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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 \(\gamma^{2}\).

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Most popular questions from this chapter

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 \(\beta\) 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 \(\boldsymbol{Y}^{\prime}\left[\boldsymbol{I}-\boldsymbol{X}\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} \boldsymbol{X}^{\prime}\right] \boldsymbol{Y} / \sigma^{2}\)

Show that the square of a noncentral \(T\) random variable is a noncentral \(F\) random variable.

Often in regression the mean of the random variable \(Y\) is a linear function of \(p\) -values \(x_{1}, x_{2}, \ldots, x_{p}\), say \(\beta_{1} x_{1}+\beta_{2} x_{2}+\cdots+\beta_{p} x_{p}\), where \(\boldsymbol{\beta}^{\prime}=\left(\beta_{1}, \beta_{2}, \ldots, \beta_{p}\right)\) are the regression coefficients. Suppose that \(n\) values, \(\boldsymbol{Y}^{\prime}=\left(Y_{1}, Y_{2}, \ldots, Y_{n}\right)\) are observed for the \(x\) -values in \(\boldsymbol{X}=\left[x_{i j}\right]\), where \(\boldsymbol{X}\) is an \(n \times p\) design matrix and its ith row is associated with \(Y_{i}, i=1,2, \ldots, n .\) Assume that \(Y\) is multivariate normal with mean \(\boldsymbol{X} \boldsymbol{\beta}\) and variance-covariance matrix \(\sigma^{2} \boldsymbol{I}\), where \(\boldsymbol{I}\) is the \(n \times n\) identity matrix. (a) Note that \(Y_{1}, Y_{2}, \ldots, Y_{n}\) are independent. Why? (b) Since \(\boldsymbol{Y}\) should approximately equal its mean \(\boldsymbol{X} \boldsymbol{\beta}\), we estimate \(\boldsymbol{\beta}\) by solving the normal equations \(\boldsymbol{X}^{\prime} \boldsymbol{Y}=\boldsymbol{X}^{\prime} \boldsymbol{X} \boldsymbol{\beta}\) for \(\boldsymbol{\beta}\). Assuming that \(\boldsymbol{X}^{\prime} \boldsymbol{X}\) is non- singular, solve the equations to get \(\hat{\boldsymbol{\beta}}=\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} \boldsymbol{X}^{\prime} \boldsymbol{Y}\). Show that \(\hat{\boldsymbol{\beta}}\) has a multivariate normal distribution with mean \(\boldsymbol{\beta}\) and variance-covariance matrix $$ \sigma^{2}\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} $$ (c) Show that $$ (\boldsymbol{Y}-\boldsymbol{X} \boldsymbol{\beta})^{\prime}(\boldsymbol{Y}-\boldsymbol{X} \boldsymbol{\beta})=(\hat{\boldsymbol{\beta}}-\boldsymbol{\beta})^{\prime}\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)(\hat{\boldsymbol{\beta}}-\boldsymbol{\beta})+(\boldsymbol{Y}-\boldsymbol{X} \hat{\boldsymbol{\beta}})^{\prime}(\boldsymbol{Y}-\boldsymbol{X} \hat{\boldsymbol{\beta}}) $$ say \(Q=Q_{1}+Q_{2}\) for convenience. (d) Show that \(Q_{1} / \sigma^{2}\) is \(\chi^{2}(p)\). (e) Show that \(Q_{1}\) and \(Q_{2}\) are independent. (f) Argue that \(Q_{2} / \sigma^{2}\) is \(\chi^{2}(n-p)\). (g) Find \(c\) so that \(c Q_{1} / Q_{2}\) has an \(F\) -distribution. (h) The fact that a value \(d\) can be found so that \(P\left(c Q_{1} / Q_{2} \leq d\right)=1-\alpha\) could be used to find a \(100(1-\alpha)\) percent confidence ellipsoid for \(\beta\). Explain.

Given the following observations associated with a two-way classification with \(a=3\) and \(b=4\), compute the \(F\) -statistic used to test the equality of the column means \(\left(\beta_{1}=\beta_{2}=\beta_{3}=\beta_{4}=0\right)\) and the equality of the row means \(\left(\alpha_{1}=\alpha_{2}=\alpha_{3}=0\right)\), respectively. $$ \begin{array}{ccccc} \hline \text { Row/Column } & 1 & 2 & 3 & 4 \\ \hline 1 & 3.1 & 4.2 & 2.7 & 4.9 \\ 2 & 2.7 & 2.9 & 1.8 & 3.0 \\ 3 & 4.0 & 4.6 & 3.0 & 3.9 \\ \hline \end{array} $$

In Exercise 9.2.1, show that the linear functions \(X_{i j}-X_{. j}\) and \(X_{. j}-X\).. are uncorrelated. Hint: Recall the definition of \(\bar{X}_{. j}\) and \(\bar{X}_{. .}\) and, without loss of generality, we can let \(E\left(X_{i j}\right)=0\) for all \(i, j\)

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