I am working through the proof for partitioning sum of squares.
Could someone please explicitly explain how one goes from line 14 to 15 in the following PDF:


That is, how one moves between these lines -

\begin{align} \rm{SSTO} &= \sum_{i=1}^r \sum_{j=1}^{n_i} (Y_{ij}-\overline{Y}_{..})^2 \\ &= \sum_{i=1}^r \sum_{j=1}^{n_i} [(Y_{ij}-\overline{Y}_{i.})+(\overline{Y}_{i.}-\overline{Y}_{..})]^2 \end{align}

I'm actually reading through Casella and Berger's "Statistical Inference" (pg 536), but the same jump occurs in both. Sorry, I was never the best at algebra... would someone please explain this.

  • $\begingroup$ Do you mean the bit in brackets where it says $\sum_{j=1}^{n_i} (Y_{ij}-\overline{Y}_i) =0$ ? If so, this is because the definition of $\overline{Y}_i$ is $\frac{1}{n_i} \sum_{j=1}^{n_i} Y_{ij}$ and so you are adding up $n_i$ copies of $\overline{Y}_i$ and then subtracting $n_i$ copies of $\overline{Y}_i$, so you end up with zero. We can elaborate further if this is what you want. But there are no line numbers in the pdf, so I'm not sure! $\endgroup$
    – Flounderer
    Mar 18, 2013 at 0:30
  • $\begingroup$ Thanks for your edit of the tags. I have edited your question to add the mathematics of the two lines you're concerned about. Please double check it's right. $\endgroup$
    – Glen_b
    Mar 18, 2013 at 1:22

1 Answer 1


It just involves adding and subtracting a term:

$(a-b)^2 = (a -m+m -b)^2 = [(a-m)+ (m-b)]^2 $

Since everything is squared, it's really just doing this:

$(a-b) = (a -m+m -b) = [(a-m)+ (m-b)] $

but where $a = Y_{ij}$, $b=\overline{Y}_{..}$ and $m = \overline{Y}_{i.}$

This "add and subtract the same thing inside a sum of squares" is a standard trick, you see it all over.

  • $\begingroup$ Thank you. I mean the 1st and second line that you graciously added here. $\endgroup$
    – Patrick
    Mar 18, 2013 at 1:08
  • $\begingroup$ See my edit above $\endgroup$
    – Glen_b
    Mar 18, 2013 at 1:12
  • $\begingroup$ OK Glen; I greatly appreciate it. I love math now, but historically have never been good at it. Cheers! $\endgroup$
    – Patrick
    Mar 18, 2013 at 1:34
  • $\begingroup$ Is there some further explanation or discussion needed in the answer at all, or do you think it's been covered now? $\endgroup$
    – Glen_b
    Mar 18, 2013 at 1:42

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