Suppose that $X_1, \ldots, X_n$ are mutlivariate normal with equal correlation $\rho$ and each of them are marginally distributed as $N(0,1)$. Let $X_{(1)}, \ldots, X_{(n)}$ be the corresponding order statistics. The distribution of $X_{(1)}$ and $X_{(n)}$ are easily found. What about the distribution of the other order statistics? Can anyone give some reference on this?

Thank you. Hanna

  • $\begingroup$ C.W.Helstrom's book Statistical Theory of Signal Detection (1967?) has some discussion of the equal correlation case and $X_{(n)}$ though not of the order statistics as I recall. Just remember that for any (not necessarily multivariate normal) equicorrelated equivariance random variables, $$\operatorname{var}\left(\sum_{i=1}^n X_i\right)=n\sigma^2+n(n-1)\rho\sigma^2 \geq 0$$ shows that $\rho \geq -\frac{1}{n-1}$. $\endgroup$ Mar 27 '13 at 2:38

As shown in Tong, Y. L. (1990). Multivariate normal distribution. Springer-Verlag., ch. 6, for the setup described in the question and for non-negative correlation coefficient $\rho\in [0,\;1)$, the distribution function (cdf) and the density of an order statistic $X_{(i)}$ are (where $\phi()$ and $\Phi()$ are the standard normal pdf and cdf)

$$G_{(i)}(x) = \int_{-\infty}^{\infty}F_{(i)}\left(\frac{x+\sqrt{\rho}z}{\sqrt{1-\rho}}\right)\phi(z)dz$$

and differentiating,

$$g_{(i)}(x) = \int_{-\infty}^{\infty}\frac 1{\sqrt{1-\rho}}f_{(i)}\left(\frac{x+\sqrt{\rho}z}{\sqrt{1-\rho}}\right)\phi(z)dz$$


$$f_{(i)}(y) = \frac{n!}{(i-1)!(n-i)!}[\Phi(y)]^{i-1}[\Phi(-y)]^{n-i}\phi(y)$$ and $$F_{(i)}(y) = \sum_{j=i}^n {n \choose j}[\Phi(y)]^{j}[\Phi(-y)]^{n-j}$$

i.e $f_{(i)}(y)$ and $F_{(i)}(y)$ are the pdf and cdf of the order statistic $(i)$ from an i.i.d. standard normal random sample.

For the corresponding results when the correlation coefficient is negative, the author refers to the book "Order Statistics", by H.A. David & H.N. Nagaraja ch. 5 (now in its 3d edition, 2003).

  • $\begingroup$ Does $G_{(i)}(x)$ or $g_{(i)}(x)$ simplify to a different form if $i=n$ or $i=1$? $\endgroup$
    – Frey
    Apr 24 '17 at 0:30
  • $\begingroup$ @Frey Obviously the expressions are simplified to a degree, but I am not aware of "closed-form expressions" even in these cases (we are talking about correlated variable here), although the OP argues that "they are easily found". $\endgroup$ Apr 25 '17 at 1:30

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