how to prove SST follows a chi-square distribution with (m−1) degrees of freedom

Question

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

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?

Answer 1

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 $SST/\sigma^2 \sim \chi_{k-1}^2$.