I have that $R^{2} = 1 - \frac{\text{RSS}}{\sum_{i=1}^{n}(Y_{i}-\bar{Y})^{2}}$. Also, $\text{RSS}= {\sum_{i=1}^{n}(Y_{i}-\hat{Y_{i}})^{2}}$ for the simplest linear model with only the intercept term. I also know that $\frac{1}{n}\sum_{i=1}^{n}(Y_{i}-\bar{Y})^{2}$ is the total variance for the intercept only model and that $\frac{\text{RSS}}{\frac{1}{n}{\sum_{i=1}^{n}(Y_{i}-\bar{Y})^{2}}}$ is approximately $\frac{\text{var. of model}}{\text{variance}}$.

However I still don't get why $R^{2}$ is the proportion of total variance of the data explained by the model.


2 Answers 2


There is an error in your equations, $RSS = \sum(Y_i - \hat{Y}_i)^2$

Maybe it would help not looking at so many equations to understand.

RSS is the sum of the residual variance, basically the sum of all the variance that the model can't explain.

Therefore $\frac{RSS}{\sum{(Y_i - \bar{Y})^2}}$ is $\frac{unexplained \ variance}{Sum \ of \ all \ variance}$


$1- \frac{unexplained \ variance}{Sum \ of \ all \ variance} = \frac{Sum \ of \ all \ variance - unexplained \ variance}{Sum \ of \ all \ variance} = \frac{explained \ variance}{Sum \ of \ all \ variance} $

Does this help?

  • $\begingroup$ Hi, could you please explain what you mean by the 'model can't explain'? $\endgroup$ Feb 8, 2017 at 19:56
  • $\begingroup$ So the residuals are the errors left after fitting, for each $y_i$. If the residual for $y_0$ is 0, then the model has perfectly predicted $y_0$ (in the observed data) and thus can completely explain/predict $y_0$. If the residual is not 0, then there is some variance in $y_0$ that the model is not explaining. Therefore, $\sum{(y_i - \hat{y}_i})$ is the sum of the predicted values, minus the actual values, so the sum of what is not explained by the model $\endgroup$ Feb 8, 2017 at 20:05
  • $\begingroup$ In your definition you have $\frac{1}{n}$ as a normalizing term for your variance. When considering proportions of variance you should either include that in both the numerator and denominator or neither, otherwise one may be orders of magnitude larger than the other. Try considering your definition of total variance without the normalizing term. $\endgroup$ Feb 9, 2017 at 20:13

We have $TSS = \sum_i (Y_i - \bar{Y})^2,\ RSS = \sum_i(Y_i - \hat{Y}_i)^2,\ ESS = \sum_i(\hat{Y}_i - \bar{Y})^2$

$TSS$ - total variance, $RSS$ - residual variance, $ESS$ - regression variance

From ANOVA identity we know that

$$TSS = RSS + ESS$$

So we have $R^2 = 1 - \frac{RSS}{TSS} = \frac{ESS}{TSS}$. From last equation you can clearly see that $R^2$ states how much "variance" is explained by the regression

  • $\begingroup$ Ah, I never really covered ANOVA in class so that's why I don't have it in my original post $\endgroup$ Feb 8, 2017 at 19:39

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.