# Finding $L_n$ so that $\lim_{n \rightarrow \infty} P(L_n<\rho<1)=\alpha$

Let $$(X_1,Y_1),(X_2,Y_2),..,(X_n,Y_n)$$ be iid pairs of random variables with $$E(X_1)=E(Y_1)$$, $$\text{Var}(X_1)=\text{Var}(Y_1)=1$$,and $$\text{cov}(X_1,Y_1)=\rho \in(-1,1)$$. Given $$\alpha>0$$ , obtain a statistic $$L_n$$ which is a function of $$(X_1,Y_1),(X_2,Y_2),..,(X_n,Y_n)$$ such that $$\lim_{n \rightarrow \infty} P(L_n<\rho<1)=\alpha$$.

Since the sample distribution of the correlation coefficient is quite hard to work with I am really not being able to crack this one. Can anyone help?

Let $$Z_i=X_i-Y_i$$, so that $$Z_1,Z_2,\ldots,Z_n$$ are i.i.d variables with zero mean and variance $$2-2\rho$$.

So by classical CLT,

$$\frac{\sqrt n\overline Z}{\sqrt{2-2\rho}}\stackrel{L}\longrightarrow N(0,1)\,,$$

where $$\overline Z=\frac{1}{n} \sum\limits_{i=1}^n Z_i=\overline X-\overline Y$$.

Using this pivot, you can readily obtain an asymptotic confidence interval for $$\rho$$.

In particular, you can start from $$\frac{\sqrt n\overline Z}{\sqrt{2-2\rho}}\ge z_{\alpha}$$

to get a lower bound for $$\rho$$ with limiting probability $$\alpha$$.

Here $$z_{\alpha}$$ is of course the $$(1-\alpha)$$th fractile of $$N(0,1)$$.