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What is the difference between the summations using $i<j$ and $i\neq j$ in the formula below: $$\sigma^{2}(\boldsymbol{w})=\sum_{i} \tilde{w}_{i}^{2}+2 \sum_{i<j} \tilde{w}_{i} \tilde{w}_{j} \rho_{i, j}=\sum_{i} \tilde{w}_{i}^{2}+\rho(\boldsymbol{w}) \sum_{i \neq j} \tilde{w}_{i} \tilde{w}_{j}$$ Screenshot here.

  1. Are both summations operationally equivalent?
  2. If so, why break consistency and have two competing representations?
  3. Which summation is more correct, or which to use for which situations?

One of the non-best answers here seem to apply, but not sure how in my case.

develarist
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1 Answers1

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$\sum_{i<j}$ sums over all the possible pairs $(i,j)$ for which $i<j$ holds. Similarly, $\sum_{i\neq j}$ sums over all the possible pairs $(i,j)$ for which $i\neq j$ holds.

For example, if $i$ and $j$ can take values in $\{1,2,3\}$, then

$$\sum_{i<j}a_{i,j}=a_{1,2}+a_{1,3}+a_{2,3},$$

whereas

$$\sum_{i\neq j}a_{i,j}=a_{1,2}+a_{1,3}+a_{2,1}+a_{2,3}+a_{3,1}+a_{3,2}.$$

If the summand is symmetric, i.e., $a_{i,j}=a_{j,i}$ holds for all $i$ and $j$, these two quantities are related by

$$\sum_{i\neq j}a_{i,j}=2\sum_{i<j}a_{i,j}.$$

Sangchul Lee
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    is it more correct to have double summation symbol, or should they be combined into one: $\sum_{i\neq j}^N w_i w_j$ or $\sum_{i=1}^N \sum_{i\neq j}^N w_i w_j$? and are they equivalent – develarist Nov 10 '20 at 13:05
  • @develarist, Both sound equally correct to me, so it might be better to choose a more economical convention. But if you feel that the double summation notation seems ambiguous, it might be good to write $$\sum_{i,j \mathop{:} i\neq j} w_i w_j.$$ By doing so, we can clarify the summation variables. – Sangchul Lee Nov 10 '20 at 13:35