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I am working on simple linear regression with the normal error assumption $Y_i=\beta_0+\beta_1 X_i+\epsilon_i$, $\epsilon_i \sim N(0,\sigma^2)$ where the estimated mean response is $\hat{Y}_i=b_0+b_1 X_i$ with $b_0= \bar{Y}-b_1\bar{X}$ and $b_1=\frac{\sum\limits_i(X_i-\bar{X})(Y_i-\bar{Y})}{\sum\limits_i(X_i-\bar{X})^2}$.

In the derivation of the distribution of the quantity $\frac{(b_1-\beta_1)}{s[b_1]}$, where $s[b1]$ is the standard error in $b_1$, I have encountered the claim that $$ \frac{SSE}{\sigma^2(n-2)}\sim \frac{\chi^2_{(n-2)}}{n-2} $$ where SSE is the sum of the square residual error $SSE=\sum\limits_i(Y_i-\hat{Y}_i)^2$.

This seems to me to imply that $\frac{Y_i-\hat{Y}_i}{\sigma}\sim N(0,1)$. However, I have not been able to prove the above facts.

Can anybody help me understand why the above quantity is $\chi^2_{(n-2)}$ distributed? THANKS!

eyehearyou
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1 Answers1

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Note that $b_0 \sim N( \beta_0, \sigma^2_{b_0} )$ and $b_1 \sim N( \beta_1, \sigma^2_{b_1} )$, hence $ \hat{Y}_i = b_0 + b_1X_i|Y_i \sim N(Y_i, \sigma^2) $. However, note that $\{ \hat{Y}_i\}_{i=1}^n$ are not independent as from the normal equations you have $\sum_{i=1}^n\hat{Y}_i= \sum_{i=1}^n X_i\hat{Y}_i = 0$, hence the degrees of freedom of the $\chi^2$ distribution is not $n$, but $n-2$.

V. Vancak
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