change in $R^2$ linear regression

Question

I find this question about linear regression, "I have a n sample $(X,Y)_1,...,(X,Y)_n$, I do a linear regression based on those data (I've guessed that the predictor is X), and I calculate the $R^2$.

Now I created a 2n sample based on those data, my sample is $(X,Y)_1,... ,(X,Y)_n, ((X,Y)_{n+1}=(X,Y)_1),...((X,Y)_{2n}=(X,Y)_n)$. Basically, in order to build this 2n sample, I use the previous data that I have. Now the question is how the $R^2$ have changed if I do a linear regression ?

I would say that the $R^2$ doesn't change, using the formula $R^2= \frac{\text{explained variance}}{\text{total variance}}$, As we are just adding the same data. If we add regressors I know that the $R^2$ increases (overfitting that why we should measure of the goodness of the fit such that the adjusted $R^2$, or the AIC criterion)

Can you tell me if I have missed something ? I hope I have made myself clear

Answer 1

It doesn't change. Take in account that the determination coefficient $R^2$ is the square of the linear correlation coefficient $r$. Thus, if the correlation coefficient doesn't change, then $R^2$ won't change. Now the proof that $r$ won't change:

Given that $r$ is the correlation coefficient of the first sample and $r'$ the correlation coefficient of the second sample, we have to prove that $r=r'$. We have the following: $$ r'=\frac{\sum_{i=1}^{2n}{(x_i-\bar{X})(y_i-\bar{Y})}}{2n} $$ But because we just repeat the first half of the data, then: $$ r'=\frac{2\sum_{i=1}^{n}{(x_i-\bar{X})(y_i-\bar{Y})}}{2n}=\frac{\sum_{i=1}^{n}{(x_i-\bar{X})(y_i-\bar{Y})}}{n}=r $$

Proved!