Asymptotic joint distribution of sample mean and sample variance I have a square integrable strictly stationary time series $(r_t)$. Suppose that $(r_t)$ satisfies certain conditions such that
$$\sqrt{T}(\bar{r}_T-\mu_r) \rightsquigarrow N(0,\sigma_1^2)$$
$$\sqrt{T}(s_T^2-\gamma_r(0)) \rightsquigarrow N(0,\sigma_2^2)$$
$\mu_r := E[r_t]$ and $\gamma_r(0) := E[(r_t-\mu_r)^2]$. Here $\bar{r}_T$ denotes the sample mean, i.e. $\bar{r}_T = \frac{1}{T}\sum_{t=1}^Tr_t$. $s_T^2$ is the sample autocovariance at lag $0$, i.e.
$s_T^2 = \frac{1}{T}\sum_{t=1}^T(r_t-\bar{r}_T)^2$.
I want to find the limit distribution of
$$\sqrt{T}\begin{pmatrix}\bar{r}_T-\mu_r \\ s_T^2-\gamma_r(0)\end{pmatrix}$$
For this purpose I think the Cramer-Wold device is suitable. So let $a_1,a_2 \in \mathbb{R}$ such that either $a_1 \neq 0$ or $a_2\neq 0$. Then I need to find the asymptotic distribution of 
$$\sqrt{T}\begin{pmatrix}a_1 & a_2\end{pmatrix}\begin{pmatrix}\bar{r}_T-\mu_r \\ s_T^2-\gamma_r(0)\end{pmatrix}$$
I manipulated the expression above a bit but I am kind of stuck. I am also not sure whether this is the way to go. I would appreciate some help.
I realized that for my purposes the limit distribution of 
$$\sqrt{T}\begin{pmatrix}\bar{r}_T-\mu_r \\ \bar{r^2}_T-E[r_t^2]\end{pmatrix}$$
is also an acceptable answer.
Here $\bar{r^2}_T = \frac{1}{T}\sum_{t=1}^Tr_t^2$
 A: The problem I will try to solve is the asymptotic joint distribution of
$$\sqrt{T}\begin{pmatrix}\bar{r}_T-\mu_r \\ \bar{r^2}_T-E[r_t^2]\end{pmatrix}$$
I will write this as 
$$ \sqrt{n}\begin{pmatrix}\bar{X}_n-\mu \\ \bar{Y}_n-(\sigma^2 + \mu^2)\end{pmatrix}$$
where $X_1, \dots, X_n, \dots$ are i.i.d., for each $n$ we define $Y_n := X_n^2$, and then we write $\mathbb{E}X_1 =: \mu$ and $Var(X_1) =: \sigma^2$, so that $\mathbb{E}[Y_1] = \mathbb{E}[X_1^2] = Var(X_1) + (\mathbb{E}X_1)^2 = \sigma^2 + \mu^2$. This corresponds also to writing the indexing variable as $n$ instead of $T$ and the RV's as $X$ instead of $r$.
I will write it this way solely because I am more comfortable with this and want to avoid unnecessary mistakes due to discomfort with notation.
Anyway, it follows then by the multivariate central limit theorem that
$$ \sqrt{n}\begin{pmatrix}\bar{X}_n-\mu \\ \bar{Y}_n-(\sigma^2 + \mu^2)\end{pmatrix}  \overset{D}{\to} \mathscr{N}(0, \Sigma)$$
where $\Sigma$ is the matrix:
$$\begin{pmatrix} Var(X_1) & Cov(X_1, Y_1) \\ Cov(X_1, Y_1) & Var(Y_1) \end{pmatrix} = \begin{pmatrix} Var(X_1) & Cov(X_1, X_1^2) \\ Cov(X_1, X_1^2) & Var(X_1^2) \end{pmatrix}$$ $$ = \begin{pmatrix} \sigma^2 & \mathbb{E}(X_1^3) - \mathbb{E}(X_1)\mathbb{E}(X_1^2) \\ \mathbb{E}(X_1^3) - \mathbb{E}(X_1)\mathbb{E}(X_1^2) & \mathbb{E}(X_1^4) - (\mathbb{E}(X_1^2))^2 \end{pmatrix} $$ $$= \begin{pmatrix} \sigma^2 & \mathbb{E}(X_1^3) - \mu(\sigma^2 + \mu^2) \\ \mathbb{E}(X_1^3) - \mu(\sigma^2 + \mu^2) & \mathbb{E}(X_1^4) - (\sigma^2 + \mu^2)^2 \end{pmatrix}  \,.$$
To get from here to the joint asymptotic distribution of the sample mean and variance, first use the identity that the sample variance, $S_n^2$, is equal to $\bar{Y}_n - (\bar{X}_n)^2$. Therefore if we define:
$$ g: \begin{pmatrix} z_1 \\ z_2  \end{pmatrix} \mapsto \begin{pmatrix} z_1 \\ z_2 - z_1^2 \end{pmatrix} $$ this should give us the final result we want if we apply the Delta Method.
Please double check my work, but I think this gives that the asymptotic covariance matrix is:
$$\begin{pmatrix}\sigma^2  & \mathbb{E}(X_1 - \mu)^3 \\ \mathbb{E}(X_1 - \mu)^3 & \mathbb{E}(X_1 - \mu)^4 - \sigma^4 \end{pmatrix}$$
Here are some related links:


*

*Statistics SE: asymptotic distribution of sample variance of non-normal sample  

*Asymptotic distribution of variance in important special cases  

*Example 5.9: How to get from the first joint asymptotic distribution to the second  

*How to find the covariance of sample mean and sample variance $Cov(M,S^2)$ for Poisson distribution?
In the case that the $X_i$ are normal, one can show that $S_n^2$ and $\bar{X}_n$ are independent, likewise that $\mathbb{E}[(X_1 - \mu)^4] = 3\sigma^4$, which leads to everything simplifying considerably. It seems that people as a result are often tempted to ignore the non-normal case (e.g. here).
