For i.i.d sequence of random vectors $(X_i,Y_i), i=1,...,n)$ show that $\sqrt{n}(\bar{XY}- \bar X\bar Y -E(XY)+E(X)E(Y)) \to_d N(0,\alpha^2)$$,
assume all moments are finite. Also find an expression for $\alpha^2$
I know that
$\sqrt{n}((\bar X,\bar Y) - (E(X),E(Y)))\to_d N(0,\Sigma)$ by central limit theorem for random vectors.
And I know that similarly I should have
$\sqrt{n}(\bar{XY} - E(XY))\to_d N(0,\sigma_{xy}^2)$
But I dont believe I have a way to combine these to get the result. And also I dont believe there is a function of $(\bar X,\bar Y)$ that could give me $\bar{XY}$ to use delta method. So Im not sure what to do from here.
Similar to your last question, multivariate CLT tells us under finite fourth moments of $X_i,Y_i$,
$$W_i\equiv(X_i,Y_i,X_iY_i)'\text{ iid}\implies \sqrt n (\frac{1}{n}\sum_i W_i-E[W_1])\to_d N(0,V(W_1)).$$
Letting $\hat \sigma_{XY}$ and $\sigma_{XY}$ respectively denote the sample and population covariances between $X_i$ and $Y_i$, write
$$\sqrt n(\hat \sigma_{XY}-\sigma_{XY})=\sqrt n \left(g\left(\frac{1}{n}\sum_i W_i\right)-g\left(E[W_1]\right)\right)$$
where $$g:\mathbb{R}^3\to\mathbb{R}, g(a,b,c)=c-ab$$ and use delta method.