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I'm very very sorry for the banality of my question but I've never gotten through this before. Could you please tell me what I am supposed to do when facing two sigma such as in the covariance formula? Should I sum the results obtained by each or rather multiply them? Thanks guys.

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Could you give an example so that your question becomes clearer? – M Turgeon Jun 4 '12 at 16:43
As stated above an example is the formula for the variance. It looks like this ∑∑... where the first sigma goes from i=1 to n and the second from j=1 to r . Thanks @m-turgeon for answering! – ocram Jun 4 '12 at 16:58
What are you summing? In some particular cases, we can simplify. – Davide Giraudo Jun 4 '12 at 17:13
Salut @davide , ce n'est pas un cas particulier qui m'intéresse ici. en effet je voulais juste savoir comment faut-il se comporter face à un double sigma. dois-je sommer ou bien multiplier les résultats obtenus ? O preferivi in italiano? Grazie in anticipo. – ocram Jun 4 '12 at 17:22
Non importa (anche se qui la linguia ufficiale è l'inglese, dunque continuo in inglese). If you can write $a_{ij}=b_i\cdot c_j$, then you just multiply the sum over $i$ with the sum over $j$. But is doesn't need to be the case in general. A good way to think about double summations is summation over an array. You can first sum with respect to the rows, then with respect to the columns (and also switch the two sums). – Davide Giraudo Jun 4 '12 at 17:58

This notation has the following meaning: consider real numbers $a_{ij}$, where $i$ ranges from 1 to 3, and $j$, from 1 to 2. We thus have the following equality:

$$\sum_{i=1}^3 \sum_{j=1}^2 a_{ij}=\sum_{i=1}^3(a_{i1}+a_{i2})=(a_{11}+a_{12})+(a_{21}+a_{22})+(a_{31}+a_{32}).$$

There is nothing more to it. However, as Davide pointed out, sometimes you can simplify the computations - but this highly depends on the nature of the number $a_{ij}$ (for example, they might satisfy a recurrence relation).

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