0

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

Eric Towers
  • 67,037
Michael
  • 37
  • Welcome to MSE. Please type your post, rather than posting images, when possible. Images can't be browsed, and aren't accessible to those using screen readers. If you need help formatting math on this site, here's a tutorial. – saulspatz Nov 04 '19 at 15:24
  • ok, thank you! . – Michael Nov 04 '19 at 15:26

2 Answers2

0

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*}

Eric Towers
  • 67,037
0

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 $X_{1,2}+X_{1,3}+X_{1,4}+X_{2,3}+X_{2,4}+X_{3,4}$

JMoravitz
  • 79,518