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Chapter 5: Problem 1

Let \(\bar{X}_{n}\) denote the mean of a random sample of size \(n\) from a distribution that is \(N\left(\mu, \sigma^{2}\right) .\) Find the limiting distribution of \(\bar{X}_{n}\).

Short Answer

Expert verified
The limiting distribution of \(\bar{X}_{n}\) as \(n\) approaches infinity is \(N(\mu, 0)\), which means for large \(n\), \(\bar{X}_{n}\) practically equals the population mean \(\mu\).

Step by step solution

01

Understanding the problem

Firstly, you need to understand that \(\bar{X}_{n}\) represents the mean of a random sample of size \(n\) from a normal distribution \(N\left(\mu, \sigma^{2}\right)\). The goal is to find the limiting distribution, i.e., as \(n\) approaches infinity.
02

Applying Central Limit Theorem

Secondly, remember that according to the Central Limit Theorem (CLT), for a large enough sample size, the distribution of the sample mean will approximate a normal distribution, no matter the shape of the population distribution. Since our population distribution is normally distributed, we can apply the CLT directly to find the limiting distribution of the sample mean.
03

Finding Mean and Variance

As per CLT, the sample mean will also follow a normal distribution whose mean is equal to the population mean (\(\mu\)) and variance is equal to the population variance (\(\sigma^{2}\)) divided by the sample size \(n\). Hence, the distribution of \(\bar{X}_{n}\) will be \(N(\mu, \sigma^{2}/n)\)
04

Limiting Distribution

As \(n\) approaches infinity, the variance of the sampling distribution of the mean will approach zero (because \(\sigma^{2}/n\) -> 0 as \(n\) -> infinity). Therefore the limiting distribution of \(\bar{X}_{n}\) as \(n\) approaches infinity is \(N(\mu, 0)\). This essentially means that for large \(n\), \(\bar{X}_{n}\) is pretty much constant and equal to \(\mu\).

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