# Central Limit Theorem | ISI MStat 2018 PSB Problem 7

This problem based on Central Limit Theorem gives a detailed solution to ISI M.Stat 2018 PSB Problem 7, with a tinge of simulation and code.

## Problem

Suppose $\left(X_{1}, Y_{1}\right), \ldots,\left(X_{n}, Y_{n}\right)$ is a random sample from a bivariate normal distribution with $\mathrm{E}\left(X_{i}\right)=\mathrm{E}\left(Y_{i}\right)=0, Var\left(X_{i}\right)=Var\left(Y_{i}\right)=1$
and unknown $Corr\left(X_{i}, Y_{i}\right)=\rho \in(-1,1),$ for all $i=1, \ldots, n .$ Define $W_{n}=\frac{1}{n} \sum_{i=1}^{n} X_{i} Y_{i}$ a) Is $W_{n}$ an unbiased estimator of $\rho ?$ Justify your answer.
(b) For large $n,$ obtain an approximate level $(1-\alpha)$ two-sided confi-
dence interval for $\rho,$ where $0<\alpha<1$.

### Prerequisites

• Probability Theory (Expectation, Variance, Covariance, Correlation Coefficient)
• Unbiased Estimator
• Moments of Univariate Normal Distribution
• Bivariate Normal Distribution and a Different Definition
• Central Limit Theorem

## Solution

(a)

Just compute the $E(W_{n}$).

$E(W_{n})$ = $\frac{1}{n} \sum_{i=1}^{n} E(X_{i} Y_{i})$ = $\frac{1}{n} \sum_{i=1}^{n} \rho = \rho$.

$\rho = E(X_{i} Y_{i}) - E(X_{i})E(Y_{i}) \overset{E(X_{i}) = E(Y_{i}) = 0}{=} E(X_{i} Y_{i})$.

So, $W_{n}$ is unbiased for $\rho$.

(b)

Observe that $\left(X_{i}, Y_{i}\right)$ and $\left(X_{j}, Y_{j}\right)$ are independent sample and therefore iid.

So, $\left(X_{i}Y_{i}\right)$ and $\left(X_{j}Y_{j}\right)$ are also iid.

Hence, computing the limiting distribution of $W_{n}$, flashes in our minds, the Central Limit Theorem. So, let's dig into it. But, for that we need the following:

• $E(W_{n}) = \rho$
• $Var(W_{n}) = \frac{1}{n} \sum_{i=1}^{n} Var(X_{i}Y_{i})$ = $\frac{1}{n^2} \sum_{i=1}^{n} Var(X_{1}Y_{1}) = \frac{Var(X_{1}Y_{1})}{n} \overset{Why? Think}{=} \frac{E(({X_{1}Y_{1})^2)}}{n}$

So, how to calculate the ${E((X_{1}Y_{1})^2)}$. For that

Two random variables $X$ and $Y$ are said to be jointly normal if they can be expressed in the form $X = aU + bV, Y = cU + dV$, where $U$ and $V$ are independent standard normal random variables.

Alternate Definition of Bivariate Normal

Why do we need this? Because, $X$ and $Y$ are not independent and they have a correlation coefficient between them.

Assume, $(X, Y)$ ~ $(X_1, Y_1)$.

Exercise: Using the above result, prove that $Y$ can be written as $Y = \rho X + \sqrt{(1-\rho^2)}V$, where $V$ ~ N(0,1) and $V$ is independent of $X$.

$Y^2 = \rho^2X^2 + (1-\rho^2)V^2 + 2\rho\sqrt{(1-\rho^2)}XV$

$E(X^2Y^2) = E(\rho^2 X^4 + (1-\rho^2)X^2V^2 + 2\rho\sqrt{(1-\rho^2)}X^3V ) = \\ \rho^2E(X^4) + (1-\rho^2)E(X^2V^2) = \rho^2E(X^4) + (1-\rho^2)E(X^2)E(V^2) = 3\rho^2 + (1-\rho^2) = 1 + 2\rho^2$.

Exercise: Justify the above steps, using the independence of $X$ and $V$.

We used the fact that $E(X^4) = 3$ if $X$ ~ N(0,1). Instead of computing the whole we will use the fact that $E(Z) = n$ and $Var(Z) = 2n$ if $Z$ ~ ${{\chi}_n}^2$.

Exercise: Prove that $E(X^4) = 3$ if $X$ ~ N(0,1) using the above hint that $X^2$ ~ ${{\chi}_1}^2$.

The final result, we got is the following:

$Var(W_{n}) = \frac{1 + 2\rho^2}{n}$.
$E(W_{n}) = \rho$.

Now use Central Limit Theorem.

$\frac{\sqrt{n}(W_{n} - \rho)}{\sqrt{1 + 2\rho^2}} \to N(0, 1)$

Therefore, $P( |\frac{\sqrt{n}(W_{n} - \rho)}{\sqrt{1 + 2\rho^2}}| \leq z_{\alpha / 2} ) = (1-\alpha)$.

So, $P(\left[W_{n} - z_{\alpha / 2} \left(\frac{\sqrt{1 + 2\rho^2}}{\sqrt{n}}\right) \leq \rho \leq W_{n} +z_{\alpha / 2} \left( \frac{\sqrt{1 + 2\rho^2}}{\sqrt{n}} \right) \right]) = (1-\alpha)$. Now, you have to square it to get a confidence interval for $\rho^2$.

But, we can use variance stablizing transformation (pivotal method).

Observe that $f(x) = \int \frac{1}{\sqrt{1+2u^2}} = ln|x+\sqrt{\frac{1}{2} + x^2} |$, which is an increasing and hence bijective function.

${\sqrt{n}(f(W_{n}) - f(\rho))} \to N(0, c)$. Calculate this constanc $c = f'(\rho)^2.{\sqrt{1 + 2\rho^2}}$

Now, try to find a confidence interval for $f(\rho)$ based on this. Then take the inverse of $f(x)$ to get a confidence interval for $\rho$.

## A Computational and Simulation Dimension

N <- 2000 # Number of random samples
# Target parameters for univariate normal distributions
v = NULL
rho <- 0.5
mu1 <- 0; s1 <- 1
mu2 <- 0; s2 <- 1
mu <- c(mu1,mu2) # Mean
sigma <- matrix(c(s1^2, s1*s2*rho, s1*s2*rho, s2^2),
2) # Covariance matrix
library(MASS)
for (i in 1:1000) {
bvn1 <- mvrnorm(N, mu = mu, Sigma = sigma ) # from MASS package
W = bvn1[,1]*bvn1[,2]
Wbar = mean(W)
v = c(v, Wbar)
}
hist(v, freq =  F)
sigma2 = sqrt(1 + 2*rho^2)/sqrt(N)
x = seq(0.4, 0.6, 0.00001)
curve(dnorm(x, rho, sigma2), from = 0, col = "red", add = TRUE)

This problem was a bit more mathematical and technical, but still, I hope that the simulation along with the proofs gave you a good reading experience. Stay Tuned!

### 3 comments on “Central Limit Theorem | ISI MStat 2018 PSB Problem 7”

1. Anupriya says:

The confidence interval contains the unknown parameter i.e. correlation coefficient(row). How?

1. Srijit Mukherjee says:

First of all, this is a large sample confidence interval. With respect to the large sample, the expectation of Wn when n large goes to rho. Hence, we are seeing the confidence interval around the mean. Just, expand that out. You will get the expression. See I have added a new portion. Thanks for your doubt. Stay tuned.

2. Ishan says:

Why is the variance of X1Y1 = E(X1Y1)^2?
E(X1Y1) is rho and not zero
Variance of X1Y1 should be 1+rho^2.

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