Covariance matrix of $(\bar{X}, \bar{X^2})$

Question

I found a pdf that claims that asymptotically: $(\frac{\sum X_i}{N}, \frac{\sum X_i^{2}}{N})$ converges in distribution to a bivariate random variable with mean $(\mu_1,\mu_2)$ and covariance matrix $A$: $$A= \begin{bmatrix} \mu_2-\mu_1^2 & \mu_3-\mu_1\mu_2 \\ \mu_3-\mu_1\mu_2 & \mu_4-\mu_2^2 \\ \end{bmatrix}$$ where $\mu_k$ are the population's moments. The thing that I can't seem to do on my own is how to find the covariance, I can't understand how it is $\mu_3-\mu_1\mu_2$. Thank you.

Answer 1

In your source it is not the covariance of $\frac{1}{N}\sum_{i=1}^{N}X_{i}$ and $\frac{1}{N}\sum_{i=1}^{N}X_{i}^{2}$ that is calculated, but the covariance of $\frac{\sqrt{N}}{N}\sum_{i=1}^{N}X_{i}=N^{-\frac{1}{2}}\sum_{i=1}^{N}X_{i}$ and $\frac{\sqrt{N}}{N}\sum_{i=1}^{N}X_{i}^{2}=N^{-\frac{1}{2}}\sum_{i=1}^{N}X_{i}^{2}$.

Other essential info (that you did not mention) is that the $X_i$ are iid with $\mathbb EX_1^4<\infty$.

Covariance is bilinear and symmetric. Further if $X,Y$ are independent and their covariance exists, then it equals $0$.

Applying this we find:

$$\begin{aligned}\mathsf{Cov}\left(N^{-\frac{1}{2}}\sum_{i=1}^{N}X_{i},N^{-\frac{1}{2}}\sum_{i=1}^{N}X_{i}^{2}\right) & =N^{-1}\mathsf{Cov}\left(\sum_{i=1}^{N}X_{i},\sum_{i=1}^{N}X_{i}^{2}\right)\\ & =N^{-1}\sum_{i=1}^{N}\sum_{j=1}^{N}\mathsf{Cov}\left(X_{i},X_{j}^{2}\right)\\ & =N^{-1}\sum_{i=1}^{N}\mathsf{Cov}\left(X_{i},X_{i}^{2}\right)\\ & =\mathsf{Cov}\left(X_{1},X_{1}^{2}\right)\\ & =\mathsf{E}X_{1}X_{1}^{2}-\mathsf{E}X_{1}\mathsf{E}X_{1}^{2}\\ & =\mu_{3}-\mu_{1}\mu_{2} \end{aligned} $$where $\mu_i:=\mathsf{E}X^i$ for $i=1,2,3,4$