4
$\begingroup$

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?

$\endgroup$

1 Answer 1

6
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.