what is result the double sum(sigma) with assume n=3?

Question

I'm gonna calculate the double sum on this screenshot with n=3

$$ E[X] = E \left[ \sum_{i=1}^{n-1} \sum_{j=i+1}^n X_{ij} \right] $$

what is the result I get it?! and how to calculate it?

thanks

Answer 1

In general,

$$\sum\limits_{i=1}^{n-1}\sum\limits_{j=i+1}^n X_{i,j}$$

will be the sum of all $X_{i,j}$ where $1\leq i<j\leq n$ of which there will be $\binom{n}{2}$ such terms. So, for example with $n=1000$ you would have each of the terms like $X_{237,560}$ and $X_{2,980}$ where the first index is smaller than the second index and both within the specified range but will not have terms like $X_{500,1}$ where the first index is larger than the second nor will you have terms like $X_{15000,-10}$ where the indices are out of bounds.

In the special case of $n=3$ that would be $X_{1,2}+X_{1,3}+X_{2,3}$ while for $n=4$ you would have $X_{1,2}+X_{1,3}+X_{1,4}+X_{2,3}+X_{2,4}+X_{3,4}$

Answer 2

With $n = 3$, you have \begin{align*} E[X] &= E \left[ \sum_{i=1}^2 \sum_{j=i+1}^3 X_{ij} \right] \\ &= E \left[ \left. \sum_{j=i+1}^3 X_{ij} \right|_{i=1} + \left. \sum_{j=i+1}^3 X_{ij} \right|_{i=2} \right] \\ &= E \left[ \sum_{j=2}^3 X_{1j} + \sum_{j=3}^3 X_{2j} \right] \\ &= E \left[ \left( \left. X_{1j} \right|_{j=2} + \left. X_{1j} \right|_{j=3} \right) + \left( \left. X_{2j} \right|_{j=3} \right) \right] \\ &= E \left[ \left( X_{12} + X_{13} \right) + \left( X_{23} \right) \right] \\ &= E \left[ X_{12} + X_{13} + X_{23} \right] \text{.} \end{align*}