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I'm working through a proof involving the sum of covariances but the notation is tripping me up. What does it mean when you are taking a summation over the index $i < j$? For instance $\sum_{i < j}\mathrm{Cov}(X_{i},X_{j})$ where Cov is covariance.

I'm sure it's nothing complicated I just want to make sure I understand.

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You run over all $j$ and for each $j$ you run over all $i$ such that $i < j$. For instance is $\sum_{i < j} i = 1+2+3+\cdots+n$ if $j$ can be maximal $n$. In particular this means here that you have $\sum_{i<j} Cov(x_i,x_j) = \sum_{i<1} Cov(x_i,x_1) + \sum_{i<2} Cov(x_i,x_2) + \sum_{i<3} Cov(x_i,x_3)+ \cdots +\sum_{i<n} Cov(x_i,x_n) $ if $j$ is maximal $n$. –  André Oct 20 '12 at 16:04
    
In general $\sum_{i<j} = \sum_{j=1}^n \sum_{i=1}^j$ –  André Oct 20 '12 at 16:13
    
This is an old post. If your question is solved, maybe you'd like to pick an answer you'd like best. Regards. –  FrenzY DT. Dec 4 '12 at 17:03
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2 Answers 2

The notation means that you sum over all allowed $i$ and $j$ such that $i < j$. The notation is compact and protects you from errors that may come from doing the end-points incorrectly.

Here is an example. Suppose that $1\leq i \leq m$ and $1\leq j \leq n$ where $m$ and $n$ are not the same. If $m < n$, the sum becomes

$$\sum_{i < j} a_{ij} = \sum_{j = i+1}^n \sum_{i=1}^{m}a_{ij}.$$

If $n < m$ we have

$$\sum_{i < j} a_{ij} = \sum_{j = i+1}^n \sum_{i=1}^{n-1}a_{ij}.$$

Formally, these are two different expressions and to determine which one you are using can take a few seconds here but it could take a lot more time for more complicated sums.

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An Illustration

Suppose you have $i\in I=\{1,2,\cdots,5\}$ and $j\in J=\{1,2,\cdots,6\}$. The following picture shows you which of the terms $\sum\limits_{i<j} a_{i,j}$ include.

The dashed like is i=j. The region is i<j. All index-pairs falling in this region is summed.

According to the picture, if you choose a certain value of $i$, then $j$ should at least be $i+1$, otherwise the condition $i<j$ isn't met (falling on or below the line $i=j$).

In conclusion, $$\sum\limits_{i<j} a_{i,j} = \sum_{i} \sum_{\text{All $j$'s that's $>i$}} a_{i,j}.$$

If you sum $j$ last, you get $$\sum\limits_{i<j} a_{i,j} = \sum_{j} \sum_{\text{All $i$'s that's $<j$}} a_{i,j}.$$

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