CLT and Confidence Limits | ISI MStat 2016 PSB Problem 8

This is a problem from ISI MStat Examination 2016. This primarily tests the student's knowledge in finding confidence intervals and using the Central Limit Theorem as a useful approximation tool.

The Problem:

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

(a) Show that there exists a function \( c(\rho) \) such that :

\(\lim_{n \rightarrow \infty} P(\sqrt{n}(\bar{X}-\bar{Y}) \le c(\rho) )=\Phi(1)\) , where \( \Phi \) is the cdf of \( N(0,1) \)

(b)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
\(\lim_{n \rightarrow \infty} P(L_n<\rho<1)=\alpha \) .

Prerequisites:

(a)The Central Limit Theorem and Convergence of Sequence of RVs

(b)Idea of pivots and how to obtain confidence intervals.

Solution:

(a) See that \( E(\bar{X}-\bar{Y})=0 \).

See that \( \text{Var}(\bar{X}-\bar{Y})=\frac{\sum_{i=1}^{n} \text{Var}(X_i-Y_i)}{n^2} =\frac{2(1-\rho)}{n} \)

Take \( c(\rho)=2(1-\rho) \).

By the Central Limit Theorem,

see that \( P(\frac{\sqrt{n}(\bar{X}-\bar{Y})}{\sqrt{2(1-\rho)}} \le 1) \rightarrow \Phi(1) \) as \(n \rightarrow \infty \)

(b) Let \( Z_i = X_i - Y_i \)

\(Z_i\)'s are iid with \( E(Z_i)=0 \) and \(V(Z_i)=2(1- \rho) \)

Again, by CLT, \( \frac{\sqrt{n} \bar{Z}}{\sqrt{2-2\rho}} \stackrel{L}\longrightarrow N(0,1) \)

Use this as the pivot to obtain an asymptotic confidence interval for \( \rho \).

See that \( P(\frac{\sqrt{n} \bar{Z}}{\sqrt{2-2\rho}}) \ge \tau_{\alpha})=\alpha \) , where \( \tau_\alpha \) : upper \(\alpha\) point of \(N(0,1) \).

Equivalently , you can write, \( P( \rho \ge 1- \frac{n \bar{Z}^2 }{2 \tau_{\alpha}^2} ) =\alpha \) , as \( n \rightarrow \infty \).

Thus , \(L_n=1- \frac{n \bar{Z}^2 }{2 \tau_{\alpha}^2} \).

Food For Thought:

Suppose \(X_1,..,X_n \) are equi-correlated with correlation coefficient \( \rho \).

Given, \( E(X_i)= \mu \) , \(V(X_i) = \sigma_i ^2 \), for \(i=1,2,..,n\).

Can you see that \( \rho \ge -\frac{1}{\frac{(\sum \sigma_i)^2}{\sum \sigma_i ^2} -1 } \) ?

It's pretty easy right? Yeah, I know 😛 . I am definitely looking forward to post inequalities more often .

This may look trivial but it is a very important result which can be used in problems where you need a non-trivial lower bound for \( \rho \) .

Well,well suppose now \(Y_i=X_i - \bar{X} \).

Can you show that a necessary and sufficient condition for \(X_i\) , \( \{ i=1,2,..,n \} \) to have equal variance is that \( Y_i \) should be uncorrelated with \( \bar{X} \)?

Till then Bye!

Stay Safe.

Correlation of two ab(Normals) | ISI MStat 2016 PSB Problem 6

This problem is an interesting application of the moment generating function of normal random variable to see how the correlation behaves under monotone function. This is the problem 6 from ISI MStat 2016 PSB.

Problem

Suppose that random variables \(X\) and \(Y\) jointly have a bivariate normal distribution with \(\mathrm{E}(X)=\mathrm{E}(Y)=0, {Var}(X)={Var}(Y)=1,\) and
correlation \(\rho\). Compute the correlation between \(e^{X}\) and \(e^{Y}\).

Prerequisites

Solution

\(M_X(t) = E(e^{tX})\) is called the moment generating function.

Now, let's try to calculate \( Cov(e^X, e^Y) = E(e^{X+Y}) - E(e^X)E(e^Y)\)

For, that we need to have the following in our arsenal.

\( \sigma^2 = Var(X+Y) = Var(X) + 2Cov(X,Y) + Var(Y) = 1 + 2\rho + 1 = 2(1+\rho) \).

Now observe the following:

Important Observation

\( Cor(e^X, e^Y)\) and \(Cor(X,Y) = \rho \) always have the same sign. Can you guess why? There is, in fact, a general result, which we will mention soon.

Now, we are left to calculate \(Var(e^X) = (Var(e^Y)\).

\(Var(e^X) = E(e^{2X}) - (E(e^{X}))^2 = M_X(2) - M_X(1)^2 = e^{\frac{4}{2}} - (e^{\frac{1}{2}})^2 = e^2 - e^1 = Var(e^Y)\).

Therefore, \( Cor(e^X, e^Y) = \frac{e(e^{\rho} - 1)}{e^2 - e^1} =\frac{e^{\rho} - 1}{e - 1} \).

Observe that the mininum correlation of \(e^X\) and \(e^Y\) is \(\frac{-1}{e}\).

Back to the important observation

\( Cor(e^X, e^Y)\) and \(Cor(X,Y) = \rho \) always have the same sign. Why is this true?

Because, \( f(x) = e^x\) is an increasing function. So, if \(X\) and \(Y\) are positively correlated then, as \(X\) increases, \(Y\) also increases in general, hence, \(e^X\) also increases along with \(e^Y\) hence, the result, which is quite intuitive.

Observe that in place of \( f(x) = e^x \) if we would have taken, any increasing function \(f(x)\), this will be the case. Can you prove it?

Research Problem of the day ( Is the following true? )

Let \(f(x)\) be an increasing function of \(x\), then