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

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 $$i$$?

• I would have thought $(\bar{Y}, Y_i - \bar{Y})$ has a bivariate normal distribution. In that case, zero correlation implies independence as the joint density would be the product of the marginal densities Apr 15, 2021 at 10:24