# Ordered Weighted Averaging operator clarification

I was just lookning at the difinition of the OWA operator. It is defined as:

$F(a_1, ..., a_n) = \sum_{j=1}^n w_j b_j$ where $b_j$ is the $j$th largest of the $a_i$.

The part that strikes me odd is "$b_j$ is the $j$th largest". I am not mathematician, so I wanted to know if saying that something is "$j$th largest" considered an ok practice? Strikes me as not being very mathematics-like...

Is there an alternative definition, that does not use "English" language?

Just curious!

P.S.: I have no idea what to tag this question with, so I'd appreciate if someone could re-tag it appropriately.

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It is ok practice as long as the meaning is clear and unambiguous. It means that there is a permutation $i\mapsto k_i$ of $\{1,2,\ldots,n\}$ such that $a_{k_1}\geq a_{k_2}\geq\cdots\geq a_{k_n}$, and then by definition, $b_j=a_{k_j}$.

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Oh, I see, so this just gets rid of a lot of sub-sub-scripts? – drozzy Jan 19 '12 at 4:02
That is one thing it does. Another possible advantage over the more detailed description is that $b_j$ is uniquely determined, but the particular choice of $k_i$ is not when some of the $a_i$s repeat. Example: If you start with $(3,4,4,7)$, then the permutation could be $1\mapsto 4, 2\mapsto 2, 3\mapsto 3, 4\mapsto 1$ or $1\mapsto 4, 2\mapsto 3, 3\mapsto 2, 4\mapsto 1$, so whether $k_2=2$ or $k_2=3$ isn't well defined. However, in either case there is no question that $b_2=4$. – Jonas Meyer Jan 19 '12 at 4:08
Excuse my dumb question, but what is a permutation? Looking here, on mathworld I don't see no little arrows ;-( – drozzy Jan 19 '12 at 4:12
Informally it means a reordering. So the different permutations of $(1,2,3)$ include for example $(1,3,2)$, $(2,3,1)$, and $(3,1,2)$. More formally it means a function that is bijective, both injective and surjective. Functions can be denoted with arrows: $a\mapsto f(a)$ means that $f(a)$ is the value of the function when $a$ is the input. When this notation is applied to the permuation $(3,2,1)$ of the ordered triple $(1,2,3)$, we have a function $f:\{1,2,3\}\to\{1,2,3\}$ such that $f(1)=3$, $f(2)=2$, and $f(3)=1$. Or, with arrow notation, $1\mapsto 3, 2\mapsto 2$, and $3\mapsto 1$. – Jonas Meyer Jan 19 '12 at 4:17
But in your first comment you start with $(3,4,4,7)$ and have a permutation: $1\mapsto 4$. But $1$ is not in $(3,4,4,7)$... – drozzy Jan 19 '12 at 17:13