I am new to statistics and trying to prove that in simple linear regression, the $SS_{\text{Regression}}/\sigma^2$ can be expressed as the square of a standard normal. ( Where $SS_{\text{Regression}} = \sum_{1}^n(\hat Y_i - \bar Y)^2$ is the sum of squares due to regression)

I tried searching for Cochran's theorem but the theorem with heavy matrix theory is heavy for me right now. Is there a simpler proof for the case of simple linear regression?

Help is deeply appreciated!

  • $\begingroup$ stats.stackexchange.com/q/362590/119261 $\endgroup$ Mar 27 at 6:56
  • $\begingroup$ @StubbornAtom This article seems to show that the $MS_{\text{Residuals}}$ follows a $n-2$ chi square distribution. Whereas actually, my query is for the distribution of $MS_{\text{Regression}}$ $\endgroup$
    – MathMan
    Mar 27 at 16:04

Consider the model $$y_i=\alpha+\beta x_i+\varepsilon_i\quad,\,i=1,2,\ldots,n$$

The least square estimators of $\alpha,\beta$ are then given by

$$\hat\alpha=\bar y-\hat\beta \bar x\qquad,\qquad\hat\beta=\frac{s_{xy}}{s_{xx}}$$

where $$s_{xy}=\sum_{i=1}^n (x_i-\bar x)(y_i-\bar y) \quad \text{ and } \quad s_{xx}=\sum_{i=1}^n (x_i-\bar x)^2$$

When the errors $\varepsilon_i$ are distributed as independent $N(0,\sigma^2)$ variables, we have

$$\frac{\hat\beta \sqrt{s_{xx}}}{\sigma}\sim N\left(\beta \sqrt{s_{xx}},1 \right) \tag{1}$$


$$\hat y_i=\hat\alpha+\hat\beta x_i=\overline y+\hat\beta(x_i-\overline x)$$

Therefore, $$\frac{SSR}{\sigma^2}=\frac1{\sigma^2}\sum_{i=1}^n (\hat y_i-\overline y)^2=\left(\frac{\hat\beta \sqrt{s_{xx}}}{\sigma}\right)^2$$

From $(1)$, it follows that under normality of errors, $SSR/\sigma^2$ has a noncentral $\chi^2$ distribution with $1$ degree of freedom and noncentrality parameter $\beta^2 s_{xx}$. It has a (central) $\chi^2_1$ distribution provided $\beta=0$, i.e.,

$$\frac{SSR}{\sigma^2} \stackrel{\beta=0}\sim \chi^2_1$$

So it is not true in general that $SSR/\sigma^2$ can be expressed as the square of a standard normal variable; it is true only if the slope coefficient $\beta$ is zero.


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