Define for real valued random variable $X\in L^p$, the $p$-statistic
$$X_p:=\arg\min_{c\in \mathbb R}E[|X-c|^p].$$
For example $X_1$ is the median of $X$, $X_2$ is the mean of $X$ and also $X_\infty$ is the midpoint of the range of $X$.
Let $X,Y\in L^\infty$ be two real valued random variables so that $X_p=Y_p$ for all $p>0$. Then is it true that $X=Y$ in distribution? What if we assume some regularity on $X,Y$?
This seems like an analogue to moment characterization.
No, it looks like many different sufficiently symmetric distributions with enough concentration at $0$ will have $\arg\min_c E[|X-c|^p] = 0$ for all $p > 0$.
Concrete family of examples: if $$ X = \begin{cases} 0 & \text{w.prob $2/3$} \\ t & \text{w.prob $1/6$} \\ -t & \text{w.prob $1/6$} , \end{cases} $$ then, regardless of our choice of $t > 0$, for all $p > 0$ we have $0 = \arg\min_c E[|X-c|^p]$. We can also replace $2/3$ with anything greater than $1/2$.
Proof sketch: by symmetry and monotonicity, we can suppose $0 \leq c \leq t$ in the minimization. \begin{align} E[|X-c|^p] &= \frac{1}{6} (t + c)^p + \frac{2}{3} c^p + \frac{1}{6} (t-c)^p . \end{align} For all $p \geq 1$, this expression is strictly increasing in $c$, implying it is minimized at $c=0$.
For all $0 < p < 1$, the expression is concave in $c$ (the second derivative is negative), so it is minimized at one of the endpoints, and $c=0$ results in a value of $\frac{1}{3}t^p$, while $c=t$ results in a larger value of at least $\frac{2}{3}t^p$.
