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.



Do these $L^p$ type statistics characterize distributions?
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 $\frac{2}{3}t^p$.


