/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 The proof of the Central Limit T... [FREE SOLUTION] | 91Ó°ÊÓ

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The proof of the Central Limit Theorem requires calculating the moment generating function for the standardized mean from a random sample of any distribution, and showing that it approaches the moment generating function of the standard normal distribution. Here we look at a particular case of the Laplace distribution, for which the calculation is simpler. a. Letting \(X\) have pdf \(f(x)=\frac{1}{2} e^{-|x|},-\infty

Short Answer

Expert verified
The MGF of the standardized mean approaches the standard normal MGF as the sample size increases.

Step by step solution

01

Verify the PDF of the Laplace Distribution

First, we need to verify that the given PDF \(f(x) = \frac{1}{2} e^{-|x|}\) is a valid PDF for the Laplace distribution. This involves ensuring that it integrates to 1 across the entire real line. Compute the integral \[\int_{-\infty}^{\infty} \frac{1}{2} e^{-|x|} dx = 1.\]Split the integral at 0 and compute each part:\[\int_{-\infty}^{0} \frac{1}{2} e^{x} dx + \int_{0}^{\infty} \frac{1}{2} e^{-x} dx = \left[ \frac{1}{2} e^{x} \right]_{-\infty}^{0} + \left[ -\frac{1}{2} e^{-x} \right]_{0}^{\infty} = 1.\]This confirms that the given function is a valid PDF.
02

Calculate the MGF of the Laplace Distribution

The moment generating function (MGF) is given by the integral \[M_X(t) = E[e^{tX}] = \int_{-\infty}^{\infty} e^{tx} \cdot \frac{1}{2} e^{-|x|} dx.\]Split the integral into two parts at zero:\[\int_{-\infty}^{0} e^{tx} \cdot \frac{1}{2} e^{x} dx + \int_{0}^{\infty} e^{tx} \cdot \frac{1}{2} e^{-x} dx.\]This gives:\[M_X(t) = \frac{1}{2} \left( \frac{1}{1-t} + \frac{1}{1+t} \right) = \frac{1}{1-t^2},\]provided \(-1 < t < 1\).
03

Define Standardized Mean for a Sample

Let \(X_1, X_2, ..., X_n\) be a random sample from the Laplace distribution. A standardized mean \(Y\) is given by \[Y = \sqrt{n} \left(\bar{X} - 0\right)/1,\]where \(\bar{X}\) is the sample mean. Thus, the standardized mean is simply \[Y = \sqrt{n} \bar{X}.\]
04

Derive the MGF of the Standardized Mean

Given that \(Y = \sqrt{n} \bar{X}\), we have:\[M_Y(t) = \left[M_X\left(\frac{t}{\sqrt{n}}\right)\right]^n.\]Substitute the MGF from step 2:\[M_Y(t) = \left( \frac{1}{1-(\frac{t}{\sqrt{n}})^2} \right)^n = \left(\frac{1}{1-\frac{t^2}{n}}\right)^n.\]
05

Show the Limit Matches Standard Normal MGF

Apply the limit for large \(n\), using the hint:\[\lim_{n \to \infty} \left(1+\frac{t^2}{n}\right)^{-n} = e^{-t^2}.\]Therefore, for large \(n\),\[\left(1-\frac{t^2}{n}\right)^{-n} \approx e^{t^2},\]and thus:\[M_Y(t) \to e^{t^2} = e^{\frac{t^2}{2}}.\]Thus, the limit of \(M_Y(t)\) approaches the MGF of a standard normal variable \(e^{\frac{t^2}{2}}\) as \(n\to\infty\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Laplace Distribution
The Laplace Distribution is a continuous probability distribution named after Pierre-Simon Laplace. It is used in various fields due to its property of modeling data with a central peak and heavy tails. The probability density function (PDF) of a Laplace distribution is given by:
  • \( f(x) = \frac{1}{2} e^{-|x|} \), for \(-\infty < x < \infty\).
This is a double-exponential distribution, which means it has an exponential decaying chance for values as they deviate from its mean. In this description, the distribution is symmetric around zero, which acts as its mean and mode.

When checking if a given function fits the Laplace distribution, we verify that the integral of its PDF over the entire real line equals 1. This shows it can be considered a valid probability distribution. The distribution is characterized by a sharp peak at the center (at the mean) and symmetric exponential decay on both sides. This shape is particularly useful in scenarios where outliers or extreme values are expected more frequently, such as in finance or error modeling.
Moment Generating Function
The Moment Generating Function (MGF) is a crucial tool in probability theory used to summarize all the moments of a random variable. Essentially, it offers a way to encapsulate the distribution's shape and characteristics succinctly. For a random variable \(X\), with PDF \(f(x)\), the MGF, \(M_X(t)\), is defined by:
  • \( M_X(t) = E[e^{tX}] = \int_{-\infty}^{\infty} e^{tx} f(x) \, dx \).
This function can often simplify the process of finding expected values of powers of \(X\).

In the context of the Central Limit Theorem, the MGF simplifies the task of proving the theorem, as it allows us to work algebraically with exponential functions. For the Laplace distribution, the MGF is calculated by splitting the integral of \(e^{tx}\cdot\frac{1}{2}e^{-|x|}\) into two parts at zero and evaluating separately, ensuring convergence for \(-1 < t < 1\). This calculated MGF then helps in establishing a connection with other standard distributions when imposing limiting conditions.
Standard Normal Distribution
The Standard Normal Distribution is a foundational element in statistics. It is a normal distribution with a mean of 0 and a standard deviation of 1. Its probability density function is expressed as:
  • \( \phi(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \).
This is often referred to as the \(Z\)-distribution and forms the bedrock for many statistical methodologies, including hypothesis testing and confidence interval construction.

A critical application of the Standard Normal Distribution is in the Central Limit Theorem, which states that under certain conditions, the sum (or average) of many independent, identically distributed random variables approximates a normal distribution, irrespective of the original distribution's shape.

In exercises involving the Central Limit Theorem, such as transforming a Laplace-distributed variable into a standardized normal variable, the MGFs are utilized to show how the distribution of the variable(s) approaches that of a standard normal. This is achieved by demonstrating that the MGF of the standardized mean approaches the MGF of a standard normal variable \(e^{t^2/2}\) as the sample size becomes large. This convergence showcases the underlying power and universality of the Central Limit Theorem across different densities.

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Most popular questions from this chapter

Suppose the proportion of rural voters in a certain state who favor a particular gubernatorial candidate is \(.45\) and the proportion of suburban and urban voters favoring the candidate is .60. If a sample of 200 rural voters and 300 urban and suburban voters is obtained, what is the approximate probability that at least 250 of these voters favor this candidate?

Two airplanes are flying in the same direction in adjacent parallel corridors. At time \(t=0\), the first airplane is \(10 \mathrm{~km}\) ahead of the second one. Suppose the speed of the first plane \((\mathrm{km} / \mathrm{h})\) is normally distributed with mean 520 and standard deviation 10 and the second plane's speed, independent of the first, is also normally distributed with mean and standard deviation 500 and 10 , respectively. a. What is the probability that after \(2 \mathrm{~h}\) of flying, the second plane has not caught up to the first plane? b. Determine the probability that the planes are separated by at most \(10 \mathrm{~km}\) after \(2 \mathrm{~h}\).

Carry out a simulation experiment using a statistical computer package or other software to study the sampling distribution of \(X\) when the population distribution is lognormal with \(E[\ln (X)]=3\) and \(V[\ln (X)]=1\). Consider the four sample sizes \(n=10,20,30\), and 50 , and in each case use 500 replications. For which of these sample sizes does the \(X\) sampling distribution appear to be approximately normal?

Suppose the sediment density \((\mathrm{g} / \mathrm{cm})\) of a randomly selected specimen from a region is normally distributed with mean \(2.65\) and standard deviation \(.85\) (suggested in "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants,"Water Res., 1984: 1169-1174). a. If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most \(3.00\) ? Between \(2.65\) and \(3.00\) ? b. How large a sample size would be required to ensure that the first probability in part (a) is at least .99?

Show that \(F_{p, v_{1}, v_{2}}=1 / F_{1-p, v_{2}, v_{1}}\).

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