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 $\mu.$

A tricky part of this question is that the median sounds like a good estimator and mathematically it is not so clear what the mean is gonna improve.
So imagine the following simpler case:

*

*Say we have a sample of size $X_1, X_2, \dots , X_n$ where each $X_i$ follows (independently) a normal distribution $X_i \sim N(\theta,1)$ and we want to estimate $\theta$.

*We could make an estimate $\hat{\theta} = X_1$ and it is obvious that this is not a sufficient statistic. The other $n-1$ values can provide information about $\theta$ as well. The statistic $X_1$ is not sufficient.

*A sufficient statistic occurs when the distribution of the data is independent of the parameter $\theta$ conditional on the sufficient statistic. This is not true for $X_1$. For example, the distribution of $\bar{X}$ conditional on $X_1$ is not independent from $\theta$

