/*! 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 1 Let \(X\) be a geometric random ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(X\) be a geometric random variable with parameter \(p\). Find the maximum likelihood estimator of \(p\) based on a random sample of size \(n\).

Short Answer

Expert verified
The maximum likelihood estimator of \( p \) is \( \frac{n}{\sum_{i=1}^{n} x_i} \).

Step by step solution

01

Understand the Geometric Distribution

The probability mass function (PMF) of a geometric random variable \(X\) with parameter \(p\) is given by \( P(X = k) = (1-p)^{k-1}p \), where \( k = 1, 2, 3, \dots \). It models the number of Bernoulli trials needed to get one success.
02

Write the Likelihood Function

For a random sample \( x_1, x_2, \ldots, x_n \) from a geometric distribution, the likelihood function \( L(p) \) is the product of the PMFs: \[ L(p) = \prod_{i=1}^{n} (1-p)^{x_i - 1} p = p^n (1-p)^{\sum_{i=1}^{n} (x_i - 1)} = p^n (1-p)^{\sum_{i=1}^{n} x_i - n} \].
03

Take the Logarithm of the Likelihood Function

The log-likelihood function \( \ell(p) \) is: \[ \ell(p) = \log(p^n (1-p)^{\sum_{i=1}^{n} x_i - n}) = n \log(p) + (\sum_{i=1}^{n} x_i - n) \log(1-p) \].
04

Differentiate the Log-Likelihood Function

Find the derivative of \( \ell(p) \) with respect to \( p \): \[ \frac{d\ell}{dp} = \frac{n}{p} - \frac{\sum_{i=1}^{n} x_i - n}{1-p} \].
05

Set the Derivative to Zero for Optimization

Set \( \frac{d\ell}{dp} = 0 \) to find the critical points:\[ \frac{n}{p} - \frac{\sum_{i=1}^{n} x_i - n}{1-p} = 0 \].
06

Solve for \( p \)

Rearrange the equation and solve for \( p \):\[ \frac{n}{p} = \frac{\sum_{i=1}^{n} x_i - n}{1-p} \].Cross-multiply to get:\[ n(1-p) = p(\sum_{i=1}^{n} x_i - n) \].Simplify and solve for \( p \):\[ n - np = p \sum_{i=1}^{n} x_i - np \],\[ n = p \sum_{i=1}^{n} x_i \],\[ p = \frac{n}{\sum_{i=1}^{n} x_i} \].

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 91Ó°ÊÓ!

Key Concepts

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

Geometric Distribution
The geometric distribution is a fascinating concept in probability theory. It represents a scenario where we are interested in the number of trials it takes to achieve the first success in a series of Bernoulli trials.
Each of these trials is independent and has the same probability of success, denoted as \(p\).
The outcome that we are focused on is the first occurrence of success.

In the case of the geometric distribution, the probability mass function (PMF) helps us determine the likelihood of achieving this first success on the \(k\)-th trial. The formula for this is expressed as \(P(X = k) = (1-p)^{k-1}p\), where \(k\) can be any positive integer, \(k = 1, 2, 3, \ldots\)
.
  • The expression \((1-p)^{k-1}\) captures the probability of observing failures in the first \(k-1\) trials.
  • The part \(p\) reflects the probability of success on the \(k\)-th trial.
Understanding this distribution model sheds light on phenomena such as waiting times or trial consumption before achieving a desired outcome.
Probability Mass Function
A probability mass function (PMF) is a key concept used to describe a discrete random variable. The PMF provides specific probabilities of occurrence for each possible value of the random variable.
In essence, it links every outcome of a random variable to a probability between 0 and 1.

For a geometric distribution, as mentioned earlier, the PMF is given by \(P(X = k) = (1-p)^{k-1}p\).
  • The PMF must satisfy two main rules:
  • Each probability is non-negative; \(P(X = k) \geq 0\).
  • When summed over all possible values, the probabilities must add up to one; \(\sum P(X = k) = 1\).
This function is crucial in determining the likelihood of specific outcomes when dealing with discrete probability distributions, like the geometric distribution.
Log-Likelihood Function
The log-likelihood function is a transformation of the likelihood function, which is used extensively in statistics, especially in the context of Maximum Likelihood Estimation (MLE).
The likelihood function, \(L(p)\), is the joint probability of observing a given sample.
By taking the logarithm of this function, we obtain the log-likelihood, \(\ell(p)\), which often simplifies the mathematics of finding an estimator.

The logarithm turns products into sums, which are much easier to handle analytically.
In the case of the geometric distribution, the log-likelihood for a sample \(x_1, x_2, \ldots, x_n\) is:
\[ \ell(p) = n \log(p) + \left(\sum_{i=1}^{n} x_i - n\right) \log(1-p) \]
Maximizing this log-likelihood function with respect to \(p\) helps to estimate the parameter \(p\) efficiently and effectively.
Differentiation in Statistics
Differentiation is a mathematical technique used to find the rate at which a function is changing at any given point.
In statistics, differentiation plays a crucial role, especially when optimizing functions such as the log-likelihood function.
Finding the derivative of the log-likelihood function allows us to locate maximum or minimum points, which is vital for Maximum Likelihood Estimation (MLE).

For the geometric distribution problem, the derivative of the log-likelihood function with respect to \(p\) is:
\[ \frac{d\ell}{dp} = \frac{n}{p} - \frac{\sum_{i=1}^{n} x_i - n}{1-p} \]
This derivative is crucial in finding the value of \(p\) that maximizes the likelihood equation. By setting this derivative equal to zero, we can determine the optimal parameter estimate.
This process of differentiation and solving for critical points forms the foundation of many statistical estimation techniques.

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

A random sample of size \(n=16\) is taken from a normal population with \(\mu=40\) and \(\sigma^{2}=5\). Find the probability that the sample mean is less than or equal to \(37 .\)

Scientists at the Hopkins Memorial Forest in westem Massachusetts have been collecting meteorological and environmental data in the forest data for more than 100 years. In the past few years, sulfate content in water samples from Birch Brook has averaged \(7.48 \mathrm{mg} / \mathrm{L}\) with a standard deviation of \(1.60 \mathrm{mg} / \mathrm{L}\) (a) What is the standard error of the sulfate in a collection of 10 water samples? (b) If 10 students measure the sulfate in their samples, what is the probability that their average sulfate will be between 6.49 and \(8.47 \mathrm{mg} / \mathrm{L} ?\) (c) What do you need to assume for the probability calculated in (b) to be accurate?

You plan to use a rod to lay out a square, each side of which is the length of the rod. The length of the rod is \(\mu\). which is unknown. You are interested in estimating the area of the square, which is \(\mu^{2}\). Because \(\mu\) is unknown, you measure it \(n\) times, obtaining observations \(X_{1}, X_{2} \ldots, X_{n}\). Suppose that each measurement is unbiased for \(\mu\) with variance \(\sigma^{2}\) (a) Show that \(\bar{X}^{2}\) is a biased estimate of the area of the square. (b) Suggest an estimator that is unbiased.

Suppose that the random variable \(X\) has a log normal distribution with parameters \(\theta=1.5\) and \(\omega=0.8\). A sample of size \(n=15\) is drawn from this distribution. Find the standard error of the sample median of this distribution with the bootstrap method using \(n_{n}=200\) bootstrap samples.

Suppose that \(\Theta_{i}\) and \(\Theta_{2}\) are unbiased estimators of the parameter \(\theta .\) We know that \(V\left(\Theta_{1}\right)=10\) and \(V\left(\Theta_{2}\right)=4\). Which estimator is better and in what sense is it better? Calculate the relative efficiency of the two estimators.

See all solutions

Recommended explanations on Math Textbooks

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.