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usul
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No, it looks like many different sufficiently symmetric distributions with enough concentration at $0$ will have $E[|X-c|^p] = 0$$\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 $\frac{2}{3}t^p$.

No, it looks like many different sufficiently symmetric distributions with enough concentration at $0$ will have $E[|X-c|^p] = 0$ 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 $\frac{2}{3}t^p$.

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 $\frac{2}{3}t^p$.

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usul
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No, it looks like many different sufficiently symmetric distributions with enough concentration at $0$ will have $E[|X-c|^p] = 0$ 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 $\frac{2}{3}t^p$.