Assuming a simple linear regression model , $n_1$ points are sampled at $X_1$ and $n_2$ at $X_2$ and let $\bar{Y_1} , \bar{Y_2}$ be the averages at $ X_1 , X_2$ respectively. The problem that I am struggling with is to show that that the regression line with least squares estimates of parameters passes through the points $(X_1,\bar{Y_2}),(X_2,\bar{Y_2})$.

I have tried putting in $X_1$ into the equation and hoping to get $\bar{Y_1}$ back to prove that the point lies on the line but I end up with a term looking like $$\frac{n_1\bar{Y_1}+n_2\bar{Y_2}}{n_1+n_2}+ \frac{n_1(X_1-\bar{X})\bar{Y_1}+n_2(X_2-\bar{X})\bar{Y_2}}{\sum(X_i-\bar{X})^2} $$

I'm not sure if I've made a mistake here or that this is actually reducible to $\bar{Y_1}$ and I just haven't noticed how.

  • $\begingroup$ Please add the self-study tag $\endgroup$
    – Peter Flom
    Dec 3, 2019 at 12:21

1 Answer 1


To find out if a point lies on the line, we can plug the values in for x and y just like in regular algebra.

Recall the simple regression line formula is: $$\hat{y} = \hat{\theta}_0 + \hat{\theta}_1 x$$

Plugging in $(\bar{x}, \bar{y})$, we get: $$\bar{y} = \hat{\theta}_0 + \hat{\theta}_1 \bar{x}$$

We can substitute $\hat{\theta}_0 = \bar{y} - \hat{\theta}_1 \bar{x}$ to get: $$\bar{y} = \bar{y} - \hat{\theta}_1 \bar{x} + \hat{\theta}_1 \bar{x}$$

Which simplifies to: $$\bar{y} = \bar{y}$$

The equation is valid, which means the point $(\bar{x}, \bar{y})$ lies on the line.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.