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.
Prove that the sample median is an unbiased estimator
-
0There's got to be a short proof based on symmetry. – 2012-03-12
2 Answers
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.
-
2It 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? – 2012-03-12
-
0@DilipSarwate Thanks for the comment. Indeed, normality assumption was not used. I will edit the post to reflect that. – 2012-03-12
-
0You don't need normality to prove this. Definitely it's a weaker assertion than what can be proved by the same methods. – 2012-03-12
-
0@Sasha: Thanks a lot ! – 2012-03-13
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 $\operatorname{E}(\operatorname{median}(X_1,\ldots,X_n))$ $=\operatorname{E}(\operatorname{median}(Y_1+\mu,\ldots,Y_n+\mu))$ $=\operatorname{E}(\mu + \operatorname{median}(Y_1,\ldots,Y_n))=\mu$.
-
5The 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 $m = E[\text{median}(Y_1,\ldots,Y_n)] = E[\text{median}(Z_1,\ldots,Z_n)] = E[-\text{median}(Y_1,\ldots,Y_n)] = -m$. – 2012-03-12
-
0@MichaelHardy: Thanks a lot. – 2012-03-13