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.
- Are both summations operationally equivalent?
- If so, why break consistency and have two competing representations?
- 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.
$\sum_{i<j}$ 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
$$\sum_{i\neq j}a_{i,j}=2\sum_{i<j}a_{i,j}.$$