$\epsilon =\sum (y_i-\beta x_i)$

$\beta = \frac {\sum x_iy_i}{\sum x_i^2}$

$\sum y_i - \frac {\sum x_iy_i}{\sum x_i^2}\sum x_i=0$?

Is the sum equal zero or not? I only know that $\sum y_i =n\bar y$ and $\sum x_i =n\bar x$ but what to do with the fraction term?

  • $\begingroup$ Juggling the terms around you should be able to get a relevant consequence of that equality. You can then try to work out if that consequence holds in general or not. $\endgroup$
    – cangrejo
    Oct 18, 2017 at 12:59

1 Answer 1


The residuals are perpendicular to the regressor vectors $x_{j,i}$. That means for each $j$ we have: $$\sum_{i=1}^{i=n} x_{j,i} \epsilon_i = 0$$

So if one of the regressors vectors, say $x_{1,i}$, is an intercept term (that means $x_{1,i} = 1$, or a constant instead of 1, for all $i$) then we always have: $$\sum_{i=1}^{i=n} \epsilon_i = 0$$ which means that then the residuals sum to zero.

If this is not the case (as is your case), then the residuals do not necessarily sum to zero.

You will get:

$$\sum \epsilon_i = \sum y_{i} - \beta_1 \sum x_{1,i} - \beta_2 \sum x_{2,i} -... - \beta_{k-1} \sum x_{k-1,i} - \beta_k \sum x_{k,i}$$

In your case when there is only a single regressor

$$\sum \epsilon_i = \sum y_{i} - \beta_1 \sum x_{1,i}$$

which is only zero iff

$$\frac{\sum y_{i}}{\sum x_{i}}= \frac{\sum x_{i} y_i }{\sum x_{i}^2}$$

(where we applied a division, possibly by zero, so also the case $\sum x_{i} = \sum y_{i} = 0$, counts)


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