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Suppose $X_1,X_2,\ldots,X_n$ are i.i.d random variables with an absolutely continuous distribution.

We say the observation $X_i$ has rank $R_i$ if $$X_i=X_{(R_i)}\quad,\,i=1,2,\ldots,n,$$

where $X_{(k)}$ is the $k$-th order statistic.

I am looking for the correlation between $X_i$ and $R_i$ for each $i=1,\ldots,n$.

Let us assume that $X_i\sim F$, where the distribution function $F$ is known. The difficulty I am facing is that I do not know the joint distribution of $(X_i,R_i)$, which is required for finding $E(X_iR_i)$ in the expression for the covariance. But I suspect that the correlation can be derived regardless.

We can find the mean and variance of $X_i$ and $R_i$ separately once we have the distribution $F$ at hand. But how can we find the covariance?

I know that the conditional distribution $[(X_1,\ldots,X_n)\mid X_{(1)},\ldots,X_{(n)}]$ has the form

$$P\left[X_1=x_1,\ldots,X_n=x_n\mid X_{(1)}=x_{(1)},\ldots,X_{(n)}=x_{(n)}\right]=\frac{1}{n!}\mathbf1_{(x_1,\ldots,x_n)\in A},$$

where $A$ is the set of $n!$ realisations of $(x_{(1)},\ldots,x_{(n)})$.

And that the rank vector $(R_1,R_2,\ldots,R_n)$ is also distributed as

$$P(R_1=r_1,\ldots,R_n=r_n)=\frac{1}{n!}\mathbf 1_{(r_1,\ldots,r_n)\in B},$$

where $B$ is the set of $n!$ realisations of $(1,2,\ldots,n)$.

Any hints on how to proceed would be great.

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I will give a hint. The key concept is exchangeability, meaning that the random vector $(X_1, \dotsc, X_n)$ has the same distribution as $(X_{\pi 1}, \dotsc, X_{\pi n})$ for all permutations $\pi$ of $(1,2,\dotsc, n)$. Then you can check that the vector of ranks $(R_1, \dotsc, R_n)$ also will be exchangeable. Exchangeability is a generalization of iid, so will generalize your eventual result.

We need something more: even the distribution of the $n$ pairs $$ \left( (\begin{smallmatrix} X_1\\R_1\end{smallmatrix}), \dotsc, (\begin{smallmatrix} X_n\\R_n\end{smallmatrix}) \right) $$ is exchangeable. (Then of course we need to assume first exist).

Now calculate: (for some $j$ between 1 and $n$) \begin{align} \DeclareMathOperator{\E}{\mathbb{E}} & \sum_\pi \E X_{\pi j} R_{\pi j} \\ = {} & \E \sum_\pi X_{\pi j} R_{\pi j} \\ = {} & \E \sum_{r=1}^n \sum_{\pi\colon R_{\pi j=r}} X_{\pi j} R_{\pi j} \\ = {} & (n-1)! \sum_{r=1}^n \E X_{\pi j} r \\ = {} & (n-1)! \mu \frac{n (n+1)}{2} \end{align} where $\mu$ is the common expectation of the $X_i$. You should be able to conclude.

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  • $\begingroup$ Thanks. Could you tell me why the $(n-1)!$ comes in the 3rd step? $\endgroup$ – StubbornAtom Nov 8 '18 at 17:06
  • $\begingroup$ Because the sum over n! permutations is divided i to n classes, each one fixing the rank of variable j, each one having (n-1)! elements $\endgroup$ – kjetil b halvorsen Nov 8 '18 at 17:29
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    $\begingroup$ No. By exchangeability, all the expectations summed over are equal, so the sum is that common expectation times $n!$. So the expectation is $\mu \times \frac{n+1}{2}$, which you can see equals $\E X_j \times \E R_j$. It will follow that the correlation is zero. $\endgroup$ – kjetil b halvorsen Nov 9 '18 at 13:30
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    $\begingroup$ I have been thinking about this. There is this similar question on Math.SE where the answer suggests that the correlation can be non-zero. I will get back to you when I make some progress. $\endgroup$ – StubbornAtom Nov 16 '18 at 16:04
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    $\begingroup$ The answer given there cannot be correct, see my comment there. $\endgroup$ – kjetil b halvorsen Nov 16 '18 at 16:31

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