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 $\mu.$

Let us assume that $n = 2k+1$ is odd, and let $J_i$ be a ternary variable showing whether $X_i$ is equal to the median $M$, below it or above it. That is,
$$
J_i = 
\begin{cases}
1 & \text{$X_i > M$} \\
0 & \text{$X_i = M$} \\
-1 &  \text{$X_i < M$}
\end{cases}
$$
If $M$ is sufficient, then $(M,J_1,\dots,J_n)$ has to be sufficient as well. So it is enough to show that $(M,J_1,\dots,J_n)$ is not sufficient. What does sufficiency mean? That the distribution of $(X_1,\dots,X_n)$ given $(M,J_1,\dots,J_n))$ is independent of the parameter, $\mu$ in this case. That this independence fails in this case is intuitive, but a bit cumbersome to write down.
With probability 1, exactly one of $J_1,\dots,J_n$ can be equal to 0, and the rest divided equally among +1 and -1 (since $N(\mu,1)$ is continuous w.r.t. the Lebesgue measure the chance of seeing repeated values in a finite sequence of independent draws from it is zero.)
Without loss of generality, let us condition on
\begin{align}
A = \{M = m,\;\; J_1 = 0, \;\;&J_2 = +1, \dots, J_{k+1} = +1, \\ &J_{k+2} = -1, \dots, J_{2k+1} = -1\},
\end{align}
that is, $X_1$ is the median, the next $k$ observations are below the median and the next $k$ after are above the median.
Then, $X_1 = m$ is completely determined, and by independence $X_2,\dots,X_{k+1}$ are i.i.d. from $N(\mu,1)$ truncated to $(m,\infty)$ which we denote as $N(\mu,1; m, \infty)$. Similarly, $X_{k+2},\dots,X_{2k+1}$ are i.i.d. from $N(\mu,1)$ truncated to $(-\infty,m)$ which we denote as $N(\mu,1; -\infty, m)$. Clearly, the distribution of $(X_1,\dots,X_n)$ given $A$ depends on $\mu$ and hence sufficiency fails.
If you want to write it down, technically,
$$
(X_1,\dots,X_n) \, |\, A  \;\;\sim \;\; \delta_{m} \otimes \Bigl(\prod_{i=1}^k N(\mu,1; m, \infty) \Bigr) \otimes \Bigl(\prod_{i=1}^k N(\mu,1; -\infty, m) \Bigr)
$$
where $\delta_m$ is the point mass measure at $m$, and $\otimes$  and $\prod$ denote products of measures. You can also consider other combinations of values for $\{J_i\}$ and it would be similar (you just get a different permutation of the same terms), but not necessary to establish the failure of sufficiency (one combination is enough).
You can also answer the question about ancillarity. A statistic is ancillary if its distribution does not depend on the parameter $\mu$.
