/*! 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 19 Suppose that \(X_{1}, X_{2}, \ld... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 19

Suppose that \(X_{1}, X_{2}, \ldots, X_{n}\) are i.i.d. \(N\left(\mu, \sigma^{2}\right)\). a. If \(\mu\) is known, what is the mle of \(\sigma ?\) b. If \(\sigma\) is known, what is the mle of \(\mu ?\) c. In the case above \((\sigma \text { known), does any other unbiased estimate of } \mu\) have smaller variance?

Short Answer

Expert verified
a. \(\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2\). b. \(\hat{\mu} = \bar{X}\). c. No, \(\bar{X}\) has the smallest variance.

Step by step solution

01

Identify the Problem For Part (a)

We need to find the maximum likelihood estimator (MLE) of \(\sigma\) given that \(\mu\) is known. The data \(X_{1}, X_{2}, \ldots, X_{n}\) are independent and identically distributed normal random variables with mean \(\mu\) and variance \(\sigma^2\).
02

Write the Likelihood Function For Part (a)

The likelihood function for \(X_1, X_2, \ldots, X_n\) is given by \[L(\sigma^2) = \prod_{i=1}^n \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(X_i - \mu)^2}{2\sigma^2}\right).\]
03

Simplify the Likelihood and Log-Likelihood For Part (a)

The log-likelihood function becomes:\[\log L(\sigma^2) = -n \log(\sqrt{2\pi\sigma^2}) - \frac{1}{2\sigma^2} \sum_{i=1}^n (X_i - \mu)^2.\]
04

Differentiate the Log-Likelihood For Part (a)

Differentiate \(\log L(\sigma^2)\) with respect to \(\sigma^2\) and set it to zero to find the MLE:\[\frac{d}{d\sigma^2}\log L(\sigma^2) = -\frac{n}{2\sigma^2} + \frac{1}{2\sigma^4} \sum_{i=1}^n (X_i - \mu)^2 = 0.\]
05

Solve for MLE For Part (a)

Solving the equation from the previous step, we find:\[\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2.\] This is the MLE of \(\sigma^2\) when \(\mu\) is known.
06

Identify the Problem For Part (b)

Now, we must find the MLE of \(\mu\) given that \(\sigma\) is known. The data \(X_1, X_2, \ldots, X_n\) still follow a normal distribution.
07

Write the Likelihood Function For Part (b)

The likelihood function is:\[L(\mu) = \prod_{i=1}^n \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(X_i - \mu)^2}{2\sigma^2}\right).\]
08

Differentiate the Log-Likelihood For Part (b)

Taking the derivative of the log-likelihood with respect to \(\mu\):\[\frac{d}{d\mu}\log L(\mu) = \frac{1}{\sigma^2} \sum_{i=1}^n (X_i - \mu).\] Setting this to zero, we solve for \(\mu\).
09

Solve for MLE For Part (b)

From the previous step, setting the derivative to zero gives:\[\sum_{i=1}^n X_i - n\mu = 0 \quad \Rightarrow \quad \hat{\mu} = \frac{1}{n}\sum_{i=1}^n X_i.\] This is the MLE of \(\mu\) when \(\sigma\) is known.
10

Analyze Variance For Part (c)

In this step, we verify if there exists a better unbiased estimator for \(\mu\) than \(\hat{\mu} = \bar{X}\). The mean \(\hat{\mu}\) is equal to \(\mu\) and the variance \(\text{Var}(\bar{X}) = \frac{\sigma^2}{n}\) is the smallest possible variance for an unbiased estimator.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Unbiased Estimator
An unbiased estimator is a statistical tool we use to estimate a parameter of a population based on sample data. What makes an estimator "unbiased" is that, on average, it hits the true value of the parameter it is estimating. This means the expected value of the estimator equals the parameter.
Consider the case where we have a set of observations, say, measurements of heights, and we want to determine the average height in the population. The sample mean, often denoted as \( \bar{X} \), is a common estimator used to estimate the true population mean, \( \mu \). For the sample mean to be unbiased, we need the expected value \( E(\bar{X}) \) to be equal to the population mean \( \mu \).
  • If \( \mu \) is known, the estimator \( \bar{X} \) is unbiased for estimating \( \mu \) because \( E(\bar{X}) = \mu \).
  • If \( \sigma \) is known, there might be questions about alternative estimators. However, the sample mean \( \bar{X} \) still proves to be unbiased.
Variance of Estimators
Variance is a measure of how much the estimates of the parameter vary across different samples. For an unbiased estimator, we are interested in finding one with the smallest possible variance because it indicates more precise estimates.
Consider an estimator \( \bar{X} \) used to estimate the parameter \( \mu \). The variance of this estimator, when assuming normal distribution with known variance \( \sigma^2 \), is given by \( \text{Var}(\bar{X}) = \frac{\sigma^2}{n} \), where \( n \) is the sample size.
  • This formula tells us that as the sample size increases, the variance of the estimator decreases, leading to more reliable estimates.
  • When evaluated in part (c) of the exercise, \( \bar{X} \) was confirmed to have the smallest possible variance for an unbiased estimator of \( \mu \), reinforcing its efficiency over other potential estimators.
Normal Distribution
The normal distribution, a cornerstone of statistics, is often used to model real-world variables like heights, test scores, or measurement errors. It's characterized by its bell-shaped curve, and two parameters: the mean (\( \mu \)) and the variance (\( \sigma^2 \)).
In the context of maximum likelihood estimation (MLE), normal distribution assumptions help simplify computations. Observations \( X_1, X_2, ..., X_n \), assumed to come from a normal distribution, facilitate using MLE to find estimates for unknown parameters like \( \mu \) and \( \sigma^2 \).
  • The normal distribution assumes all data points are identically and independently distributed (i.i.d.), which is vital for applying MLE effectively.
  • For example, in part (a) and (b) of the exercise, knowing that our data follows a normal distribution simplifies finding MLE for \( \sigma \) and \( \mu \), respectively, because normal equations have nice mathematical properties.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Laplace's rule of succession. Laplace claimed that when an event happens \(n\) times in a row and never fails to happen, the probability that the event will occur the next time is \((n+1) /(n+2) .\) Can you suggest a rationale for this claim?

Suppose that \(X_{1}, X_{2}, \ldots, X_{n}\) are i.i.d. random variables on the interval [0,1] with the density function $$f(x | \alpha)=\frac{\Gamma(3 \alpha)}{\Gamma(\alpha) \Gamma(2 \alpha)} x^{\alpha-1}(1-x)^{2 \alpha-1}$$ where \(\alpha>0\) is a parameter to be estimated from the sample. It can be shown that $$\begin{aligned} E(X) &=\frac{1}{3} \\ \operatorname{Var}(X) &=\frac{2}{9(3 \alpha+1)} \end{aligned}$$ a. How could the method of moments be used to estimate \(\alpha ?\) b. What equation does the mle of \(\alpha\) satisfy? c. What is the asymptotic variance of the mle? d. Find a sufficient statistic for \(\alpha .\)

Suppose that \(X_{1}, X_{2}, \ldots, X_{n}\) are i.i.d. random variables on the interval [0,1] with the density function $$f(x | \alpha)=\frac{\Gamma(2 \alpha)}{\Gamma(\alpha)^{2}}[x(1-x)]^{\alpha-1}$$ where \(\alpha>0\) is a parameter to be estimated from the sample. It can be shown that $$\begin{aligned} E(X) &=\frac{1}{2} \\ \operatorname{Var}(X) &=\frac{1}{4(2 \alpha+1)} \end{aligned}$$ a. How does the shape of the density depend on \(\alpha ?\) b. How can the method of moments be used to estimate \(\alpha ?\) c. What equation does the mle of \(\alpha\) satisfy? d. What is the asymptotic variance of the mle? e. Find a sufficient statistic for \(\alpha .\)

Show that the gamma distribution is a conjugate prior for the exponential distribution. Suppose that the waiting time in a queue is modeled as an exponential random variable with unknown parameter \(\lambda,\) and that the average time to serve a random sample of 20 customers is 5.1 minutes. A gamma distribution is used as a prior. Consider two cases: (1) the mean of the gamma is 0.5 and the standard deviation is \(1,\) and (2) the mean is 10 and the standard deviation is \(20 .\) Plot the two posterior distributions and compare them. Find the two posterior means and compare them. Explain the differences.

Find a very new shiny penny. Hold it on its edge and spin it. Do this 20 times and count the number of times it comes to rest heads up. Letting \(\pi\) denote the probability of a head, graph the log likelihood of \(\pi .\) Next, repeat the experiment in a slightly different way: This time spin the coin until 10 heads come up. Again, graph the log likelihood of \(\pi\).
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.