Derivation of $SSE =S_{yy} - S_{xy}^2/S_{xx}$

Question

I have seen this result and I am trying to figure out how to derive it from $SSE = \sum(Y_i - \hat{Y})^2$. I know that $r = \frac{s_{xy}}{\sqrt{s_{xx}s_{yy}}} $ and I have seen online to use and substitute in $ S_{xy}^2=r^2s_{xx}s_{yy}$ but I am not sure where to do that or how to represent $SSE$ as a sum of squares.

Answer 1

the key point is, in regression analysis: $$r^2 = \frac{SSR}{SST} = \frac{L_{xy}^2}{L_{xx}L_{yy}} = \frac{S_{xy}^2}{S_{xx}S_{yy}}$$ hence to prove: $$SSE = L_{yy} - \frac{L_{xy}^2}{L_{xx}}$$ is to prove: $$SSE = L_{yy} - \frac{r^2 L_{xx} L_{yy}}{L_{xx}}$$ $$SSE = L_{yy} - r^2 L_{yy}$$ $$SSE = (1 - r^2) L_{yy}$$ $$SSE = (1 - \frac{SSR}{SST}) L_{yy}$$ $$SSE = \frac{SST - SSR}{SST} L_{yy}$$ by $$SST = SSR + SSE$$ hence $$SSE = \frac{SSE}{SST} L_{yy} $$ notice that $$SST = L_{yy} = \sum_{i=1}^{n} (y_i - \bar y)^2$$ hence $$SSE = SSE$$ $$Q.E.D$$

also be aware of the difference between $L_{yy}$ and $S_{yy}$ $$S_{yy} = \frac{1}{n-1}\sum_{i=1}^{n} (y_i - \bar y)^2$$ $$S_{xy} = \frac{1}{n-1}\sum_{i=1}^{n} (y_i - \bar y) (x_i - \bar x)$$ however $$L_{yy} = \sum_{i=1}^{n} (y_i - \bar y)^2$$ $$L_{xy} = \sum_{i=1}^{n} (y_i - \bar y) (x_i - \bar x)$$ which means $$L_{yy} = (n - 1)S_{yy}$$ $$L_{xx} = (n - 1)S_{xx}$$ $$L_{xy} = (n - 1)S_{xy}$$ but $$r^2 = \frac{L_{xy}^2}{L_{xx}L_{yy}} = \frac{S_{xy}^2}{S_{xx}S_{yy}}$$ holds since you can get rid of all coefficents