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 $i$?

  • $\begingroup$ 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 $\endgroup$
    – Henry
    Apr 15, 2021 at 10:24

1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.