1
$\begingroup$

I am working my way through a textbook, and it contains the following identity: $$\frac{1}{|C_k|} \sum_{i,i'\in C_k}\sum_{j=1}^p (x_{ij}-x_{i'j})^2 = 2\sum_{i\in C_k}\sum_{j=1}^p(x_{ij}-\bar{x}_{kj})^2$$ where $$\bar{x}_{kj} = \frac{1}{|C_k|}\sum_{i \in C_k}x_{ij}$$

I'm having difficulty proving this to be true. I've tried expanding both sides and transforming one side into the other, but I think I'm having difficulty with some of the notation. Can someone help me construct a proof of this, and maybe explain the notation along the way? I've never seen a summation like the first sigma on the left-hand side of the equation.

$\endgroup$
  • $\begingroup$ Could you provide more context to the problem, like what $C_k$ is, and here I'm assuming the $x_{ij}$'s are in $\mathbb R$? $\endgroup$ – hausdork Mar 18 '17 at 2:31
  • $\begingroup$ This is from a chapter about k means clustering. $C_k$ is a cluster of observations, so I suppose the notation is saying that i and i' are in the kth cluster, as that would make sense. However, I'm still very thrown off by how to approach turning the left side into the right. I've also tried starting on the right side, but I haven't gotten very far with this. I assume $x_{ij}$'s a real $\endgroup$ – ping Mar 18 '17 at 2:46

Your Answer

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

Browse other questions tagged or ask your own question.