/*! 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 7 Let the independent random varia... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let the independent random variables \(Y_{1}, Y_{2}, \ldots, Y_{n}\) have, respectively, the probability density functions \(N\left(\beta x_{i}, \gamma^{2} x_{i}^{2}\right), i=1,2, \ldots, n\), where the given numbers \(x_{1}, x_{2}, \ldots, x_{n}\) are not all equal and no one is zero. Find the maximum likelihood estimators of \(\beta\) and \(\gamma^{2}\).

Short Answer

Expert verified
The maximum likelihood estimators of \( \beta \) and \( \gamma^{2} \) are the solutions to the first order conditions obtained by differentiating the log-likelihood function.

Step by step solution

01

Form the log-likelihood function

Since we know that given \(Y_{i}\) follow a Normal distribution with mean \( \beta x_{i} \) and variance \( \gamma^{2}x_{i}^2 \), the probability density function can be written as: \(f_{Yi}(y) = \frac{1}{\sqrt{2\pi\gamma^2x_i^2}}e^{\frac{-(y-\beta x_i)^2}{2\gamma^2x_i^2}}\). The likelihood function is the product of the individual densities: \(L(\beta,\gamma^2 | y) = \prod_{i=1}^{n}f_{Yi}(y_i)\). Taking the logarithm, we arrive at the log-likelihood function.
02

Compute the partial derivatives

To find the MLE of \( \beta \) and \( \gamma^{2} \), we need to compute the derivative of the log-likelihood with respect to \( \beta \) and \( \gamma^{2} \), which gives the first order condition.
03

Solve the partial derivative equations

To find the maximum likelihood estimators, equate these partial derivatives to zero. The solutions for \( \beta \) and \( \gamma^{2} \) are the maximum likelihood estimators.

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!

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

The driver of a diesel-powered automobile decided to test the quality of three types of diesel fuel sold in the area based on mpg. Test the null hypothesis that the three means are equal using the following data. Make the usual assumptions and take \(\alpha=0.05\). \(\begin{array}{llllll}\text { Brand A: } & 38.7 & 39.2 & 40.1 & 38.9 & \\ \text { Brand B: } & 41.9 & 42.3 & 41.3 & & \\\ \text { Brand C: } & 40.8 & 41.2 & 39.5 & 38.9 & 40.3\end{array}\)

Show that \(\sum_{i=1}^{n}\left[Y_{i}-\alpha-\beta\left(x_{i}-\bar{x}\right)\right]^{2}=n(\hat{\alpha}-\alpha)^{2}+(\hat{\beta}-\beta)^{2} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}+\sum_{i=1}^{n}\left[Y_{i}-\hat{\alpha}-\hat{\beta}\left(x_{i}-\bar{x}\right)\right]^{2}\)

Suppose \(\boldsymbol{Y}\) is an \(n \times 1\) random vector, \(\boldsymbol{X}\) is an \(n \times p\) matrix of known constants of rank \(p\), and \(\boldsymbol{\beta}\) is a \(p \times 1\) vector of regression coefficients. Let \(\boldsymbol{Y}\) have a \(N\left(\boldsymbol{X} \boldsymbol{\beta}, \sigma^{2} \boldsymbol{I}\right)\) distribution. Discuss the joint pdf of \(\hat{\boldsymbol{\beta}}=\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} \boldsymbol{X}^{\prime} \boldsymbol{Y}\) and \(\boldsymbol{Y}^{\prime}\left[\boldsymbol{I}-\boldsymbol{X}\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} \boldsymbol{X}^{\prime}\right] \boldsymbol{Y} / \sigma^{2}\)

Let the \(4 \times 1\) matrix \(\boldsymbol{Y}\) be multivariate normal \(N\left(\boldsymbol{X} \boldsymbol{\beta}, \sigma^{2} \boldsymbol{I}\right)\), where the \(4 \times 3\) matrix \(\boldsymbol{X}\) equals $$ \boldsymbol{X}=\left[\begin{array}{rrr} 1 & 1 & 2 \\ 1 & -1 & 2 \\ 1 & 0 & -3 \\ 1 & 0 & -1 \end{array}\right] $$ and \(\boldsymbol{\beta}\) is the \(3 \times 1\) regression coefficient matrix. (a) Find the mean matrix and the covariance matrix of \(\hat{\boldsymbol{\beta}}=\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} \boldsymbol{X}^{\prime} \boldsymbol{Y}\). (b) If we observe \(\boldsymbol{Y}^{\prime}\) to be equal to \((6,1,11,3)\), compute \(\hat{\boldsymbol{\beta}}\).

Fit \(y=a+x\) to the data $$ \begin{array}{l|lll} \mathrm{x} & 0 & 1 & 2 \\ \hline \mathrm{y} & 1 & 3 & 4 \end{array} $$ by the method of least squares.

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.