I came across a paper that uses the abbreviation "p.e.":

Khatri and Mardia, The Von Mises-Fisher Matrix Distribution in Orientation Statistics. 1976.

It's in Section 7 on page 105. I'm including a short excerpt:

Using the density function of $K^{\frac{1}{2}}XX'K^{\frac{1}{2}} = S$ given in Khatri (1975a), we find that the p.e. $X$ given $XX'=I_n$ is

What does the abbreviation "p.e." mean in this sentence?


1 Answer 1


It means probability element.

This paragraph is taken from the following book chapter: Order Statistics

If $X$ is such that the probability $X\leq x$ is $F(x)$, or briefly if $$ Pr(X\leq x)=F(x), $$

then we say that $X$ is a random variable which has the cdf $F(x)$. If $F(x)$ has a continuous derivative $f(x)$, then $f(x)dx$ is called the probability element of $X$, and $f(x)$ the probability density function (pdf) of $X$.

I think it would have something to do with measure theory as well. Going over a Probability Theory course might help too.

  • $\begingroup$ So does this mean that the paper should have referred to "the p.e. of $X$" instead of "the p.e. $X$"? $\endgroup$
    – amcnabb
    Commented Mar 27, 2013 at 22:16
  • 1
    $\begingroup$ @amcnabb Sounds like so. The paper your were reading used the notation p.e. on page 97 as well: "... we find that the joint p.e. (probability element) of Y and X1 is ... ". $\endgroup$
    – nil
    Commented Mar 27, 2013 at 22:39
  • $\begingroup$ I missed the earlier reference. Thanks for finding it. $\endgroup$
    – amcnabb
    Commented Mar 28, 2013 at 2:51

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