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It is easy to prove SSE/variance follows a chi-square distribution with (n-m) degrees of freedom. But how to prove SSTreament/variance follows a chi-square distribution with (m−1) degrees of freedom? I can prove it by using moment generationg functions since I know the mgf of SSTotal/Variance and SSE/variance, then I can get the mgf of SSTreament/variance in reverse. But it is very easy to see SSE/variance follows a a chi-square distribution enter image description here

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but it's not easy to do it for SSTreament/Variance.

Is the moment generating function the only way to work out SSTreament/variance follows a a chi-square distribution?

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    $\begingroup$ Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. $\endgroup$ – gung Jan 23 '16 at 1:32
  • $\begingroup$ @gung I can't transfer it to chi squaire distribution $\endgroup$ – whoisit Jan 23 '16 at 1:37
  • $\begingroup$ Thank you for adding the tag. Please be sure you've read the wiki at the link, it contains our policy for this type of question. Please update the text of your question to include what you understand thus far, what you've tried & where you're stuck. $\endgroup$ – gung Jan 23 '16 at 1:42
  • $\begingroup$ Neither SSE nor SST follow a chi-square distribution, and the included image doesn't state that SST follows a chi-square distribution. It says something slightly different from that. Note that the null is assumed true, so the proof is actually quite similar to that for the SSE, but easier (because there's only one $\mu$). $\endgroup$ – Glen_b Jan 23 '16 at 1:58
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    $\begingroup$ For SST, the distribution applies under the null (as I already mentioned and as is explicitly stated in the image), so $\mu_1=\mu_2=\mu_3...$. There's no mixing, they're all the same distribution. $\endgroup$ – Glen_b Jan 23 '16 at 4:51
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Your proof via MGF is great actually. However, I can't think out a smarter direct proof than MGF, I can just provide a indirect method here.

Since you've already known the decomposition as following: \begin{align} SS = \sum_{i=1}^k\sum_{j=1}^{n_i}(X_{ij} - \bar{X}_{..})^2 = SSE + SST \end{align} and the general assumption (I changed some of your notations here): \begin{align} X_{ij} = \mu + \alpha_i + \epsilon_{ij}(1\le i \le k, 1\le j \le n_i) \end{align} where $\epsilon_{ij} \sim N(0,\sigma^2)$. If we are given a more specific conditions \begin{align} \alpha_1 = \alpha_2 = \cdots = \alpha_k = \alpha \end{align} then all $X_{ij}$ are from the same distribution $N(\mu+ \alpha, \sigma^2)$. So $SS$ is just a sample variance of size $N = \sum_{i=1}^k n_i$, which means $SS/\sigma^2 \sim \chi_{N-1}^2$. Noted here that $SST$ and $SSE$ are independent. Combined with $SSE/\sigma^2 \sim \chi_{N-k}^2$, which you've already known, and the additivity of Chi-distribution, we conclude here that $SST/\sigma^2 \sim \chi_{k-1}^2$.

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