I am trying to find a book or something that explains the use of a inequality in the index notation of a double summation like in this example:

$$\mathop{\sum\sum}_{i\leq j}Cov(X_i,X_j).$$

What does this mean? Does not seem intuitive for me a like a normal double summation like:


where one just have to sum this way:


Thank you very much.


3 Answers 3


Suppose $i$ and $j$ are indices which take values $1, \dots, m$, and for each $i$ and $j$, we have a number $a_{ij}$. Note that $(i, j) \in \{1, \dots, m\}\times\{1, \dots, m\}$. If we were to sum over all possible values of $(i, j)$ we would have

$$\sum_{(i, j) \in \{1, \dots, m\}\times\{1, \dots, m\}}a_{ij}$$

which could also be written as

$$\sum_{i \in \{1, \dots, m\}}\sum_{j\in\{1, \dots, m\}}a_{ij}$$

or more commonly


Sometimes, we are not interested in all possible pairs of indices, but only those pairs which satisfy some condition. In the example you are looking at, the pairs of indices of interest are $(i, j) \in \{1, \dots, m\}\times\{1, \dots, m\}$ such that $i \leq j$. One way to denote the sum over all such pairs of indices is

$$\sum_{\substack{(i, j) \in \{1, \dots, m\}\times\{1, \dots, m\}\\ i \leq j}}a_{ij}$$

but this is rather cumbersome. It would be much more helpful if we could write it as a double sum as above. To do this, note that we can list all suitable pairs of indices, by first fixing $i \in \{1, \dots, m\}$ and then allowing $j$ to vary from $i$ to $m$ (as these are the only possible values of $j$ with $i \leq j$). Doing this, we obtain the double sum


Note, we could also have fixed $j \in \{1, \dots, m\}$ and then allowed $i$ to vary from $1$ to $j$ (as these are the only possible values of $i$ with $i \leq j$). Doing this, we obtain an alternative double sum


The notation that you are asking about is yet another way to express the sum. That is,

$$\mathop{\sum\sum}_{i \leq j}a_{ij} = \sum_{\substack{(i, j) \in \{1, \dots, m\}\times\{1, \dots, m\}\\ i \leq j}}a_{ij} = \sum_{i=1}^m\sum_{j=i}^ma_{ij} = \sum_{j=1}^m\sum_{i=1}^ja_{ij}.$$

  • $\begingroup$ Thank you very much, very clarifying with the other ways of rewritting the summation, if you let me ask. And what would be another way to write this summation: $\mathop{\sum\sum}_{i < j}Cov(X_i,X_j)$ where $i$ cannot be equal $j$ anymore, thanks $\endgroup$ Aug 30, 2014 at 2:28
  • $\begingroup$ If you fix $i \in \{1, \dots, m\}$, what are the possible values of $j$ such that $i < j$? $\endgroup$ Aug 30, 2014 at 2:51
  • $\begingroup$ I would say that could be this: $\mathop{\sum\sum}_{i < j}Cov(X_i,X_j) = \sum_{i=1}^m\sum_{j=i+1}^m Cov(X_i,X_j)$ as you did in the last example $\endgroup$ Aug 30, 2014 at 3:07
  • 1
    $\begingroup$ That's correct. Well done. $\endgroup$ Aug 30, 2014 at 3:27
  • 1
    $\begingroup$ @cpage: I would write $$\mathop{\sum\sum}_{i \neq j}a_{ij}$$ in the first case and in the second you can just write $$\sum_{i=1}^ma_{ii}.$$ $\endgroup$ Jan 31, 2019 at 22:51

It means sum over the range of $j$ and then sum over the range of $i$ with the condition that $i$ is smaller or equal to $j$. Written explicitly like $$\sum_{j=1}^n\sum_{i=1}^jf$$


Well, let $n > 0$ be an integer and $(X_{j})_{1}^{n}$ a sequence of random variables. Then the sum in question equals $$\sum_{1}^{n}cov(X_{1}, X_{j}) + \sum_{2}^{n}cov(X_{2}, X_{j}) + \cdots + cov(X_{n}, X_{n}).$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.