summation
I'm confused about this notation, can someone please explain how I can write
$2 \sum_{1\leq j < k \leq N} P(F_j F_k)$ as a double sum. Would it be $\sum_{j=1}^k \sum_{k=n}^N P(F_j F_k)$?
Also I am just confused about what a double sum would actually mean, I understand that you would sum over the values twice but can someone please explain more in detail? Thank you
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
$$\sum_{i=1}^m\sum_{j=1}^ma_{ij}.$$
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
$$\sum_{i=1}^m\sum_{j=i}^ma_{ij}.$$
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
$$\sum_{j=1}^m\sum_{i=1}^ja_{ij}.$$
The notation that you are asking about is yet another way to express the sum. That is,
$$\mathop{\sum\sum}_{i \leq j}a_{ij} = \sum_{\substack{(i, j) \in \{1, \dots, m\}\times\{1, \dots, m\}\\ i \leq j}}a_{ij} = \sum_{i=1}^m\sum_{j=i}^ma_{ij} = \sum_{j=1}^m\sum_{i=1}^ja_{ij}.$$
The best way to understand it is to draw the set of couples $(i,j)$ such that
$$k\le j\le i\le n$$ as this
The cases where there are the stars are the set of desired couples and for example the first equality is to sweep these couples by rows: $$\underbrace{(a_{k,k})}_{\text{first row}}+(a_{k+1,k}+a_{k+1,k+1})+\cdots+\underbrace{(a_{n,k}+a_{n,k+1}+\cdots+a_{n,n})}_{\text{last rows}}$$
Best Answer
A double sum is a sum of the form $$\sum_{j=a}^b \sum_{k=c(j)}^{d(j)} f(j,k) = \sum_{j=a}^b \left(\sum_{k=c(j)}^{d(j)} f(j,k)\right)$$ I used the notation $c(j),d(j)$ because $c$ and $d$ may very well depend on $j$, but they don't have to.
There are more ways you can approach this.
We want one variable, say $j$, to be independent, and let it independently take all the values it can possibly have that satisfy the condition $1\le j < k \le N$. We will then adjust $k$ so that this condition is actually met.
Therefore, let $j$ vary across all values that it can possibly have. These are $1\le j \le N-1$, hence the $\sum_{j=1}^{N-1}$ symbol. Then, let $k$ take all values such that the condition $j < k \le N$. A different way to write this inequality is $j+1 \le k \le N$. Hence, your sum is equal to $$2\sum_{j=1}^{N-1} \sum_{k=j+1}^N P(F_jF_k)$$ To expand this sum, first write out all terms of the inner sum, and then add them up according to the outer sum, like $$2\sum_{j=1}^{N-1} (P(F_jF_{j+1})+P(F_jF_{j+2})+...+P(F_jF_N)) =$$ $$= 2[(P(F_1F_2)+P(F_1F_3)+...+P(F_1F_N)) +$$ $$+(P(F_2F_3)+P(F_2F_4)+...+P(F_2F_N))+...+$$ $$+P(F_{N-1}F_N)]$$ Notice that each successive term of the outer sum has less and less terms of the inner sum (the last one has only one term).
Let $k$ vary across all values that it can possibly have. These are $2\le k \le N$. Then, let $j$ take all values such that the condition $1\le j < k$ is met. A different way to write this inequality is $1 \le j \le k-1$. Hence, your sum is equal to $$2\sum_{k=2}^{N} \sum_{j=1}^{k-1} P(F_jF_k)$$
The difference between these two approaches is which variable you assign "independence" to.