here is my question. The setting is below:
$(X_1,Y_1),\ldots,(X_n,Y_n)$: random samples in a 2-dim'l distribution, $EX_1=EY_1=0,\:EX_1^2=EY_1^2=1,\:EX_1Y_1=\rho,\:EX_1^4<+\infty,\:EY_1^4<+\infty$.
I want to prove that $p\lim_{n\rightarrow\infty}\sqrt n\bar X\bar Y=0$. (converges in probability)
First, I have tried to use the following statement:
If $EX_n\rightarrow a,\:Var(X_n)\rightarrow 0$ as $n\rightarrow\infty$, then $p\lim_{n\rightarrow\infty}X_n=a$.
So, I computed $E[\sqrt n\bar X\bar Y]$. Since $(X_i,Y_i)\:(i=1,\ldots,n)$ are random samples, $X_i$ and $Y_j$ are independent for distinct $i,j$. It implies that $EX_iY_j=(EX_i)(EY_j)=0$ for distinct $i,j$ by assumption. Thus, $$ E[\sqrt n\bar X\bar Y]=\sqrt n\cdot\frac{1}{n^2}\cdot\bigg(\sum_{i=1}^n EX_iY_i+\sum_{i\neq j}EX_iY_j\bigg)=\frac{\rho}{\sqrt n}\rightarrow 0 $$ as $n\rightarrow\infty$. However, I cannot show that $Var(\sqrt n\bar X\bar Y)\rightarrow 0$ as $n\rightarrow\infty$. I have tried to prove it similarly but I cannot get the value $EX_1^2Y_1^2$. In fact, if I got the value, I could not solve this problem.
Could you give me a hint or another way to prove? Thanks in advance.