Let $\mathbf{X}=(X_1,X_2,\ldots,X_{2k})$ be a sample from normal distribution $\mathcal{N}(0,1)$.

Prove that distribution of sample median (for even sample) is symmetric around 0.

$me(\mathbf{X}) = \frac{X_{k:2k}+X_{(k+1):2k}}{2}$, where $X_{i:2k}$ is $i$-th order statistic.

I was trying to use formula $f_{X_{k:n}}(x) =\frac{n!}{(k-1)!(n-k)!}[F_X(x)]^{k-1}[1-F_X(x)]^{n-k} f_X(x)$, but without any success (maybe there is a way to do it differently than finding $me(\mathbf{X})$ distribution?

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    $\begingroup$ This is one case where proving a generalization of the statement might be easier than proving a special case, as well as giving more insight: can you show that the conclusion is true when the normal distribution is replaced by any symmetric distribution? $\endgroup$
    – whuber
    Jan 29 '14 at 21:42
  • $\begingroup$ Well the problem is that I have a problem with that as well - I was trying to use formula for joint distribution $(X_{k:2k}, X_{(k+1):2k})$, then use $P\{X+Y\leq \alpha\} = \int_{-\infty}^\infty \int_{v=-\infty}^{v=\alpha - u}f_{X,Y}(u, v)\,\mathrm dv\,\mathrm du.$ but I have no idea how to compute that (I will try to figure something out) $\endgroup$
    – whoiswho
    Jan 29 '14 at 22:27
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    $\begingroup$ Go back to basics on this one. Suppose $E$ is the set of $\mathbf{X}$ for which the median lies in some small interval $[x,x+dx]$. Let $-E$ be the set $\{-\mathbf{X}\ |\ \mathbf{X}\in E\}.$ (1) What can you say about the medians of the datasets in $E$? (2) What is the relationship between the probabilities of $E$ and $-E$? (3) Note that a distribution $F$ is symmetric if and only if ${\Pr}_F((a,b])={\Pr}_F([-b,-a))$ for all $a\lt b$. $\endgroup$
    – whuber
    Jan 29 '14 at 22:35
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    $\begingroup$ For the median $M:=(X_{(k)}+X_{(k+1)})/2$, your goal is to prove that $P(M\leq -t)=P(M>t)$, for every $t\in\mathbb{R}$, which is equivalent, in terms of the distribution function of $M$, to $F_M(-t)=1-F_M(t)$. $\endgroup$
    – Zen
    Jan 29 '14 at 23:04
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    $\begingroup$ The statement is also true for an odd sample size (assuming the original distribution is symmetric about 0). $\endgroup$
    – Henry
    Jan 29 '14 at 23:56

To those comfortable with the mathematics of probability and random variables the following argument is a one-liner, because you will take for granted most of the manipulation (and will find it obvious). To cover all bases, though, I provide the details.

The setting--and a generalization

Let $X = (X_1, X_2, \ldots, X_n)$ be a vector-valued random variable with a distribution $F$ that is "symmetric" about the origin in the sense that for all events $E\subset\mathbb{R}^n,$

$${\Pr}_F(E) = {\Pr}_F(-E).$$

The notation "$-E$" refers to $\{-x\ |\ x \in E\}$. No assumptions are made about the parity of $n$ or about independence of the components $X_i$. Note that the conditions of the problem--viz, $n$ even and $X_i$ iid Normal$(0,1)$--are a special case.

Observe--because this is the crux of the matter--that median$(-X)$ = $-$median$(X)$ no matter what $X$ may be. (For this to be true, it is essential that we define the median of an even number of elements to be the average of their two middle values.)

The one-line proof

The distribution of the median is symmetric because the distribution of $X$ is symmetric and, as we just observed, the median commutes with the symmetry operation $X \to -X,$ QED.

The details

We are asked to show that $f(X)$ = median$(X)$ has a symmetric distribution. To this end, let $D\subset\mathbb{R}$ be measurable. The chance that $f(X)$ lies in $D$ is--by definition--given by

$${\Pr}_F(f(X)\in D) = {\Pr}_F(X\in f^{-1}(D)) = {\Pr}_F(f^{-1}(D))$$

The first equality is valid because $f$ is a measurable function. To prove symmetry, we need to deduce that ${\Pr}_F(f^{-1}(D)) = {\Pr}_F(f^{-1}(-D)).$

Using the definitions and the key observation (which justifies the third equality below), notice that

$$\eqalign{ f^{-1}(D) &= \{X\in\mathbb{R}^n\ |\ f(X)\in D\} \\ &= \{X\in\mathbb{R}^n\ |\ -f(X)\in -D\} \\ &= \{X\in\mathbb{R}^n\ |\ f(-X)\in -D\} \\ &= \{-X\in\mathbb{R}^n\ |\ f(X)\in -D\} \\ &=-f^{-1}(-D). }$$

The symmetry of $F$, applied to the event $E = f^{-1}(D),$ implies the second equality below:

$${\Pr}_F(f^{-1}(D)) = {\Pr}_F(-f^{-1}(-D)) = {\Pr}_F(f^{-1}(-D)).$$

But the latter is exactly the chance that the median lies in $-D$, proving the median has a symmetric distribution.

  • $\begingroup$ You claim that since $f$ is measurable then ${\Pr}_F(f(X)\in D) = {\Pr}_F(X\in f^{-1}(D))$ but isnt this only true if $f$ is 1-1 and onto? $\endgroup$
    – alpastor
    Mar 1 '17 at 0:56
  • $\begingroup$ @n.e., no, this is just the definition of $f^{-1}(D)$. $\endgroup$
    – whuber
    Mar 1 '17 at 1:22

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