# How to interpret a sum with two indices?

Could anyone please break the following down for me?

$$\sum_{i,t}y_{i,t-2}e_{it}=0$$

I am familiar with $\sum_{i}$ and also $\sum_i\sum_t$. But I have not seen the above before. I will appreciate if someone could explain the first equation. Thanks in advance.

$\displaystyle\sum_{i,t}$ means the same as $\displaystyle\sum_i \sum_t$. In the second notation, a specific summation order is given, whereas in the first one there isn't.

So the first notation is only appropriate if the order of summation doesn't matter. For example in the finite case.

Missing are the index sets for $i$ and $t$ respectively. The notation $e_{it}$ indicates that a multiplication of index $i$ and index $t$ is possible.

If e.g. $\{5,7\}$ it the index set of $i$ and $\{2,8\}$ is the index set of $t$ then the LHS equals $$y_{5,2}e_{10}+y_{5,8}e_{40}+y_{7,2}e_{14}+y_{7,8}e_{56}$$

• Possible, but not likely! If that's what the writer meant s/he should say so. – Ethan Bolker Mar 3 '18 at 13:31
• @EthanBolker what do you mean then? The multiplication of indices maybe? Btw, it is not unlikely that the writer has been sloppy. If a comma is used to discern indices in the first factor then consistency demands that it also used in the second factor. – drhab Mar 3 '18 at 13:33
• I've never encountered a sum like that where the terms came from multiplying indices. Adding, yes. So if someone were multiplying I'd expect a warning in words rather then relying on the reader to parse indices in a formally correct but very unorthodox way. – Ethan Bolker Mar 3 '18 at 13:37

While Klirk offers a definition that is useful for this particular case, I think it is more instructive to interpret this more generally:

$\displaystyle\sum_{i,j}$ essentially means is shorthand for $\displaystyle\sum_{(i,j)\in D\times D'}$ where $D$ and $D'$ are the 'omitted' domains over which the summation is performed (often, $D=D'$), i.e. there is a term for each pair from $D\times D'$.

This more general interpretation becomes useful when $D$ and $D'$ are defined depending on each-other in a non-trivial way. While that can still be reduced to a form of $\displaystyle\sum_i \sum_j$, the latter form can become 'ugly', compare: $\displaystyle\sum_{i,j: \text{$i+j>0$}}$ with $\displaystyle\sum_{i}\sum_{j: i+j>0}$. The first is clearer, as it directly shows what domain we work on and doesn't require the convention that empty sets sum to $0$ (which can occur if the domain for $i,j$ is $\{-2,-1,0,1,2\}$, for instance).