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.

(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?

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    $\begingroup$ I tried {1,2,9} as an experiment, and found an approximate answer for that case: .894 + 1594 Beta[1/8, 64]. I didn't find it easy. $\endgroup$
    – user44143
    Sep 20, 2017 at 0:37
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    $\begingroup$ @MattF. Could you elaborate a bit on how you find that approximation? Thanks! $\endgroup$ Sep 20, 2017 at 2:20
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    $\begingroup$ I used trial and error with beta's to get the needed value of 7 for $(X(3)-X(2)) / (X(2)-X(1))$, and then scaled linearly to get the 1, 2 and 9. And before that I used trial and error to find a good family of distributions to fit parameters. $\endgroup$
    – user44143
    Sep 20, 2017 at 3:55
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    $\begingroup$ If your original question was answered to your satisfaction, I suggest making new posts for any follow-up questions (giving a link to this question, of course). $\endgroup$
    – j.c.
    Sep 20, 2017 at 15:33
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    $\begingroup$ @user3026001 it's not too late to do so this time $\endgroup$ Sep 20, 2017 at 18:45

1 Answer 1


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.

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 $b - a = (x_3 - x_1)^3 / (3 (x_3 - x_2) (x_2 - x_1)) > 0$.

  • $\begingroup$ Nice argument! I would like to ask a follow-up question. Please the edited question. $\endgroup$ Sep 20, 2017 at 15:15
  • $\begingroup$ @user3026001: The restriction discussed above is clearly not sufficient. You can easily find other inequalities of a similar kind by finding other linear combinations of $F_j$ with coefficients that add up to zero that are negative functions of $F \in [0,1]$. Does this family of linear conditions give an "if and only if" condition? I would not be surprised if this was the case. Can one write down these conditions in a systematic way? I have no idea. $\endgroup$ Sep 20, 2017 at 21:16
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    $\begingroup$ @user3026001: Regarding the construction, apparently it is sufficient to consider discrete distributions. A systematic way could be the following: consider random variables satisfying $\mathbb{P}(X=a_j)=\tfrac{1}{k}$ for $j=1,2,\ldots,k$ and $a_1<a_2<\ldots<a_k$. Then expectations of the order statistics $x_1,x_2,\ldots,x_n$ are linear combinations of $a_1,a_2,\ldots,a_k$ (with fairly explicit coefficients). You can solve it for the unknown $a_1,a_2,\ldots,a_k$, remembering that $a_1<a_2<\ldots<a_k$. If no such solution exists, increase $k$ and try again. $\endgroup$ Sep 20, 2017 at 21:23

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