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I need to prove why we have the following result: When: $Y_i=\beta_0+\epsilon_i$, then: $\sum\frac{\epsilon_i^2}{\sigma^2}\sim \chi ^2(n-1)$

Thanks :)

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You've omitted a fair amount of information. If, as in many contexts in which this notation is used, $\varepsilon_1,\ldots,\varepsilon_n$ are normally distributed with expected value $0$ and variance $\sigma^2$, and are mutually independent, and $\beta_0$ is constant, i.e. not random, then the result is false. In that case, you'd have $$ \sum_{i=1}^n \frac{\varepsilon_i^2}{\sigma^2} \sim\chi^2_n. \tag{1} $$ Now let $\hat\beta_0$ be the sample mean $(Y_1+\cdots+Y_n)/n$, as opposed to the population mean $\beta_0$, and let $\hat\varepsilon_i$ be the residual $\hat\varepsilon_i=Y_i-\hat\beta_0$, as opposed to the error $\varepsilon=Y_i-\beta_0$. Then one has $$ \sum_{i=1}^n \frac{\hat\varepsilon_i^2}{\sigma^2} \sim\chi^2_{n-1}. \tag{2} $$ The reason for $(1)$ is that the $\chi^2_n$ distribution is the distribution of the sum of squares of $n$ independent random variables with a standard normal distribution.

The proof of $(2)$ is more involved: write $$ \begin{bmatrix} Y_1 \\ \vdots \\ Y_n\end{bmatrix} = \hat\beta_0\begin{bmatrix} 1 \\ \vdots \\ 1 \end{bmatrix} + \begin{bmatrix} Y_1-\hat\beta_0 \\ \vdots \\ Y_n-\hat\beta_0\end{bmatrix}. $$ The vector on the left has a distribution that is rotationally symmetric.

The second vector in the sum on the right side of $(2)$ is constrained to lie in the space in which the sum of the components is $0$. It is the orthogonal projection of the vector on the left side of $(2)$ onto that space. It inherits the rotational symmetry, but it should be apparent that its expected value is the zero vector.

Since it's distribution is rotationally symmetric and its in an $(n-1)$-dimensional subspace and has expected value $0$, the distribution of the sum of squares of its entries is the same as the distribution of the squares of the norms of $n-1$ independent random variables each normally distributed with expected value $0$ and variance $\sigma^2$, so the distribution you get is $\chi^2_{n-1}$.

That's a somewhat terse answer, but you should get clear about what the statement says and which statements are true and which are not before you try to understand why.

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Thank you, this was all I needed. – ChiGuy Feb 15 '13 at 13:10
@ChiGuy : Could you "accept" the answer in that case? – Michael Hardy Feb 15 '13 at 17:08
@ChiGuy Plz read the comment just above this. – kaka Jul 29 '13 at 15:57

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