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Consider the probability density function $$ f(x)=\frac{1}{\theta^{2}} x e^{-x / \theta}, \quad 0 \leq x<\infty, \quad 0<\theta<\infty $$ Find the maximum likelihood estimator for \(\theta .\)

Short Answer

Expert verified
The MLE for \(\theta\) is \(\hat{\theta} = \frac{1}{2n} \sum_{i=1}^{n} x_i \).

Step by step solution

01

Understand the Problem

We are asked to find the maximum likelihood estimator (MLE) for \( \theta \) given the probability density function (pdf) \( f(x) = \frac{1}{\theta^{2}} x e^{-x / \theta} \). This implies we need to find the value of \( \theta \) that maximizes the likelihood given a set of observed data \( x_1, x_2, ..., x_n \).
02

Write the Likelihood Function

The likelihood function for a set of independent observations \( x_1, x_2, ..., x_n \) based on the given pdf is:\[ L(\theta) = \prod_{i=1}^{n} \frac{1}{\theta^{2}} x_i e^{-x_i / \theta} = \left( \frac{1}{\theta^{2}} \right)^n \prod_{i=1}^{n} x_i \prod_{i=1}^{n} e^{-x_i / \theta} \]This can be simplified as:\[ L(\theta) = \frac{1}{\theta^{2n}} \prod_{i=1}^{n} x_i \cdot e^{- \sum_{i=1}^{n} x_i / \theta} \]
03

Take the Log-Likelihood

To simplify, we take the natural logarithm of the likelihood function. The log-likelihood \( \ell(\theta) \) is:\[ \ell(\theta) = \ln \left( \frac{1}{\theta^{2n}} \right) + \ln \left( \prod_{i=1}^{n} x_i \right) - \frac{\sum_{i=1}^{n} x_i}{\theta} \]\[ \ell(\theta) = -2n \ln(\theta) + \sum_{i=1}^{n} \ln(x_i) - \frac{\sum_{i=1}^{n} x_i}{\theta} \]
04

Differentiate the Log-Likelihood

Differentiate \( \ell(\theta) \) with respect to \( \theta \) to find its critical points:\[ \frac{d\ell}{d\theta} = -\frac{2n}{\theta} + \frac{\sum_{i=1}^{n} x_i}{\theta^2} \]
05

Solve for Critical Points

Set \( \frac{d\ell}{d\theta} = 0 \) to find \( \theta \):\[ -\frac{2n}{\theta} + \frac{\sum_{i=1}^{n} x_i}{\theta^2} = 0 \]Multiply through by \( \theta^2 \) to clear the fractions:\[ -2n\theta + \sum_{i=1}^{n} x_i = 0 \]Solving for \( \theta \), we find:\[ \theta = \frac{1}{2n} \sum_{i=1}^{n} x_i \]
06

Verify Maximization

Check the second derivative \( \frac{d^2\ell}{d\theta^2} \) to ensure it is a maximum.\[ \frac{d^2\ell}{d\theta^2} = \frac{2n}{\theta^2} - \frac{2\sum_{i=1}^{n} x_i}{\theta^3} \]At \( \theta = \frac{1}{2n} \sum_{i=1}^{n} x_i \), this will be negative ensuring a maximum.

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Key Concepts

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

Probability Density Function
A probability density function (PDF) is a function that describes the likelihood of a random variable taking on a particular value. In the context of continuous probability distributions, the PDF helps to define the shape of the distribution. Here, the probability density function for the variable \( x \) is \( f(x)=\frac{1}{\theta^{2}} x e^{-x / \theta} \).
This function is defined for \( x \geq 0 \) and \( \theta > 0 \). The PDF must integrate to 1 over its entire range to ensure it properly represents a probability distribution.
  • \( \frac{1}{\theta^2} \) is a scaling factor ensuring everything integrates correctly.
  • \( x e^{-x / \theta} \) shows the functional form, often seen in distributions such as the exponential.
Understanding how to manipulate this function is crucial for finding quantities like maximum likelihood estimators.
Log-Likelihood
The likelihood function is an essential part of maximum likelihood estimation. For a set of observations based on a PDF, it gives the probability of the observed data as a function of the parameter \( \theta \). The likelihood function derived from the PDF is \( L(\theta) = \frac{1}{\theta^{2n}} \prod_{i=1}^{n} x_i \cdot e^{- \sum_{i=1}^{n} x_i / \theta} \).
Taking the natural logarithm of the likelihood function simplifies calculations and transforms the product into a sum. The log-likelihood \( \ell(\theta) \) is given by:
\[ \ell(\theta) = -2n \ln(\theta) + \sum_{i=1}^{n} \ln(x_i) - \frac{\sum_{i=1}^{n} x_i}{\theta} \]
  • Logarithms turn products into sums, simplifying derivatives.
  • \( \ell(\theta) \) is used to find maximum likelihood estimates by exploring where it peaks.
Working with the log-likelihood makes the differentiation process in subsequent steps easier and more opportune.
Differentiation
Differentiation involves finding the derivative of a function concerning a variable. In maximum likelihood estimation, this technique enables us to locate critical points of the log-likelihood function, indicating potential maxima or minima.
The derivative of the log-likelihood \( \frac{d\ell}{d\theta} \) concerning \( \theta \) is:
\[ \frac{d\ell}{d\theta} = -\frac{2n}{\theta} + \frac{\sum_{i=1}^{n} x_i}{\theta^2} \]
  • This derivative expresses how \( \ell(\theta) \) changes as \( \theta \) changes.
  • Finding where \( \frac{d\ell}{d\theta} = 0 \) helps locate critical points, vital for estimating \( \theta \).
Being able to differentiate the log-likelihood function correctly is key to uncovering the estimator we need.
Critical Points
Critical points are values of a variable, in this case, \( \theta \), where the derivative of a function is zero or undefined. They represent potential maximum or minimum points of the function.
To find the critical points for \( \ell(\theta) \), we set \( \frac{d\ell}{d\theta} = 0 \):
\[ -\frac{2n}{\theta} + \frac{\sum_{i=1}^{n} x_i}{\theta^2} = 0 \]
Solving this equation, we multiply by \( \theta^2 \) to eliminate the fractions:
\[ -2n\theta + \sum_{i=1}^{n} x_i = 0 \]
Solving for \( \theta \), we find:
\[ \theta = \frac{1}{2n} \sum_{i=1}^{n} x_i \]
  • This value of \( \theta \) is believed to potentially maximize \( \ell(\theta) \).
  • Verification requires the second derivative test to confirm a maximum, ensuring \( \frac{d^2\ell}{d\theta^2} < 0 \).
Accurate identification and solution of critical points is crucial to determining the maximum likelihood estimator.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be uniformly distributed on the interval 0 to \(a\). Recall that the maximum likelihood estimator of \(a\) is \(\hat{a}=\max \left(X_{i}\right)\) a. Argue intuitively why \(\hat{a}\) cannot be an unbiased estimator for \(a\). b. Suppose that \(E(\hat{a})=n a /(n+1) .\) Is it reasonable that \(\hat{a}\) consistently underestimates \(a\) ? Show that the bias in the estimator approaches zero as \(n\) gets large. c. Propose an unbiased estimator for \(a\). d. Let \(Y=\max \left(X_{i}\right)\). Use the fact that \(Y \leq y\) if and only if each \(X_{i} \leq y\) to derive the cumulative distribution function of \(Y\). Then show that the probability density function of \(Y\) is $$ f(y)=\left\\{\begin{array}{ll} \frac{n y^{n-1}}{a^{n}}, & 0 \leq y \leq a \\ 0, & \text { otherwise } \end{array}\right. $$ Use this result to show that the maximum likelihood estimator for \(a\) is biased. e. We have two unbiased estimators for \(a\) : the moment estimator \(\hat{a}_{1}=2 \bar{X}\) and \(\hat{a}_{2}=[(n+1) / n] \max \left(X_{i}\right),\) where \(\max \left(X_{i}\right)\) is the largest observation in a random sample of size \(n\). It can be shown that \(V\left(\hat{a}_{1}\right)=a^{2} /(3 n)\) and that \(V\left(\hat{a}_{2}\right)=\) \(a^{2} /[n(n+2)]\). Show that if \(n>1, \hat{a}_{2}\) is a better estimator than \(\hat{a}\). In what sense is it a better estimator of \(a\) ?

Suppose that the random variable \(X\) has the continuous uniform distribution $$f(x)=\left\\{\begin{array}{ll}1, & 0 \leq x \leq 1 \\\0, & \text { otherwise }\end{array}\right.$$ Suppose that a random sample of \(n=12\) observations is selected from this distribution. What is the approximate probability distribution of \(\bar{X}-6\) ? Find the mean and variance of this quantity.

Data on the \(\mathrm{pH}\) of rain in Ingham County, Michigan, are as follows: $$\begin{array}{lllllllllll}5.47 & 5.37 & 5.38 & 4.63 & 5.37 & 3.74 & 3.71 & 4.96 & 4.64 & 5.11 & 5.65 & 5.39 \\\4.16 & 5.62 & 4.57 & 4.64 & 5.48 & 4.57 & 4.57 & 4.51 & 4.86 & 4.56 & 4.61 & 4.32 \\\3.98 & 5.70 & 4.15 & 3.98 & 5.65 & 3.10 & 5.04 & 4.62 & 4.51 & 4.34 & 4.16 & 4.64 \\\5.12 & 3.71 & 4.64 & & & & & & & & &\end{array}$$ What proportion of the samples has \(\mathrm{pH}\) below \(5.0 ?\)

A synthetic fiber used in manufacturing carpet has tensile strength that is normally distributed with mean 75.5 psi and standard deviation 3.5 psi. Find the probability that a random sample of \(n=6\) fiber specimens will have sample mean tensile strength that exceeds 75.75 psi.

Suppose that we have a random sample \(X_{1}, X_{2}, \ldots,\) \(X_{n}\) from a population that is \(N\left(\mu, \sigma^{2}\right) .\) We plan to use \(\hat{\Theta}=\) \(\sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2} / c\) to estimate \(\sigma^{2} .\) Compute the bias in \(\hat{\Theta}\) as an estimator of \(\sigma^{2}\) as a function of the constant \(c\).

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