ISI MStat PSB 2012 Problem 5 | Application of Central Limit Theorem

This is a very beautiful sample problem from ISI MStat PSB 2012 Problem 5 based on central limit theorem . Let's give it a try !!

Problem- ISI MStat PSB 2012 Problem 5

Let \( X_{1}, X_{2}, \ldots, X_{j} \ldots \) be i.i.d. \(N(0,1)\) random variables. Show that for any \(a>0\)

\( \lim {n \rightarrow \infty} P(\sum_{i=1}^{n} {X_i}^2 \leq a) = 0 \)

Prerequisites

Limit

Central Limit Theorem

Normal Distribution

Chi- Square Distribution

Solution :

\( X_{1}, X_{2}, \ldots, X_{j} \ldots \) are i.i.d. \(N(0,1)\) random variables .

Let \( S_n = \sum_{i=1}^{n} {X_i}^2 \) , then \( S_n \sim \chi^{2}(n) \) , where \( \chi^{2}(n) \) is Chi-Square distribution with n degrees of freedom .

Therefore , \( E(S_n)= n \) and \( Var(S_n)=2n \) .

In this type of problems obvious thing that would come to our mind is to apply Central Limit Theorem right ! Let's try to apply it .

Now by Lindeberg Levy Central Limit Theorem we can say \( \frac{S_n-E(S_n)}{\sqrt{Var(S_n)}} \) = \( \frac{S_n-n}{\sqrt{2n}} {\to }^{d} N(0,1) \) , as n approaches infinity.

So, \( \lim {n \rightarrow \infty} P(\sum_{i=1}^{n} {X_i}^2 \leq a) \)

= \( \lim {n \rightarrow \infty} P( \frac{S_n-n}{\sqrt{2n}} \le \frac{a-n}{\sqrt{2n}} ) \)

= \( \lim {n \rightarrow \infty} \Phi(\frac{a-n}{\sqrt{2n}}) \)

= \( \lim {n \rightarrow \infty} \Phi(\frac{a}{\sqrt{2n}}- \sqrt{\frac{n}{2}}) = \Phi(0- \infty) \) (Since \( \Phi(x) \) is right continuous ) \( = 0 \) .

Hence Proved .

Food For Thought

Let \( \{X_{1}: i \geq 1 \}\) be a sequence of independent random variables each having a normal distribution with mean 2 and variance 5.Then \( (\frac{1}{n} \sum_{i=1}^{n} x_{i})^{2} \) converges in probability to ?

Outstanding Statistics Program with Applications

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.

Central Limit Theorem by Simulation ( R Studio)

This post verifies central limit theorem with the help of simulation in R for distributions of Bernoulli, uniform and poisson.

Central Limit Theorem

Mathematicaly, in \(X_1, X_2, …, X_n\) are random samples taken from a popualaton with mean \(\mu\) and finte variance \(\sigma^2\) and \(\bar{X}\) is the sampe mean, then \(Z = \frac{\sqrt{n}(\bar{X}-\mu)}{\sigma} \to N(0,1) \).

Simulation

Pseudocode

N # Number of trials (population size)
n # Number of simulations
standardized_sample_mean = rep(0,n)
EX #Expectation
VarX #Variance
  for (i in 1:n){
    samp #Sample from any distribution
    sample_mean <- mean(samp) # Sample mean
    standardized_sample_mean[i] <- sqrt(N)*(sample_mean - EX)/sqrt(VarX)
#Standardized Sample Mean
  }
hist(standardized_sample_mean,prob=TRUE) #Histogram
qqnorm(standardized_sample_mean) #QQPlot

Bernoulli \(\frac{1}{2}\)

N <- 2000 # Number of trials (population size)
n <- 1000 # Number of simulations
standardized_sample_mean = rep(0,n)
EX <- 0.5
VarX <- 0.25
  for (i in 1:n){
    samp <- rbinom(1, size = N, prob = 0.05)
    sample_mean <- mean(samp) # sample mean
    standardized_sample_mean[i] <- sqrt(N)*(sample_mean - EX)/sqrt(VarX)
  }
par(mfrow=c(1,2))
hist(standardized_sample_mean,prob=TRUE)
qqnorm(standardized_sample_mean)

Uniform \((0,1)\)

N <- 2000 # Number of trials (population size)
n <- 1000 # Number of simulations
standardized_sample_mean = rep(0,n)
EX <- 0.5
VarX <- 0.25
  for (i in 1:n
){
    samp <- runif( N, 0, 1)
    sample_mean <- mean(samp) # sample mean
    standardized_sample_mean[i] <- sqrt(N)*(sample_mean - EX)/sqrt(VarX)
  }
par(mfrow=c(1,2))
hist(standardized_sample_mean,prob=TRUE)
qqnorm(standardized_sample_mean)

Poission(1)

N <- 2000 # Number of trials (population size)
n <- 1000 # Number of simulations
standardized_sample_mean = rep(0,n)
EX <- 1
VarX <- 1
  for (i in 1:n){
    samp <- rpois(N,1)
    sample_mean <- mean(samp) # sample mean
    standardized_sample_mean[i] <- sqrt(N)*(sample_mean - EX)/sqrt(VarX)
  }
par(mfrow=c(1,2))
hist(standardized_sample_mean,prob=TRUE)
qqnorm(standardized_sample_mean)

Central Limit Theorem by simulation graph

Exercise

Try for other distributions and mixtures and play around and verify yourself.

Stay Tuned! Stay Blessed!

Problem- ISI MStat PSB 2012 Problem 5

Prerequisites

Solution :

Food For Thought

Similar Problems and Solutions

Related Program

Outstanding Statistics Program with Applications

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The Problem:

Prerequisites:

Solution:

Food For Thought:

Central Limit Theorem

Simulation

Pseudocode

Bernoulli \(\frac{1}{2}\)

Uniform \((0,1)\)

Poission(1)

Exercise