Define $\bar{\bar{Y}}=\sum n_i \bar{Y}_{i.}/\sum n_i$ and $\bar{\theta}=\sum n_i\theta_i / \sum n_i$, where $Y_i \sim N(\theta,\sigma^2)$. How does can I show that $\frac{1}{\sigma^2}\sum^k_{i=1}n_i[(\bar{Y}_{i.}-\bar{\bar{Y}})-(\theta_i-\bar{\theta)}]^2 \sim \chi^2_{k-1}$ under ANOVA assumptions?
my work:
Let $\bar{U}_i=\bar{Y}_{i.}-\theta_i$, for $i=1,...,n$. So, $\bar{U}_i \sim N(0,\frac{\sigma^2}{n_i})$.
Let $\bar{\bar{U}}=\bar{\bar{Y}}-\bar{\theta}$. So, $\bar{\bar{U}} \sim N(0,\frac{\sigma^2}{\sum n_i})$.
The linear combination $\bar{U}_i-\bar{\bar{U}} \sim N(0,\sigma^2(\frac{1}{n_i}+\frac{1}{\sum n_i}))$.
Hence we can rewrite the original expression as $\frac{1}{\sigma^2}\sum^k_{i=1}n_i(\bar{U}_{i}-\bar{\bar{U}})^2$.
I feel like my work is close, but I messed up somewhere, and I need help finding my error. Due to my variance term of the distribution given by $\bar{U}_{i}-\bar{\bar{U}}$, the given expression does not appear to follow a $\chi^2_{k-1}$ distribution. Where did I mess up?
updated work:
I have that $\frac{\sum n_i \bar{U}_i^2}{\sigma^2}=\frac{\sum n_i (\bar{U}_i-\bar{\bar{U}})^2}{\sigma^2}+\frac{\sum n_i \bar{\bar{U}}^2}{\sigma^2}$, where $\frac{\sum n_i \bar{U}_i^2}{\sigma^2} \sim \chi^2_k$ and $\frac{\sum n_i \bar{\bar{U}}^2}{\sigma^2}\sim \chi^2_1$
However, I need to show that the two added terms on the right-hand side of the inequality are independent. How may I go about this?