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.
Best Answer
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.