I have this example of sufficiency:
Let $Y_1, \dots, Y_n$ be i.i.d. $N(\mu, \sigma^2)$. Note that $\sum_{i = 1}^n (y_i - \mu)^2 = \sum_{i = 1}^n (y_i - \bar{y})^2 + n(\bar{y} - \mu)^2$. Hence
$$\begin{align} L(\mu, \sigma; \mathbf{y}) &= \prod_{i = 1}^n \dfrac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{1}{2\sigma^2}(y_i - \mu)^2} \\ &= \dfrac{1}{(2\pi \sigma^2)^{n/2}}e^{-\frac{1}{2\sigma^2}\sum_{i = 1}^n (y_i - \bar{y})^2}e^{-\frac{1}{2\sigma^2}n(\bar{y} - \mu)^2} \end{align}$$
From Theorem 1, it follows that where $T(\mathbf{Y}) = (\bar{Y}, \sum_{i = 1}^n (Y_i - \bar{Y})^2)$ is a sufficient statistic for $(\mu, \sigma)$.
It then says the following:
We now show that $\bar{Y} \sim N(\mu, \frac{\sigma^2}n)$.
It is clear that
$$Y_1 + \dots + Y_n \sim N(n\mu, n\sigma^2)$$
and so
$$\bar{Y} \sim N\left( \mu, \frac{\sigma^2}n \right)$$
And then the following:
We show that $Y$ and $\sum_{i = 1}^n (Y_i - \bar{Y})^2$ are independent.
One can show that
$$\begin{align} \text{Cov}(\bar{Y}, Y_i - \bar{Y}) &= \dfrac{1}{n^2} \text{Cov} \left( \sum_{j = 1}^n Y_j, nY_i - \sum_{j = 1}^n Y_j \right) \\ &= \dfrac{1}{n^2} \left( (n - 1)\text{Var}(Y_i) - \sum_{j = 1, j \not= i}^n \text{Var}(Y_j) \right) \\ &= \dfrac{1}{n^2} ((n - 1) \sigma^2 - (n - 1)\sigma^2) \\ &= 0 \end{align}$$
Since $(\bar{Y}, Y_i - \bar{Y})$ is normally distributed and this implies $\bar{Y}$ and $Y_i - \bar{Y}$ are independent for all $i$. So $\bar{Y}$ and $(Y_1 - \bar{Y}, \dots, Y_n - \bar{Y})$ are also independent. This implies $\bar{Y}$ and $\sum_{i = 1}^n (Y_i - \bar{Y})^2$ are independent.
How is $\text{Cov}(\bar{Y}, Y_i - \bar{Y}) = \dfrac{1}{n^2} \text{Cov} \left( \sum_{j = 1}^n Y_j, nY_i - \sum_{j = 1}^n Y_j \right)$? It seems like it should be $\text{Cov}(\bar{Y}, Y_i - \bar{Y}) = \dfrac{1}{n} \text{Cov} \left( \sum_{j = 1}^n Y_j, nY_i - \sum_{j = 1}^n Y_j \right)$, no?
