However, if the CDF of $X$ looked like $1-1/(\log |x|)$ for large $x$ and $1/(\log|x|)$ for large negative $x$, I don't think the variance will exist. The variance integrand will be like $f(y)y^2\log^{-r}|y|$ for large negative $y$, which will not be $o(|y|^{-1})$ and so will not have a finite integral (and similarly in the positive direction).

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    $\begingroup$ Could you supply some rationale for why you don't think the variance will exist? BTW (it makes no difference to the argument) the coefficient in $f_n(y)$ should be $\binom{n+1}{r;1;r}=\binom{n}{r}/(n+1).$ $\endgroup$
    – whuber
    Apr 11 at 17:48

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