SST = SSE + SSR, proof

#1
Hi,

I've been trying to figure out for a long time how to derive the starting equation of the proof. I have no idea how to come up with it. It's the third equation on the picture in the attachement.

Thank you very much for your help!
 
#2
I read into this a bit last night. I believe you can add and subtract [math]\hat{Y}_i[/math]. So, the right side of line 3 becomes: [math] \sum_{i = 1}^n (\hat{Y}_i-\bar{Y}+Y_i-\hat{Y}_i)^2[/math]. I think you have a [math] \hat{Y}[/math] in place of [math] \bar{Y}[/math]. Then expand from here?

This is the first line of the proof. I referenced https://web.njit.edu/~wguo/Math644_2012/Math644_Chapter 1_part4.pdf
Maybe we can finish the proof after fixing this small mistake?
 
Last edited:

Dragan

Super Moderator
#3
It is perhaps best to use the deviation form of the regression function i.e.,

[math]y_{i}=\hat{y_{i}}+e_{i}[/math].

Squaring both sides and summing over the sample yields,

[math]\sum y_{i}^{2}=\sum \hat{y_{i}}^{2}+\sum e_{i}^{2}+2\sum \hat{y_{i}}e_{i}[/math]

[math]=\sum \hat{y_{i}}^{2}+\sum e_{i}^{2}[/math]

[math]=\hat{\beta _{1}^{2}}\sum x_{i}^{2}+\sum e_{i}^{2}[/math]

because

[math] \sum \hat{y_{i}}e_{i}=0[/math].

As such, the various sums of squares above are:

[math] \sum y_{i}^{2} [/math] is the total variation (TSS),

[math]\sum \hat{y_{i}}^{2}=\hat{\beta _{1}^{2}}\sum x_{i}^{2}[/math] is the explained sum of squares from the regression (ESS),

[math]\sum e_{i}^{2}[/math] is the residual sum of squares (RSS).

Thus, it follows that TSS=ESS+RSS.