Equality between order statistics and certain generalized means? Question
Suppose I have a random variable $X$ with CDF, $F(x)$, and I want to model an IID sample $\{X_1, \cdots, X_n \}$ of size $n$.
For any order statistic $X_{(k)}$ for this sample where $1 \leq k \leq n$, is there a corresponding order $p \in \mathbb{R}$ for the generalized mean on this sample such that $X_{(k)} = M_{p}(X_1, \cdots, X_n)$?
Effort
I've tried collecting some results that seem relevant based on my limited experience of real analysis (pun intended).
Special (limiting) cases that hold:

*

*$M_{-\infty}(X_1, \cdots, X_n) = \min \{ X_1, \cdots, X_n \} = X_{(1)}$

*$M_{\infty}(X_1, \cdots, X_n) = \max \{ X_1, \cdots, X_n \} = X_{(n)}$
There is also an order relation between generalized means of different orders: $p < q \implies M_{p}(X_1, \cdots, X_n) \leq M_{q}(X_1, \cdots, X_n)$
And the generalized mean is a continuous function almost everywhere of $p$.
I am not sure if there is a mathematical result I should invoke here, but the claim holding at the limits, the monotonicity of the generalized mean over $p$, and the almost-everywhere continuity have the look and feel of premises to go into a proof.


*

*Note that I am not asking about the reverse of finding a unique order statistic for every order $p$. That I don't believe exists b/c the orders on a countable-sized sample will not be one-to-one with the dense set $\mathbb{R}$.

*Although an algorithm for finding the relevant $p$ for any given $F(x)$ and $k$ would be fantastic, showing existence is all I am asking here.


No.
Fix a realisation $x_1,\dots,x_n$ and suppose all the values are distinct (which happens with probability 1 for continuous distributions). Consider the order statistic and the generalised means as functions of the $n$ observed values.  If $p$ is finite, the partial derivative $\partial M_p/\partial x_i$ is never zero, but the corresponding derivative of the order statistic is zero for all observations except the one equal to the order statistic. So the generalised mean and the order statistic aren't the same function of the data.
