My book says that sample median of a normal distribution is an unbiased estimator of its mean, by virtue of the symmetry of normal distribution. Please advice how can this be proved.

  • $\begingroup$ There's got to be a short proof based on symmetry. $\endgroup$ Mar 12 '12 at 22:17

Let $Z_i$, $1 \leqslant i \leqslant n$ be independent identically distributed normal variables with mean $\mu$ and variance $\sigma^2$, and let $Z_{k:n}$ denote $k$-th order statistics.

We separately consider the case of even $n$ and odd $n$.

Let $n$ be odd, i.e. $n = 2m+1$. Then the sample median corresponds to $M = Z_{m+1:2m+1}$. The probability density of this order statistics is: $$ f_{M}(x) = (m+1) \binom{2m+1}{m} f_X(x) \left( F_X(x) (1-F_X(x)) \right)^m $$ Since $F_X(x) = 1-F_X(2\mu-x)$, we clearly get $f_M(x) = f_M(2\mu -x)$ by symmetry, and therefore $$ \mathbb{E}(M) = \mathbb{E}(2 \mu -M) \implies \mathbb{E}(M) = \mu $$

Now consider the case of even $n$, i.e. $n = 2m$. Then the sample median corresponds to $M = \frac{1}{2} \left( Z_{m:2m} + Z_{m+1:2m} \right)$. The joint probability density is: $$ f_{Z_{m:2m}, Z_{m+1:2m}}(x_1,x_2) = m^2 \binom{2m}{m}f_X(x_1) f_X(x_2) \left(F_X(x_1) (1-F_X(x_2))\right) ^{m-1} [ x_1 \leqslant x_2 ] $$ Clearly, again $f_{Z_{m:2m}, Z_{m+1:2m}}(x_1,x_2)=f_{Z_{m:2m}, Z_{m+1:2m}}(2\mu - x_2,2 \mu - x_1)$ by symmetry, therefore $$ \mathbb{E}(M) = \mathbb{E}\left( \frac{ Z_{m:2m} + Z_{m+1:2m}}{2} \right) = \mathbb{E}\left( \frac{ (2\mu-Z_{m+1:2m}) + (2\mu - Z_{m:2m})}{2} \right) = \mathbb{E}(2\mu - M) $$ This again implies that $\mathbb{E}(M) = \mu$ as a consequence of the symmetry.

Added: The normality assumption was not used in the above demonstration, thus the proof holds for any continuous random variable with symmetric probability density and finite mean.

  • 2
    $\begingroup$ It does not appear that normality is needed in this proof, just the symmetry of the density function about $\mu$ and the requirement that the mean exist (and hence be equal to $\mu$). Or am I missing something? $\endgroup$ Mar 12 '12 at 21:16
  • $\begingroup$ @DilipSarwate Thanks for the comment. Indeed, normality assumption was not used. I will edit the post to reflect that. $\endgroup$
    – Sasha
    Mar 12 '12 at 21:23
  • $\begingroup$ You don't need normality to prove this. Definitely it's a weaker assertion than what can be proved by the same methods. $\endgroup$ Mar 12 '12 at 22:19
  • $\begingroup$ @Sasha: Thanks a lot ! $\endgroup$
    – preeti
    Mar 13 '12 at 21:46

Let $\mu$ be the population mean (so that's assumed to exist), and assume the distribution is symmetric and there's a density. (So those are weaker assumptions than normality, and maybe the density assumption can be dropped too.) Let $X_1,\ldots,X_n$ be the sample; let $Y_i=X_i-\mu$ for $i=1,\ldots,n$. Let $m=\operatorname{E}(\operatorname{median})=\operatorname{E}(\operatorname{median}(Y_1,\ldots,Y_n))$. By symmetry of the distribution of the $Y$s about $0$, $-m=\operatorname{E}(-{\operatorname{median}})=\operatorname{E}(\operatorname{median})$. Since $m=-m$, we must have $m=0$. Since $\operatorname{E}(\operatorname{median})=0$, we conclude $\operatorname{E}(\operatorname{median}(X_1,\ldots,X_n))$ $=\operatorname{E}(\operatorname{median}(Y_1+\mu,\ldots,Y_n+\mu))$ $=\operatorname{E}(\mu + \operatorname{median}(Y_1,\ldots,Y_n))=\mu$.

  • 6
    $\begingroup$ The density assumption can also be dropped: symmetry can be expressed as $P(Y_i \le c) = P(Y_i \ge -c)$ for all real $c$. If $Z_i = -Y_i$, $Z_i$ has the same distribution as $Y_i$ (and of course $Z_1,\ldots,Z_n$ are independent) so $m = E[\text{median}(Y_1,\ldots,Y_n)] = E[\text{median}(Z_1,\ldots,Z_n)] = E[-\text{median}(Y_1,\ldots,Y_n)] = -m$. $\endgroup$ Mar 12 '12 at 23:12
  • $\begingroup$ @MichaelHardy: Thanks a lot. $\endgroup$
    – preeti
    Mar 13 '12 at 21:45

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