Get inspired by the success stories of our students in IIT JAM 2021. Learn More 

April 3, 2020

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.


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 \(0<\alpha<1\).


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



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 \( \rho \).


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
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)
Central Limit Theorem
This gives the view of our fact that \( \rho = 0.5\) and the corresponding variance. You can play around with the values.

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. 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.

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.