Asymptotic Distribution of Covariance I've seen a lot of questions revolving around the asymptotic distribution regarding the sample variance, such as $\sqrt{n} (s^2 - \sigma^2) \xrightarrow{d} N(0, \mu_4 - \sigma^4)$, however, what would the asymptotic distribution be if that was the covariance? 
That is to say, what is the asymptotic distribution of $\sqrt{n} (\hat{\sigma}_{XY} - \sigma_{XY})$ where $\hat{\sigma}_{XY}$ is the sample covariance and $\sigma_{XY}$ is the true covariance?
We are assuming that $(X_1, Y_1), ..., (X_n, Y_n)$ are iid with joint distribution of $F(x,y)$.
 A: We assume zero-mean variables for simplicity. Then, since the sample is i.i.d, it is ergodic, and sample means converge to theoretical expected values. We must also assume that all moments, individual and joint that will appear, exist and are finite. Given these, we have
$$S_n \equiv \hat \sigma_{xy} = \frac 1n \sum_{i=1}^n x_iy_i \to_p E(XY) = \sigma_{xy}$$
Also,
$$E\left(\frac 1n \sum_{i=1}^n x_iy_i\right) = E(XY)$$
for finite samples.
Consider now the variance of this sum, writing for clarity $x_iy_i = w_i$,
$$\text{Var}\left(\frac 1n \sum_{i=1}^n w_i\right) = \frac 1 {n^2}\cdot \left[E\left(\sum w_i\right)^2 + n(n-1)\sum E(w_i)E(w_j) - \left (E\left[\sum w_i\right]\right)^2\right]$$
$$=\frac 1 {n^2} \cdot\left [ nE(w_i^2) + n(n-1)\sigma_{xy}^2- n^2 \sigma_{xy}^2 \right]= \frac 1 {n} \cdot\left [ E(x_i^2y_i^2) - \sigma_{xy}^2 \right]$$
Then we have that the standardized expression
$$\frac{S_n - E(S_n)}{\sqrt{\text{Var}(S_n)}} = \frac {\sqrt{n}(\hat \sigma_{xy} - \sigma_{xy})}{\sqrt{E(x_i^2y_i^2) - \sigma_{xy}^2}} \to_p N(0,1)$$
due to the i.i.d. / ergodicity of the sample, and existing and finite moments. It follows that 
$$\sqrt{n}(\hat \sigma_{xy} - \sigma_{xy}) \approx N\left(0, E(x_i^2y_i^2) - \sigma_{xy}^2\right) $$
