1
$\begingroup$

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.

  • 0
    This is an old post. If your question is solved, maybe you'd like to pick an answer you'd like best. Regards.2012-12-04

2 Answers 2

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.

1

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 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 isn't met (falling on or below the line $i=j$).

In conclusion, $\sum\limits_{ii$}} a_{i,j}.$

If you sum $j$ last, you get $\sum\limits_{i