# The equivalence of SSR(x2,x1)−SSR(x1) = SSE(x1)−SSE(x1,x2) in extra sum of squares in a regression

(Ugarte, Probability and Statistics with R, 2E) In the book, it reads:
$SST$: Sum of squares of Total.
$SSE$: Sum of squares of Error.
$SSR$: Sum of squares of Regression.

Marginal increases when adding x2 to a model that already contains x1 will be denoted as

$SSR(x2|x1) = SSR(x2,x1) - SSR(x1)$

which is equivalent to

$SSR(x2|x1) = SSE(x1) - SSE(x1,x2)$.

Why is there an equivalence?

What I did:

$SSR(x2|x1) = SSR(x2,x1) - SSR(x1)$ (using $SST=SSR+SSE$)

$= SST(x2,x1) - SSE(x2,x1) - [SST(x1) - SSE(x1)]$

$= SSE(x1) - SSE(x2,x1) + [SST(x2,x1) - SST(x1)]$.

So, in fact, the question turns out to be why $SST(x2,x1) - SST(x1) = 0$!

$$SST(X) = \sum_{i=1}^N(y_i - \bar{y})^2$$
so it doesn't depend on $X$.
• So, it seems the exact value of that sum is irrelevant since the cancellation happens in $SST(x2,x1)−SST(x1)$ ? Mar 6 '17 at 15:56