Is the expectation of the median the median itself?

Question

It doesn't make sense to me that |E[x] - m| = |E[X-m]|

Answer 1

$E[X-m]=E[X]-E[m]=E[X]-m$ because $m$ is a constant. For instance, if $m=5,$ then $E[X-m]=E[X-5]=E[X]-E[5]=E[X]-5.$
Similarly,
$E[X-\mu]=E[X]-\mu=\mu-\mu=0,$ a familiar fact.

Answer 2

I think the underlying issue is an apparent confusion regarding the meaning of "median" in this question. In the present context, the median refers to a property of the distribution of the random variable $X$, rather than a statistic of a sample drawn from such a distribution. Both are called "median" (for brevity) but the latter is more precisely called the sample median.

To recap, $m$ is a median of a real-valued random variable $X$ when $$ \quad \Pr[X \ge m] \ge \frac{1}{2} \quad \textrm{AND} \quad \Pr[X \le m] \ge \frac{1}{2}.$$ Presumably, the question implies that $m$ is unique, in which case equality is observed. Such a value is not stochastic: it is a deterministic quantity that depends on the parametric distribution of $X$, just as $\mu$ and $\sigma$ are (when these exist). Just because these may not be assigned a known numeric value does not mean they are random.

This is in contrast to the definition of a sample median, which is a statistic and therefore itself a random variable.