Why does $(\bar{Y}, Y_i - \bar{Y})$ being normally distributed imply that $\bar{Y}$ and $Y_i - \bar{Y}$ are independent for all $i$?

Question

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.

Why does $(\bar{Y}, Y_i - \bar{Y})$ being normally distributed imply that $\bar{Y}$ and $Y_i - \bar{Y}$ are independent for all $i$?

Answer 1

For a multivariate normal distribution, independence and uncorrelated are the same. If you only have marginal (but not joint) normality, this fails to be the case. This comes from the definition of the multivariate normal density where all of the dependence is contained in the covariance matrix.

If the covariance matrix is diagonal, then the joint density can be written as the product of the marginal densities. This is the very definition of independence.