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Is there a notation that expresses a summation over all order-independent combinations of two variables? I have a single set of objects and want to show the summation of the joint probabilities of each unordered pair of those objects. I want it to be clear I'm not counting duplicate pairs, because the order of the elements doesn't matter.

For example, if I have the set:

w = [dog, cat, bear]

I want to convey:

Z = P(dog, cat) + P(dog, bear) + P(cat, bear)

NOT

Z = P(dog, cat) + P(dog, bear) + P(cat, dog) + P(cat, bear) + P(bear, dog) + P(bear, cat)

summation

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2  
How about defining a set: "Let $C_w$ be the set of all unordered pairs $(i,j)$ where $i,j\in w$" and using that? –  Werner Jun 10 '12 at 6:09
2  
It's a math question. If you found out how it looks like it's easy to find out how to set up the formula in latex. –  vertoe Jun 10 '12 at 6:19
    
Thanks Werner, I think I'll go with that approach. –  mrjf Jun 10 '12 at 6:47
1  
@Scott: To be honest, I don't really understand your notation. I'm not familiar with \binom{A}{2} for a set A, and then shouldn't it be P(i,j)? (To add context, I'm a mathematician :-)) –  Hendrik Vogt Jun 10 '12 at 7:44
2  
You're right, I typed that quickly. A would be the set of indices, when A choose 2 just represents the set of two element subsets of A. Thinking now, wouldn't simply i<j beneath the summation accomplish what is required? –  Scott H. Jun 10 '12 at 8:22
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1 Answer

up vote 3 down vote accepted

It looks like this has already been answered in the comments. As an alternative, if order doesn't matter, then you could impose some order on your indices to use fairly standard notation:

\documentclass{article}
\usepackage{amsmath}
\usepackage{mathtools}

\begin{document}
\[
    \sum_{\mathclap{a\leq i<j\leq b}}P(w_i,w_j)
\]
\[
    \sum_{\substack{i<j\\i,j\in A}} \!P(w_i,w_j)
\]

\end{document}

If you have a range of indices then the first option might be preferable, while if you have an arbitrary set of indices (or don't like the first option) then you might prefer the second.

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Accepting. Answers in comments above also good. Thanks everyone! –  mrjf Jun 10 '12 at 20:30
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