0
$\begingroup$

In elements of statistical learning book it is given like below image, i am not able to understand how it got to second line. Can someone help ? enter image description here

$\endgroup$
1
$\begingroup$

It is an important statistical result to be able to show the following:

$$ \sum_{i=1}^n \sum_{j=1}^n (x_i - x_j)^2 = 2n \sum_{i=1}^n (x_i - \bar{x})^2$$

The proof is with double summation:

$$ \begin{eqnarray} \sum_{i=1}^n \sum_{j=1}^n (x_i - x_j)^2 &=& \sum_{i=1}^n \sum_{j=1}^n (x_i - \bar{x} + \bar{x} - x_j)^2 \\ &=& \sum_{i=1}^n \sum_{j=1}^n (x_i - \bar{x})^2 + (x_j - x_j)^2 - 2(x_i - \bar{x})(x_j - \bar{x})\\ &=& \sum_{i=1}^n \sum_{j=1}^n (x_i - \bar{x})^2 + (x_j - x_j)^2\\ &=& 2n \sum_{i=1}^n (x_i - \bar{x})^2 \end{eqnarray} $$

Where the third term in the RHS of the second line is orthogonal and sums to 0 .

$\endgroup$
  • 1
    $\begingroup$ I guess second term is x_j - X¯ $\endgroup$ – Siddharth Shakya Jun 4 '18 at 20:05
0
$\begingroup$

It's a non-obvious thing, so you'll have to chew on the math a little bit.

First thing to do: show that it is sufficient to solve this for the one dimensional case.

It's not too hard, so this is a good exercise to practice your statistics 101 skills. Give it a try. Try approaching it from both sides, and use the binomial expansion.

The key result in here is the Konig-Huygens theorem.

$\endgroup$
  • $\begingroup$ good to know that its a non obvious thing, i was like i am having to ask even obvious things nowadays $\endgroup$ – Siddharth Shakya Jun 4 '18 at 18:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.