Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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
By following the steps above, you will find the maximum likelihood estimators \(\hat{\beta}\) and \(\hat{\gamma}^{2}\) by setting the respective derivatives of the log-likelihood function to zero and solving them.

Step by step solution

01

Understanding the given information and determining the PDFs

We are given independent random variables \(Y_{1}, Y_{2}, \ldots, Y_{n}\) with respective probability density functions being normal distributions: \(N\left(\beta x_{i},\gamma^{2} x_{i}^{2}\right)\), for \(i=1,2, \ldots, n\). In the case of a normal distribution, the parameters \(\beta x_{i}\) and \(\gamma^{2} x_{i}^{2}\) represent the mean and the variance respectively.
02

Constructing the likelihood function

The likelihood function is the joint probability function of all the observations: \(L(\beta, \gamma^2) = \prod_{i=1}^{n} f_Y (y_i; \beta, \gamma^2)\). Since these observations are independent, the overall likelihood is the product of the individual likelihoods: \(L(\beta,\gamma^{2})=\prod_{i=1}^{n}\frac{1}{\sqrt{2\pi\gamma^{2}x_{i}^{2}}}e^{\frac{-(y_{i}-\beta x_{i})^{2}}{2\gamma^{2}x_{i}^{2}}}\).
03

Taking the logarithm of the likelihood function

Take the natural logarithm of the likelihood function to simplify it. This step give us the log-Likelihood function: \( l(\beta, \gamma^2) = \ln L(\beta, \gamma^2) \).
04

Taking the derivatives of the log-likelihood function

Taking the derivative of the log-likelihood function with respect to \(\beta\) and setting it equal to zero gives the maximum likelihood estimator of \(\beta\). Similarly, taking the derivative of the log-likelihood function with respect to \(\gamma^{2}\) and setting it equal to zero gives the maximum likelihood estimator of \(\gamma^{2}\).

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 Vaia!

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

(Bonferroni Multiple Comparison Procedure). In the notation of this section, let \(\left(k_{i 1}, k_{i 2}, \ldots, k_{i b}\right), i=1,2, \ldots, m\), represent a finite number of \(b\) -tuples. The problem is to find simultaneous confidence intervals for \(\sum_{j=1}^{b} k_{i j} \mu_{j}, i=1,2, \ldots, m\), by a method different from that of Scheffé. Define the random variable \(T_{i}\) by $$ \left(\sum_{j=1}^{b} k_{i j} \bar{X}_{. j}-\sum_{j=1}^{b} k_{i j} \mu_{j}\right) / \sqrt{\left(\sum_{j=1}^{b} k_{i j}^{2}\right) V / a}, \quad i=1,2, \ldots, m $$ (a) Let the event \(A_{i}^{c}\) be given by \(-c_{i} \leq T_{i} \leq c_{i}, i=1,2, \ldots, m\). Find the random variables \(U_{i}\) and \(W_{i}\) such that \(U_{i} \leq \sum_{1}^{b} k_{i j} \mu_{j} \leq W_{j}\) is equivalent to \(A_{i}^{c}\) (b) Select \(c_{i}\) such that \(P\left(A_{i}^{c}\right)=1-\alpha / m ;\) that is, \(P\left(A_{i}\right)=\alpha / m .\) Use Exercise 9.4.1 to determine a lower bound on the probability that simultaneously the random intervals \(\left(U_{1}, W_{1}\right), \ldots,\left(U_{m}, W_{m}\right)\) include \(\sum_{j=1}^{b} k_{1 j} \mu_{j}, \ldots, \sum_{j=1}^{b} k_{m j} \mu_{j}\) respectively. (c) Let \(a=3, b=6\), and \(\alpha=0.05\). Consider the linear functions \(\mu_{1}-\mu_{2}, \mu_{2}-\mu_{3}\), \(\mu_{3}-\mu_{4}, \mu_{4}-\left(\mu_{5}+\mu_{6}\right) / 2\), and \(\left(\mu_{1}+\mu_{2}+\cdots+\mu_{6}\right) / 6 .\) Here \(m=5 .\) Show that the lengths of the confidence intervals given by the results of Part (b) are shorter than the corresponding ones given by the method of Scheffé as described in the text. If \(m\) becomes sufficiently large, however, this is not the case.

Fit by the method of least squares the plane \(z=a+b x+c y\) to the five points \((x, y, z):(-1,-2,5),(0,-2,4),(0,0,4),(1,0,2),(2,1,0)\).

Using the notation of this section, assume that the means satisfy the condition that \(\mu=\mu_{1}+(b-1) d=\mu_{2}-d=\mu_{3}-d=\cdots=\mu_{b}-d .\) That is, the last \(b-1\) means are equal but differ from the first mean \(\mu_{1}\), provided that \(d \neq 0\). Let independent random samples of size \(a\) be taken from the \(b\) normal distributions with common unknown variance \(\sigma^{2}\). (a) Show that the maximum likelihood estimators of \(\mu\) and \(d\) are \(\hat{\mu}=\bar{X} . .\) and $$ \hat{d}=\frac{\sum_{j=2}^{b} \bar{X}_{. j} /(b-1)-\bar{X}_{.1}}{b} $$ (b) Using Exercise \(9.1 .3\), find \(Q_{6}\) and \(Q_{7}=c \hat{d}^{2}\) so that, when \(d=0, Q_{7} / \sigma^{2}\) is \(\chi^{2}(1)\) and $$ \sum_{i=1}^{a} \sum_{j=1}^{b}\left(X_{i j}-\bar{X}_{n}\right)^{2}=Q_{3}+Q_{6}+Q_{7} $$ (c) Argue that the three terms in the right-hand member of Part (b), once divided by \(\sigma^{2}\), are independent random variables with chi-square distributions, provided that \(d=0\). (d) The ratio \(Q_{7} /\left(Q_{3}+Q_{6}\right)\) times what constant has an \(F\) -distribution, provided that \(d=0\) ? Note that this \(F\) is really the square of the two-sample \(T\) used to test the equality of the mean of the first distribution and the common mean of the other distributions, in which the last \(b-1\) samples are combined into one.

Show that \(R=\frac{\sum_{1}^{n}\left(X_{i}-\bar{X}\right)\left(Y_{i}-\bar{Y}\right)}{\sqrt{\sum_{1}^{n}\left(X_{i}-\bar{X}\right)^{2} \sum_{1}^{n}\left(Y_{i}-Y\right)^{2}}}=\frac{\sum_{1}^{n} X_{i} Y_{i}-n \overline{X Y}}{\sqrt{\left(\sum_{1}^{n} X_{i}^{2}-n \bar{X}^{2}\right)\left(\sum_{1}^{n} Y_{i}^{2}-n \bar{Y}^{2}\right)}}\)

Two experiments gave the following results: $$ \begin{array}{cccccc} \hline \mathrm{n} & \bar{x} & \bar{y} & s_{x} & s_{y} & \mathrm{r} \\ \hline 100 & 10 & 20 & 5 & 8 & 0.70 \\ 200 & 12 & 22 & 6 & 10 & 0.80 \\ \hline \end{array} $$ Calculate \(r\) for the combined sample.

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.

Sign-up for free