Example 2.2.2 fromAn Introduction to Generalized Linear Models. Show $Y_{j k}, \bar{Y}_{j}, \hat{\beta}_{j} $ are all independent This question is related to what I have asked in my previous post:
How to derive the covariance matrix between $\bar{y}$ and $\hat{\beta_c}$ where $\hat{\beta_c}$ is the OLS estimator of a linear model?.
The example is from the book An Introduction to Generalized Linear Models by  Annette J. Dobson, Adrian Barnett(2008)
The motivating example is about birthweight and gestational age.

A fairly general model relating birthweight to gestational age is
$$\mathrm{E}\left(Y_{j k}\right)=\mu_{j k}=\alpha_{j}+\beta_{j} x_{jk}$$
where $x_{j k}$ is the gestational age of the $k$ th baby in group $j$. The intercept parameters $\alpha_{1}$ and $\alpha_{2}$ are likely to differ because, on average, the boys were heavier than the girls. The slope parameters $\beta_{1}$ and $\beta_{2}$ represent the average increases in birthweight for each additional week of gestational age. The question of interest can be formulated in terms of testing the null hypothesis $\mathrm{H}_{0}$ :$\beta_{1}=\beta_{2}=\beta($ that is, the growth rates are equal and so the lines are parallel) against the alternative hypothesis $\mathrm{H}_{1}: \beta_{1} \neq \beta_{2}$ We can test $\mathrm{H}_{0}$ against $\mathrm{H}_{1}$ by fitting two models
$$
\begin{array}{l}
\mathrm{E}\left(Y_{j k}\right)=\mu_{j k}=\alpha_{j}+\beta x_{j k} ; \quad Y_{j k} \sim \mathrm{N}\left(\mu_{j k}, \sigma^{2}\right) \\
\mathrm{E}\left(Y_{j k}\right)=\mu_{j k}=\alpha_{j}+\beta_{j} x_{j k} ; \quad Y_{j k} \sim \mathrm{N}\left(\mu_{j k}, \sigma^{2}\right)
\end{array}
$$

Then later the book suggests:
$$
\begin{aligned}
Y_{j k} & \sim \mathrm{N}\left(\alpha_{j}+\beta_{j} x_{j k}, \sigma^{2}\right) \\
\bar{Y}_{j} & \sim \mathrm{N}\left(\alpha_{j}+\beta_{j} \bar{x}_{j}, \sigma^{2} / K\right) \\
b_{j} & \sim \mathrm{N}\left(\beta_{j}, \sigma^{2} /\left(\sum_{k=1}^{K} x_{j k}^{2}-K \bar{x}_{j}^{2}\right)\right)
\end{aligned}
$$
and claimed they are all independent.
The formula for $b_j$ is provided:
$$
b_{j}=\frac{K \sum_{k} x_{j k} y_{j k}-\left(\sum_{k} x_{j k}\right)\left(\sum_{k} y_{j k}\right)}{K \sum_{k} x_{j k}^{2}-\left(\sum_{k} x_{j k}\right)^{2}}
$$
I at first thought the proof will be easily carried out using matrix formation at first, which is why I asked my previous post. However, later I realized I may use other properties of covariance operation to finish the proof.

Below shows my sketch of how I demonstrate $\hat{\beta_j}$ are independent to $\bar{Y}_{ij}, \bar{y}_j$:
Since we have assumed normal distribution of the normal term, no contrarians between the estimator indicate they are independent.
Using the fact that $\operatorname{cov}(a X, b Y)=(a b) \operatorname{cov}(X, Y)$ :
If we look at $\operatorname{cov}\left(\hat{\beta}, Y_{j k}\right)$ from the model assumption we know, $\operatorname{cov}\left(Y_{j k}, Y_{j k}\right)=$ $\operatorname{var}\left(Y_{j k}\right),$ as samples are i.i.d. the covariance of random variable goes to $0$.
Basically, if you look at $\hat{\beta}_{j}=\frac{K\left(\sum_{k} x_{j k} y_{j k}\right)-\left(\sum_{k} x_{j k}\right)\left(\sum_{k} y_{j k}\right)}{K \sum_{k} x_{j k}^{2}-\left(\sum_{k} x_{j k}\right)^{2}},$ a random variable. The denominator is a constant
$(\equiv A)$ therefore can be extracted out from the Covariance operation. Finally, the operation will reduce to form: $$\operatorname{cov}\left(\hat{\beta}, Y_{i j}\right)=\frac{K x_{j k}-K x_{j k}}{A}\times\sigma^{2}=0$$
Also, $\bar{Y}_{j}=\frac{\sum_{k} y_{j k}}{K},$ if $\hat{\beta}$ is independent from $Y_{i j},$ so will it be independent from $\bar{Y}_{j}$ as $\operatorname{cov}\left(\hat{\beta}, \bar{Y}_{j}\right)=\sum_{k} \operatorname{cov}\left(\hat{\beta}, Y_{i j}\right) / k=0$

Notice that the expression for $\bar{Y}_j$ is not given and I think $$
\bar{Y}_{j}=\frac{\sum_{k} y_{j k}}{K}
$$should be the correct way to express it. However, using this expression I am unable to show independence between $\bar{Y}_j$ and $Y_{ij}$. The covariance of the two won't go to 0. However, I feel my expression has some problems as it seems I am getting the covariance conditional on the sex. On the other hand, intuitively it doesn't make sense a group average is uncorrelated with its observation... I am unable to decern where exactly is my problem and fix it. Could someone please point out to me the correct way to demonstrate the independence of these three random variables?
 A: I'd say that (a) a self-study tag is missing, because that claim actually is an exercise :) (b) you should take another path.
Dobson (2nd ed.) & Barnett (3rd ed.) say: "Using the results of Exercise 1.4 show that..." So my hint is based on that exercise.
In general, $\sum(y_i-\bar{y})^2=\sum y_2^2-2n\bar{y}$. Replacing $y_i$ with $y_i-\mu$, whose mean is $\bar{y}-\mu$, we get:
\begin{align*}
(n-1)S^2=\sum(y_i-\bar{y})^2&=\sum((y_i-\mu)-(\bar{y}-\mu))^2\\
&=\sum(y_i-\mu)^2-n(\bar{y}-\mu)^2
\end{align*}
hence,
\begin{align*}
\sum(y_i-\mu)^2&=\sum(y_i-\bar{y})^2+n(\bar{y}-\mu)^2
\end{align*}
Dividing by $\sigma^2$:
\begin{align*}
\frac{\sum(y_i-\mu)^2}{\sigma^2}&=\frac{(n-1)S^2}{\sigma^2}+\frac{n(\bar{y}-\mu)^2}{\sigma^2}
\end{align*}
Now:
\begin{align*}
\frac{\sum(y_i-\mu)^2}{\sigma^2}&\sim\chi^2_n \\
\frac{(n-1)S^2}{\sigma^2}&\sim\chi^2_{n-1}\\
\frac{n(\bar{y}-\mu)^2}{\sigma^2}&\sim\chi^2_1
\end{align*}
A chi squared r.v. is the sum of independent standard normal variables, and the sum of $X_1\sim\chi^2_h$ and $X_2\sim\chi^2_k$ is distributed as a $\chi^2_{h+k}$ r.v. only if $X_1$ and $X_2$ are independent.
Therefore $S^2$ and $\bar{y}$ are independent random variables.
What will happen if we replace $\mu$ with $\alpha+\beta x$ and $\bar{y}$ with $a+bx$?
HTH
