Summation explanation

Question

I'm wondering how the following summation could be explained:

$\sum_{1 \leq i < j \leq K} (X_{i} - X_{j})^3$

Is this equivalent to

$\sum_{1 \leq i \leq K} \sum_{i < j \leq K}(X_{i} - X_{j})^3$

Any help would be appreciated.

Answer 1

$\sum_{1\le i < j\le K}A_{i,j}$ means add up all the $A_{i,j}$ where $i< j$ and $i \ge 1$ and $j\le K$.

To visualize this this is to add up all the:

$A_{1,2}$

$A_{1,3},A_{2,3}$

$A_{1,4},A_{2,4},A_{3,4}$

$...$

$A_{1,K},A_{2,K}, A_{3,K},.... A_{K-1, K}$.

As it is a finite sum it doesn't really matter what order you add them in and so it doesn't actually tell you what order to list these $(i,j)$ pairs in, it is very specific and clear that $\{A_{i,j}| i\ge 1; i < j; j\le K\}$ is well defined whatever order you list them in.

You could list them a:

$A_{1,2},A_{1,3},......., A_{1,K}$

$A_{2,3},A_{2,4},......., A_{2,K}$

$....$

$A_{K-1,K}$.

$\sum_{1\le i \le K} (\sum_{i<j\le K}A_{i,m})$ means to add up

$A_{1,2}, A_{1,3}, ....., A_{1,K}$

And then add up

$A_{2,3}, A_{2,4},......., A_{2,K}$

And so on.

Can you convince yourself that the sets $\{(i,j)| i\ge 1; i<j; j\le K\}$ and the set $\{(1,j)| 1<j; j\le K\} \cup \{(2,j)| 2<j; j\le K\} \cup \{(3,j)| 3<j; j\le K\}....\cup \{(K-1,K)\}$ are both the same set?

Answer 2

I always break these into sets to work better with them to see what's going on.

Observe \begin{equation}\sum_{1 \leq i < j \leq K} (X_{i} - X_{j})^3=\sum_{(i,j)\in\{(i,j)|1 \leq i < j \leq K\}} (X_{i} - X_{j})^3. \tag{1} \end{equation}

Also, \begin{equation}\sum_{1 \leq i \leq K} \sum_{i < j \leq K}(X_{i} - X_{j})^3=\sum_{(i,j)\in \{(i,j)|1\leq i \leq K \text{ and } i < j \leq K\} } (X_{i} - X_{j})^3. \tag{2}\end{equation}

So, it suffices to show $\{(i,j)|1 \leq i < j \leq K\}=\{(i,j)|1\leq i \leq K \text{ and } i < j \leq K\}$ where the expression on the left is given by $(1)$ and the expression on the right is given by $(2)$. It should be clear that these two sets are indeed equal. Thus, the summations are indeed equal.