0
$\begingroup$

Consider this question,

Let $(X_1,Y_1),(X_2,Y_2),...,(X_n,Y_n)$ be independent and identically distributed pairs of random variables with $E(X_1) = E(Y_1), Var(X_1)= Var(Y_1) = 1$, and $Cov(X_1,Y_1) = \rho \in (-1,1).$

Given $\alpha \in (0,1)$, obtain a statistic $L_n$ which is a function of $(X_1,Y_1),(X_2,Y_2),...,(X_n,Y_n)$ such that \begin{align} \displaystyle \lim_{n\to\infty}P(L_n < \rho < 1) = \alpha \end{align}

My concern is that since it is given that $\rho \in (-1,1)$, can I drop the $1$ in the equality to be proven. I mean, will the statistic $L_n$ calculated such that it satisfies \begin{align} \displaystyle \lim_{n\to\infty}P(L_n < \rho) = \alpha \end{align} would also satisfy the given equation. And if not, how should I go about this problem?

$\endgroup$
  • $\begingroup$ See math.stackexchange.com/questions/3202067/…. $\endgroup$ – StubbornAtom May 1 at 10:10
  • $\begingroup$ I had calculated it the same way once I dropped off the $1$ from the equation. However, I want to ask that can I do that? $\endgroup$ – Sanket Agrawal May 1 at 10:42
  • $\begingroup$ The question asks for a lower confidence bound for the correlation; it is of course bounded above by $1$, whether you mention it or not. $\endgroup$ – StubbornAtom May 1 at 10:46
  • $\begingroup$ Okay, got your point. Thanks $\endgroup$ – Sanket Agrawal May 1 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.