I have the following example:
Let $Y_1, \dots, Y_n$ be an i.i.d. $N(\mu, \sigma^2)$. Note that $\sum_{i = 1}^n (y_i – \mu)^2 = \sum_{i = 1}^n (y_i – \bar{y})^2 + n(\bar{y} – \mu)^2$.
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 did the authors get from $\dfrac{1}{n^2} \text{Cov} \left( \sum_{j = 1}^n Y_j, nY_i – \sum_{j = 1}^n Y_j \right)$ to $\dfrac{1}{n^2} \left( (n – 1)\text{Var}(Y_i) – \sum_{j = 1, j \not= i}^n \text{Var}(Y_j) \right)$?
Best Answer
It comes from the distributive property of covariance: $$\begin{align}C&=\text{Cov} \left( \sum_{j = 1}^n Y_j, nY_i - \sum_{j = 1}^n Y_j \right)=\operatorname{cov}\left(\sum_{j=1}^n Y_j, nY_i\right)-\operatorname{cov}\left(\sum_{j=1}^n Y_j,\sum_{k=1}^n Y_k\right)\\&= n\sum_{j=1}^n\operatorname{cov}\left(Y_j,Y_i\right)-\sum_{j=1}^n\sum_{k=1}^n \operatorname{cov}(Y_j,Y_k)\\&=n\operatorname{cov}(Y_i,Y_i)-\sum_{j=1}^n\operatorname{cov}(Y_j,Y_j)=n\operatorname{var}(Y_i)-\sum_{j=1}^n\operatorname{var}(Y_j)\\&=n\operatorname{var}(Y_i)-\operatorname{var}(Y_i)-\sum_{j=1,j\neq i}^n\operatorname{var}(Y_j)\\&=(n-1)\operatorname{var}(Y_i)-\sum_{j=1,j\neq i}^n\operatorname{var}(Y_j)\end{align}$$