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 $\mu.$

This is a great technical question.  First I want to point out that your argument

$T$ 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 $T$.

does not hold. To see it, we know that for $T = \bar{X} \sim N(\mu, \frac{1}{n})$,  whence
\begin{align}
 & \frac{f_\mu(x_1, \ldots, x_n)}{q_\mu(t)} = 
\frac{(2\pi)^{-n/2}\exp(-\frac{1}{2}\sum_{i = 1}^n(x_i - \mu)^2)}
{(2\pi n)^{-1/2}\exp(-\frac{n}{2}(t - \mu)^2)} = h(x)e^{-\frac{n}{2}(\bar{x} - t)\mu},
\end{align}
which depends on $\mu$ (you may argue that it does not depend on $\mu$ if substitute $\bar{x}$ to $t$, however, this is not what the proposed ratio formula says), but $\bar{X}$ is of course sufficient.  It looks like your "criterion" tries to match the definition of sufficiency of $T$ (Section 1.6, Theory of Point Estimation):

A statistic $T$ is said to be sufficient for $X$, or for the family $\mathcal{P} = \{P_\theta, \theta \in \Omega\}$ of possible distributions of $X$, or for $\theta$, if the conditional distribution $X$ given $T = t$ is independent of $\theta$ for all $t$.

A naive interpretation of the above definition is that the conditional density of the data $X = (X_1, \ldots, X_n)$ given $T = t$ is independent of $\theta$, which may be formally written as (note how the denominator in $(1)$ differs from that in your proposed ratio)
\begin{align}
f_{X|T; \theta}(x|t) = \frac{f_\theta(x, t)}{q_\theta(t)} \text{ is independent of $\theta$.} \tag{1}
\end{align}
However, a technical difficulty of the interpretation $(1)$ is that the "joint density" $f_\theta(x, t)$ of $(X, T)$ is degenerate (i.e., it is not a probability density on $\mathbb{R}^{n + 1}$.  In general, the conditional density defining relation $f_{X|Y = y}(x|y) = \frac{f(x, y)}{f_Y(y)}$ only makes sense when $f(x, y)$ is a valid, non-degenerate density), hence $(1)$ is actually impossible to check.  For a relevant discussion, see this question.   To solve this difficulty, some 1-1 transformation between $(X_1, \ldots, X_n)$ and $(Y_1, \ldots, Y_{n - 1}, T)$ needs to be defined and the sufficiency needs to be augmented accordingly in terms of the conditional density of $Y$ given $T$.  For details, refer to Eq (1.17) -- Eq (1.19) in Section 1.9, Testing Statistical Hypotheses.
Having clarified this, one way to show that $M$ is not a sufficient statistic for $\mu$ is to find a subset $Y$ (possibly with transformation) of $(X_1, \ldots, X_n)$, such that the conditional density of $Y$ given $T$ is well-defined (i.e., non-degenerate) and does depend on $\theta$. For simplicity and without losing of generality, assume $n = 3$.
One choice of $Y$ is $(X_{(1)}, X_{(3)}) = (\min(X_1, X_2, X_3), \max(X_1, X_2, X_3))$.  It is well known by the order statistic theory that the joint density of $(Y, M) = (X_{(1)}, X_{(3)}, X_{(2)})$ is
\begin{align}
f(x_{(1)}, x_{(3)}, x_{(2)}) = 6\varphi_\mu(x_{(1)})\varphi_\mu(x_{(2)})\varphi_\mu(x_{(3)}), \quad
x_{(1)} < x_{(2)} < x_{(3)}, \tag{2}
\end{align}
where $\varphi_\mu(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{(x - \mu)^2}{2}}$. And the marginal density of $X_{(2)}$ is (where $\Phi_\mu(x) = \int_{-\infty}^x \varphi_\mu(t)dt$)
\begin{align}
f_{X_{(2)}}(x_{(2)}) = 6\Phi_\mu(x_{(2)})(1 - \Phi_\mu(x_{(2)}))\varphi_\mu(x_{(2)}). \tag{3}
\end{align}
$(2)$ and $(3)$ then give the conditional density of $(X_{(1)}, X_{(3)})$ given $X_{(2)}$ as
\begin{align}
f_{(X_{(1)}, X_{(3)})|X_{(2)} = x_{(2)}}(x_{(1)}, x_{(3)}|x_{(2)}) = 
\frac{\varphi_\mu(x_{(1)})\varphi_\mu(x_{(3)})}
{\Phi_\mu(x_{(2)})(1 - \Phi_\mu(x_{(2)}))},
\end{align}
which depends on $\mu$.
In contrast, you can verify that
\begin{align}
f_{(X_{(1)}, X_{(3)})|\bar{X} = \bar{x}}(x_{(1)}, x_{(3)}|\bar{x}) = 
\frac{3\sqrt{3}}{\pi}\exp\left(\frac{3}{2}\bar{x}^2 - 
\frac{1}{2}(x_{(1)}^2 + x_{(3)}^2 + (3\bar{x} - x_{(1)} - x_{(3)})^2)\right),
\end{align}
which is independent of $\mu$.
