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I don't understand my lecture notes. It states that standarded residuals is by definition $$R_i=\dfrac{y_i-\hat{y}_i}{\sqrt{\dfrac{1}{n-2}\sum_{i=1}^n (y_i-\hat{y}_i)^2 }}$$ I understand that I can compute $\hat{y}$ by the ordinary least-squares method. But as we have $i$ used as a subscript of $R$ and as an index of summation, I don't understand whether it defines real numbers $R_1,\ldots,R_n$ or what. I'm familiar with notation $f_n=\sum_{i=1}^n$ but $f_i=\sum_{i=1}^n$ is new for me.

So, can anyone explain how I should understand the definition? If I just plug $i=2$ to the equation I can see that $R_2$ looks like a reasonable notation but what is $\sum_{2=1}^n$?

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So in the numerator we have $y_i$ but in the denominator we have $y_k$? – Jaakko Seppälä Oct 8 '12 at 17:54
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Sorry... i deleted my former comment by mistake. It was said that you can replace $\sum_i^n$ by $\sum_k^n$ to remove any confusion. In the numerator you therefore have the specific $y_i$, and in the denominator are all $y_k$'s for $k$ running from $1$ to $n$: $$R_1 = \frac{y_1 - \hat{y}_1}{\sqrt{\frac{1}{n-2} \left[ (y_1 - \hat{y}_1)^2 + \dotsb + (y_i - \hat{y}_i)^2 + \dotsb + (y_n - \hat{y}_n)^2 \right]}}$$ – ocram Oct 8 '12 at 18:21
The summation notation is deceiving you: an expression like "$\sum_{i=1}^n f(i)$" does not involve "$i$" at all. When you write it out fully (not as @ocram has just done, which is erroneous because "$i$" now appears free), it depends only on $f$ and $n$. This is the distinction between free and bound variables. – whuber♦ Oct 8 '12 at 18:26
True whuber, thanks. I have always avoided such situations by choosing indexes well, and this was the first time when I read a text having such a notation. – Jaakko Seppälä Oct 8 '12 at 18:31
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"residuals" is the normal English term, for future reference. Has the comment by @whuber on the two different uses of "i" cleared up the question? – Peter Ellis Oct 8 '12 at 19:25
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