Good day,

I will am currently doing some self-study on Simple Linear Regression. I understand the formula:

$$\hat{Y}_i=\beta_0+\beta_1 X_i$$

But what does it mean when the value of $\beta_1$ is 0? Does it mean that $X$ and $Y$ have no relationship?

Appreciate some advice and explanations please

  • 1
    $\begingroup$ Since you have $\hat y_i$ on the left, your right hand side should have $\hat \beta$'s. While you're editing, please remove the "e" from the end of the word 'formula'; you have only one formula so you should not use the plural. $\endgroup$ – Glen_b May 3 '14 at 10:32

If $X$ and $Y$ have no linear relationship, i.e. $\beta_1=0$, then no matter what value $X$ is, the predicted value for $Y$, $\hat{Y}$, is $\beta_0$. If you draw the regression line, it would be a flat line with the y-intercept at $\beta_0$. It might help to also think about the interpretation of the slope $\beta_1$ as the change in $\hat{Y}$ for a 1-unit change in $X$. Notice that if the slope is 0, then the value of $\hat{Y}$ does not change with $X$ (no linear relationship).

  • 2
    $\begingroup$ +1, though in respect of your last sentence, it's possible for the slope coefficient of a linear model to be 0 while there is nevertheless a relationship (for example if we fit a linear relationship, but the relationship is nonlinear in a way that's orthogonal to the linear term). I'd have said "no linear relationship" $\endgroup$ – Glen_b May 3 '14 at 10:30

@jsk 's answer is mostly correct, but perhaps needs more emphasis on the last sentence: No linear relationship. For example, if y = sin(x) and x goes from 0 to pi, then a linear regression of y on x will have a parameter estimate very very close to 0. In the real world, this might be approximated by the relationship between month of year (1 to 12) and average temperature.

There are also some U and inverse U shaped relationships. For example, the relationship between exam score and anxiety is inverse U shaped. $\beta_1$ isn't 0 in this case, but it doesn't reflect the relationship.


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