This is an open question: given a sequence of $n$ real numbers $x\_1<x\_2<\dots<x\_n$, does there always exist a probability distribution, such that $\{x\_i\}$ happens to be the $n$ *expected order statistics* of this distribution? 


In other words, can we always "reverse engineer" the distribution from its expected order statistics? Note that there is **no restrictions** on the distribution, i.e., it can be continuous, discrete, or whatever. I wonder if anything is known regarding this existence problem. 




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(Edited 09/20/2017)


When sequence $\{x\_i\}$ is unrestricted, the answer to above claim is **no**. This is shown by @Mateusz Kwaśnicki when $n=4$. However, under $n=4$, suppose $\{x\_i\}$ satisfies the condition that $3(x\_4−x\_1)⩽7(x\_3−x\_2)$, then is there a method that can construct the distribution for which $\{x\_1,x\_2,x\_3,x\_4\}$ are the expected order statistics?


In other words, suppose $\{x\_i\}$ satisfies the necessary conditions to be expected order statistics, is there a method to "reconstruct" the underlying distribution? Or is this too much to ask for?



Existence of distribution for certain order statistics
The answer is **no** when $n = 4$.


Let $X\_1, X\_2, X\_3, X\_4$ be a sample from a distribution with CDF $F$. Denote by $Y\_1, Y\_2, Y\_3, Y\_4$ the order statistics (that is, the non-decreasing rearrangement) of $X\_1, X\_2, X\_3, X\_4$, and let $x\_j = \mathbb{E} X\_j$. We will show that $3(x\_4-x\_1) \leqslant 7(x\_3-x\_2)$.


The CDFs $F\_j$ of $Y\_j$ are given by
$$\begin{aligned}
F\_1 & = F^4 , & F\_2 & = 4 F^3 - 3 F^4 , \\
F\_3 & = 6 F^2 - 8 F^3 + 3 F^4 , \qquad & F\_4 & = 4 F - 6 F^2 + 4 F^3 - F^4 .
\end{aligned}$$
Since $x\_j = \mathbb{E} Y\_j = \int\_{-\infty}^\infty t \, dF\_j(t)$, we have
$$
 3(x\_4-x\_1) - 7(x\_3-x\_2) = \int\_{-\infty}^\infty t \, dG(t) ,
$$
where
$$
 G = 3F\_4-7F\_3+7F\_2-3F\_1 = 12 F - 60 F^2 + 96 F^3 - 48 F^4 = 12 F (1 - F) (1 - 2 F)^2.
$$
Observe that $t F(t)$ and $t (1 - F(t))$ converge to zero as $t \to -\infty$ and $t \to \infty$, respectively (because the distribution with CDF $F$ has finite mean). Therefore, $t G(t)$ converges to zero as $t \to \pm \infty$. Integration by parts leads to
$$
 3(x\_4-x\_1) - 7(x\_3-x\_2) = -\int\_{-\infty}^\infty G(t) dt .
$$
However, $G = 12 F(1 - F)(1 - 2F)^2 \geqslant 0$, and so the right-hand side is non-positive, as claimed.




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Interestingly, the answer to the original question is **yes** if $n = 3$, and it is enough to consider two-point distributions. Indeed, if $x\_1 < x\_2 < x\_3$ and
$$\begin{aligned}
p & = \frac{x\_3-x\_2}{x\_3-x\_1}, & a & = \frac{3 x\_1 x\_3 - x\_1^2 - x\_1 x\_2 - x\_2^2}{3 (x\_3 - x\_2)} , & b & = \frac{x\_2^2 + x\_2 x\_3 + x\_3^2 - 3 x\_1 x\_3}{3 (x\_2 - x\_1)} ,
\end{aligned}$$
then $x\_1, x\_2, x\_3$ are expected values of the order statistics of a sample of three random variables such that $\mathbb{P}(X = a) = p$ and $\mathbb{P}(X = b) = 1 - p$. This can be verified by a direct calculation, an important point is that $b - a = (x\_3 - x\_1)^3 / (3 (x\_3 - x\_2) (x\_2 - x\_1)) > 0$.


