Why is median not a sufficient statistic? Suppose a random sample of $n$ variables from $N(\mu,1)$, $n$ odd. The sample median is $M=X_{(n+1)/2}$, the order statistic of the middle of the distribution.
How to prove that sample median is not a sufficient statistic of $\mu$? Do we need the probability density function of M?
I would like any help.
Note 1: I'm studying for the final exam of my course, and take previous exams to practice. This question appeared in a 2015 exam, that was given by other professor than mine, so it is possible that covers things that wasn't given by my professor. This question includes others two exercises:
(1) Is $M-\bar{X}$ an ancillary statistic for $\mu$?
(2) Prove that if $m$ is a unique point, such as $F_X(m)=1/2$, then $M \overset{p}{\rightarrow}m$.
I couldn't answer these two exercises, because ancillary statistics were not covered by my professor.
So, to see wheter $M$ is a sufficient statistic for $\mu$, I derived the probability density (pdf) function of $M$, that I bring below. But I think there is a more inteligent way to do, without getting the pdf first. So this is the reason I'm asking this question, to see if there is a better and faster way of solving this question.
Note 2: things I know about sufficient statistics:
$T$ is a sufficient statistic if $$ \frac{f_X(x|\theta)}{q(t|\theta)} $$ does not depend on $\theta$, where $f_X$ is the probability density function of $X$ and $q$ is the probability density function of $T$. Also, $T$ is a sufficient statistic if, and only if, there exist functions $g(t|\theta)$ and $h(x)$ such that $$f_X(x|\theta)=g(T(x)|\theta)h(x)$$ for all points of $\theta$
Note 3: The probability density function of M is
$$f_{M}(m)=\frac{n!}{\left ( \frac{n-1}{2} \right) ! \left ( \frac{n-1}{2} \right) !}[1-F_{X}(m)]^{\frac{n-1}{2}}f_{X}(m)[F_X(m)]^{\frac{n-1}{2}}$$
I divided the probability density functions of $(X_1,...,X_n)$ and $M$:
$$\frac{f_{X}(x|\mu)}{f_{M}(m)}=\frac{f_{X1}(x_{1}|\mu)\cdot ...\cdot f_{Xn}(x_{n}|\mu) }{\frac{n!}{\left ( \frac{n-1}{2} \right) ! \left ( \frac{n-1}{2} \right) !}[1-F_{X}(m)]^{\frac{n-1}{2}}f_{X}(m)[F_X(m)]^{\frac{n-1}{2}}}$$
So we have (I don't know if is wright):
$$\frac{f_{X}(x|\mu)}{f_{M}(m)}=\frac{f_{X1}(x_{1}|\mu)\cdot ...\cdot f_{X_{n-1}}(x_{n-1}|\mu) }{\frac{n!}{\left ( \frac{n-1}{2} \right) ! \left ( \frac{n-1}{2} \right) !}[1-F_{X}(m)]^{\frac{n-1}{2}}[F_X(m)]^{\frac{n-1}{2}}}$$
It seems that this expression depends of $\mu.$

One obvious way to check if a statistic is sufficient is to identify a minimal sufficient statistic (if it exists) and check if the minimal sufficient statistic is a function of your proposed statistic. Here we want to use the fact that a minimal sufficient statistic is a function of any sufficient statistic.
Take $n=3$ for example. A minimal sufficient statistic for $\mu$ is the sample mean $\overline x=\frac13(x_1+x_2+x_3)$. But $\overline x$ is not a function of the sample median $x_{(2)}$ only. You have to argue this formally.
In essence, sufficiency is concerned with data reduction (see What does it mean that a statistic $T(X)$ is sufficient for a parameter?). So $\overline x$ is sufficient means that all the information about $\mu$ in the sample can be condensed in $\overline x$. This happens to be the maximum possible data condensation in this model, which makes the sample mean minimal sufficient. Given $\overline x$, you can make inference on $\mu$. But knowing $x_{(2)}$ alone does not give you enough information about $\mu$. You also need to know $x_{(1)}$ and $x_{(3)}$ to avoid any loss of information. A numerical example might help to make this point clear.
