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I was reading some lecture notes on simple linear regression where one section said that when the slope is 0 (hence, H0: beta = 0 is actually true), (SSY - SSE)/(DFY - DFE) estimates sigma squared. This doesn't make sense to me. If beta is actually 0, then shouldn't SSY and SSE be the same value? 

  SSY = ∑(Y-Ybar)^2
  SSE = ∑(Y-Yhat)^2

However, if beta = 0, that means the best estimate (Yhat) is essentially Ybar, which in turn means

  SSY = SSE =  ∑(Y-Ybar)^2

In other words

  (SSY - SSE)/(DFY - DFE) = 0/(DFY - DFE)

Is there something that I am missing? I hope someone here can elaborate on this.

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Are you able to post a copy of the lecture notes, or link to them? – Max Oct 30 '12 at 17:01
SSY = SSE if $\hat{\beta} = 0$, which is not the same as $\beta=0$. Randomness of the residual guarantees, in the case of continuous residuals, that $\hat{\beta}$ won't actually equal 0 even when $\beta=0$.. – jbowman Oct 30 '12 at 19:50
@jbowman Thanks. But if at the same time SSE/DFE also estimates sigma^2, then my question is how is SSE/DFE = (SSY - SSE)/(DFY - DEF). – user11392 Nov 1 '12 at 22:18
Max There is no further explanation. The text is basically when beta=0, (SSY - SSE)/(DFY - DFE) estimates sigma^2. – user11392 Nov 2 '12 at 0:18

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