# Supposing joint normality, is a pair of asymptotically uncorrelated sequences also asymptotically independent?

Let's say there are two sequences of random variables $(X_n, Y_n)$ and we know that

1. For each $n$, $(X_n, Y_n)$ is normally distributed.

2. $\mathrm{cov}(X_n, Y_n) \rightarrow 0$ as $n \rightarrow \infty$.

I am wondering if we can conclude that $X_n$ and $Y_n$ are asymptotically independent (maybe not?), that is, for any Borel sets $A$ and $B$, $$\mathbb{P}(X_n \in A, Y_n \in B) - \mathbb{P}(X_n \in A)\mathbb{P}(Y_n \in B) \longrightarrow 0,\text{ as } n \rightarrow \infty$$

By the two conditions, we know that for any $\theta_1$, $\theta_2$, $$\mathbb{E}[e^{i(\theta_1 X_n + \theta_2 Y_n)}]-\mathbb{E}[e^{i\theta_1 X_n}]\mathbb{E}[e^{i\theta_2 Y_n}] \longrightarrow 0$$

Then it seems we might deduce that for any periodic and bounded functions $\phi$ and $\varphi$, $$\mathbb{E}[\varphi(X_n)\phi(Y_n)]-\mathbb{E}[\varphi(X_n)]\mathbb{E}[\phi(Y_n)] \longrightarrow 0$$

But I have no idea if we can proceed to further replace $\varphi$ and $\phi$ with indicator functions (or functions in $C_0$ space?).. Any hint will be appreciated :-)

• 1. This is false in general e.g. $X_n = Y_n \simeq \mathcal(0,n^{-1})$, $A=B=(0,+\infty)$. What you really need to assume is that the correlation coefficient converges to $0$. 2. Your notion of asymptotic independence is not very good. Say, let $X,Y$ be independent Bernoulli, $X_n = X + Y/n$, $Y_n = Y$. I would call such sequences asymptotically independent. However, $P(X_n=1,Y_n=1) - P(X_n=1)P(Y_n=1)\not\to 0$, $n\to\infty$. – zhoraster Dec 21 '16 at 15:26