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It is understood that the estimated ordinary least squares (OLS) regression line will pass through the average $x$ and $y$, but if there is no intercept it will no longer pass through ($\bar{x}$, $\bar{y}$). May I know what is the explanation for that?

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If a regression line must pass through a point $(h,k)$ then the slope which minimises the sum of squares of the residuals is $$\widehat\beta= \frac{\overline{(x - h) (y - k)}}{\overline{(x - h)^2}}$$

In simple linear regression, you can further minimise the sum of squares by choosing that point to be $(\bar x,\bar y)$, i.e. the regression line passes though the average point.

But if there is no intercept then you force that point to be $(0,0)$ and you get $\widehat\beta= \frac{\overline{xy}}{\overline{x^2}}$, so the regression line also passes through $\left(\bar x, \frac{\overline{xy}}{\overline{x^2}}\bar x\right)$ and that will usually not be the average point; the sum of squares of residuals will usually be higher than than if you had flexibility over $(h,k)$

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The reason is very similar to the reason that the residual terms will equal zero.

Why do residuals in linear regression always sum to zero when an intercept is included?

For a simple linear regression (not for OLS in general which we will show later) you have the formula:

$$y_i = \hat\alpha + \hat\beta x_{i} + \epsilon_i$$

and for the average this becomes:

$$\bar{y_i} = \alpha + \beta \bar{x_{i}} + \bar{\epsilon_i}$$

And the OLS regression with an intercept included will be such that the residual terms $\bar{\epsilon}$ are equal to zero.

The reason is that we must have the derivatives of the residual terms equal to zero

$$ \frac{\partial}{\partial \hat\alpha} \sum \epsilon_i^2 =0$$

which will lead to

$$ \begin{array}{} \frac{\partial}{\partial \hat\alpha} \sum \epsilon_i^2 &=& \frac{\partial}{\partial \alpha} \sum (y_i - \hat\alpha - \hat\beta x_{i})^2\\ &=& \sum -2 \hat\alpha (y_i - \hat\alpha + \hat\beta x_{i})\\ &=& -2 \hat\alpha \sum \epsilon_i &=& 0 \end{array}$$


Intuitively

If the line does not go through the point $\bar{x},\bar{y}$ then this will be equivalent to the sum of the errors being unequal to zero, and we will be able to improve the sum of squared residuals $\sum \epsilon_i^2$ by shifting the line up or down.

When the line goes through the point $\bar{x},\bar{y}$, then a change of the parameter $\hat{\alpha}$ does not improve the fit (the derivative is zero).

comparison


It depends on the point of view

The OLS regression curve only passes necessarily through the point $\hat{x},\hat{y}$ when you have linear regression. For OLS that is used to express a non-linear function with an intercept term you will still have the situation that the sum of residuals will be equal to zero, but you do not have anymore the relationship $$\bar{y_i} = \alpha + \beta \bar{x_{i}} + \bar{\epsilon_i}$$

An example is a quadratic cuvre

quadratic fit

However, you could express this regression of $y$ with a quadratic curve (which is seemingly a non-linear function) as a linear function of $x$ and $x^2$ by means of viewing it as multivariate linear regression. Then it will also pass through the means.

$$\bar{y_i} = \alpha + \beta_1 \overline{x_{i}} + \beta_2 \overline{x_{i}^2} + \bar{\epsilon_i}$$

because now you express it as a multidimensional case.

enter image description here

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  • $\begingroup$ +1 Very clear and insightful, thank you. $\endgroup$ – whuber Nov 11 '20 at 16:05

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