MSD of a vector

[Steven G]

I am not understanding the definition of MSD(x) below. They clearly say that x is an n-vector but then they are talking about summing over i's and j's. That part confuses me. Can someone please enlighten me possibly with an example.
Thank you,
Steven

 

[lev888]

We are summing values of an n x n array. Each value is a square of the difference of ith and jth entries of the vector.

 

[Steven G]

We are summing values of an n x n array. Each value is a square of the difference of ith and jth entries of the vector.

I am totally confused. How can an n-vector be an nxn array (I'm not saying that you are wrong). I did assume that but it seemed strange.

The passage mentioned that there will be n^2 entries. If you have an n-vector (v1, v2, ..., vn), then how do you find n differences to square? I find many more! Like (v1-v2), (v1-v3), ... (v1-vn), (v2-v3)...(v2-vn),..., (v(n-1) -vn).

What am I missing? If you have the time can you give an actual example?
THANKS

 

[lev888]

The array is not the vector.
Let's set up an n by n table. Put vector entries in the column and row headings. Put the difference of the corresponding headings in each cell. Then square each cell. Then sum all cells.

 

[lev888]

The idea is that we are looking at differences of each entry with all other entries, including itself (the diagonal will be all zeroes).

 

[HallsofIvy]

For example, let the vector be <3. 2. 1>. x1= 3 and subtracting each of the entries gives the 3- 3= 0, 3- 2= 1. 3- 1= 2. Similarly, x2 minus each of the entries is 2- 3= -1, 2- 2= -0, 2- 1= 1, and x3 minus each of the others is 1- 3= -2, 1- 2= -1. 1- 1= 0. The 9 entries in the matrix referred to are 0, 1, 2, -1, 0, 1, -2, -1, and, 0 . Of course, for each xi- xj there is a corresponding xj- xi this is an anti-symmetric matrix. The sum of all its values is 0. To avoid that we square and get 0^2+ 1^2+ 2^2+ (-1)^2+ 0^2+ 1^2+ (-2)^2+ 1^2+ 0^2= 1+ 4+ 1+ 1+ 4+ 1= 12. Dividing by n^2= 3^2= 9. the MSD is 12/9= 4/3.