Please can someone tell me why we drop the summation of i here?

$S = \sum_{i,j}(y_{ij}-\mu -\alpha_i)^2$

$\frac{dS}{d\alpha_i} = \sum_{j}(y_{ij}-\mu -\alpha_i)$

It's part of a question which is really bugging me?

Thank you for any help I really appreciate it!


Consider one particular term of the sum:

$$S_k = \sum_j (y_{kj}-\mu-\alpha_k)^2$$

Now, you have

$$ \frac{\partial S_k}{\partial \alpha_i} = \left\{\begin{matrix}-2\sum_j(y_{ij}-\mu-\alpha_i) &\text{ if } i=k\\0 &\text{ otherwise}\end{matrix}\right. $$ So when you sum over all of the $k$ values, you're left with $$ \frac{\partial S}{\partial \alpha_i} = -2\sum_j (y_{ij}-\mu-\alpha_i) $$

  • $\begingroup$ Awesome Glen, thank you. Is there a way a can improve your $\textit{reputation}$ on here? I'm new. $\endgroup$ – Kane Blackburn Apr 21 '13 at 12:19
  • $\begingroup$ By answering questions, by asking interesting questions, and things like that. More info here. $\endgroup$ – Glen O Apr 21 '13 at 12:22
  • $\begingroup$ I realise that. I was asking more whether I had to do vote for you in someway so you gain credit for the answer you've just given me. I guess you gain it automatically as soon as I tick your answer as my accepted answer? $\endgroup$ – Kane Blackburn Apr 21 '13 at 12:27
  • $\begingroup$ Ah, sorry - yes, accepting the answer, which is what the tick does, gives me 15 reputation points. Voting my answer up also gives me 10 reputation points. $\endgroup$ – Glen O Apr 21 '13 at 12:29
  • $\begingroup$ Okay once I have the 15 reputation required I will vote your answer up so that you get the further points. Thanks again. $\endgroup$ – Kane Blackburn Apr 21 '13 at 12:31

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