# Prove that distribution of sample median (for even sample) is symmetric

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?

• 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?
– whuber
Jan 29 '14 at 21:42
• 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) Jan 29 '14 at 22:27
• 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 $a\lt b$.
– whuber
Jan 29 '14 at 22:35
• 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 $F_M(-t)=1-F_M(t)$.
– Zen
Jan 29 '14 at 23:04
• The statement is also true for an odd sample size (assuming the original distribution is symmetric about 0). 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.

• 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? Mar 1 '17 at 0:56
• @n.e., no, this is just the definition of $f^{-1}(D)$.
– whuber
Mar 1 '17 at 1:22