# Linear regression: R-squared and predicted response

On a business task, I perform simple linear regression with an independent variable $X$ and a dependent variable $Y$. The result is a linear function with slope $β_1$. If $R^2$ (coefficient of determination) = 1, I conclude that an increase of 1 in $X$ will result in an increase of $β_1$ in $Y$. If $R^2$ = 0, I conclude that an increase of 1 in $X$ will on average result in an increase of 0 in $Y$.

1. Are my conclusions correct?
2. If $R^2 = 0.5$, can I conclude that an increase of 1 in $X$ will on average result in an increase of $0.5β_1$ in $Y$?
3. If 2. is incorrect, is there a way to infer such a relationship from a linear regression model («an increase of 1 in $X$ will on average result in an increase of ... in $Y$»)?

Firstly lets remember $R^2$==1-$SS_{reg}$/$SS_{tot}$. Where $SS_{reg}$ is the sum of squared errors from your model:

$Y=\beta_{1}X+\epsilon$ (Where $\epsilon$ is the error in your prediction)

And $SS_{tot}$ is the sum of squared errors when you assume $Y$ = $\bar{Y}$. This is the same as the linear model $Y=\beta_{0}+\epsilon$, otherwise known as the NULL model.

1.Your first conclusions are correct. Because $R^2$ shows how well movements in $Y$ are explained by your model $\beta_{1}X+\epsilon$. An $R^2$ of 1 means your model perfectly explains movements of $Y$ that is $SS_{reg}$=0 i.e. $\epsilon=0$. So you can conclude an increase in $X$ by 1 will definitely result in $Y$ increasing by $\beta_{1}$. Likewise an $R^2$ of 0 means your model $\beta_{1}X+\epsilon$ does no better in explaining movements in $Y$ than just assuming $Y$ will be the average $\bar{Y}$. (That is $SS_{reg}$=$SS_{tot}$). i.e. $X$ has no bearing on $Y$.

2.If $R^2$=0.5 it means your model explains 50% of the movement in $Y$ when compared to simply stating $Y$ = $\bar{Y}$. The unexplained movement in $Y$ is captured by your error term $\epsilon$, which is white noise. This means when $X$ increases by 1 $Y$ on average will increase by $\beta_{1}$ but due to the unexplained white noise you might not see this in in the data i.e. in every instance of $Y_{i}$.

3.Even where $R^2$<1 your interpretation of $\beta_{1}$ is still the average increase in $Y$ when $X$ increases by 1. It just means there is noise that has been unaccounted for in the model. Adding other predictors into the model may well change the value of $\beta_{1}$, but given the model $\beta_{1}X+\epsilon$, $\beta_{1}$ should be interpreted as the average increase in $Y$ when $X$ increases by 1. The only time this is inappropriate is when $R^2$==0 because in that case $X$ has no relationship with $Y$.

• In this business case, the individual results are not important, only the long-term average. Does this mean that I from a business point-of-view can ignore $R^2$ as long as I can show with a significance test that $R^2$ > 0 ? – matthiash Mar 16 '17 at 11:55
• $R^2$ is useful in telling you whether your model $\hat{\beta}_{1}X_{1}$ is a good predictor of $Y$ or not so I wouldn't ignore it but it doesn't change the interpretation of $\beta_{1}$, assuming $\hat{\beta}_{1}$ is statistically significant. You may want to add other predictors to your model that way you can say when holding for $X_{2}$, $X_{3}$, etc $Y$ increases on average by $\beta_{1}$ when $X_{1}$ increases by 1. At the moment your model isn't holding for any other predictors so $\hat{\beta}_{1}$ may not be a good estimation of the population $\beta_{1}$ – Morgan Ball Mar 16 '17 at 12:02
• To be clear I've used $\hat{\beta}_{1}$ in the above comment to differentiate between your estimated coefficient and the actual population coefficient $\beta_{1}$ you're trying to discover. In my answer any reference to $\beta_{1}$ is a reference to the estimated relationship between $X$ and $Y$. In reality because your not holding for other variables $\hat{\beta}_{1}$ may be a biased estimator of the population $\beta_{1}$ – Morgan Ball Mar 16 '17 at 12:09