/*! 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 14 Let the \(4 \times 1\) matrix \(... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 14

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 \(\hat{\boldsymbol{\beta}}\).

Short Answer

Expert verified
The mean matrix of \(\hat{\boldsymbol{\beta}}\) is \(\boldsymbol{\beta}\). The covariance matrix of \(\hat{\boldsymbol{\beta}}\) is given by \((\boldsymbol{X}^{\prime} \boldsymbol{X})^{-1} \sigma^{2}\). Given \(\boldsymbol{Y}^{\prime}\) equals \((6,1,11,3)\), the calculated \(\hat{\boldsymbol{\beta}}\) will be the numerical output obtained from the matrix calculations.

Step by step solution

01

Mean matrix of \(\hat{\boldsymbol{\beta}}\)

The mean matrix of \(\hat{\boldsymbol{\beta}}\), denoted as \(E(\hat{\boldsymbol{\beta}})\), is simply \(\boldsymbol{\beta}\). This happens because \(\hat{\boldsymbol{\beta}}\) is an unbiased estimator for \(\boldsymbol{\beta}\) in the context of a normal distribution.
02

Covariance matrix of \(\hat{\boldsymbol{\beta}}\)

The covariance matrix of \(\hat{\boldsymbol{\beta}}\), denoted as \(Var(\hat{\boldsymbol{\beta}})\), is given by \((\boldsymbol{X}^{\prime} \boldsymbol{X})^{-1} \sigma^{2}\). Here \(\sigma^2\) is simply the variance of the normal distribution, which is assumed to be known.
03

Compute \(\hat{\boldsymbol{\beta}}\)

With \(\boldsymbol{Y}^{\prime}\) equals \((6,1,11,3)\), the estimate of \(\boldsymbol{\beta}\) is computed by substituting into the formula \(\hat{\boldsymbol{\beta}}=\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} \boldsymbol{X}^{\prime} \boldsymbol{Y}\). First, calculate the transpose of \(\boldsymbol{X}\) and multiply it with the original \(\boldsymbol{X}\) to obtain \(\boldsymbol{X}^{\prime} \boldsymbol{X}\), then compute the inverse of that result. Multiply this inverse with the transpose of \(\boldsymbol{X}\) and finally multiply with observed \(\boldsymbol{Y}\) to obtain the estimate of \(\boldsymbol{\beta}\).

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Key Concepts

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

Covariance Matrix
When dealing with multiple random variables, it's crucial to understand not just their individual behaviors, but also how they vary together. This is where the concept of a covariance matrix becomes important. In the context of the multivariate normal distribution, the covariance matrix encapsulates the pairwise levels of variability and correlation between every pair of variables in a dataset.

Consider a scenario in which you are assessing the relationship between several financial markets. A covariance matrix can provide insights into how one market's performance might influence another. For instance, if you have three markets and you're trying to gauge their interdependence, their covariance matrix will be a 3x3 matrix containing covariance values that reflect how much the markets share in their fluctuations.

In the given exercise, the covariance matrix of the estimator \(\hat{\boldsymbol{\beta}}\) is calculated as \(Var(\hat{\boldsymbol{\beta}}) = (\boldsymbol{X}' \boldsymbol{X})^{-1} \sigma^{2}\), where \(\sigma^2\) is the variance of the normal distribution. This matrix is fundamental in understanding the precision of our estimates: smaller values on the diagonal suggest higher precision of the estimates for the corresponding variables.
Regression Coefficient
In regression analysis, the regression coefficient represents the quantitative effect one or more predictor variables have on the outcome variable. It quantifies the change in the mean of the dependent variable given a one-unit shift in the independent variable. In simpler terms, these coefficients are the values that multiply with the associated variables to predict the outcome.

Going back to the financial markets example, if you want to predict the future value of a market index based on previous performances and other economic indicators, your regression coefficients will tell you how much influence each predictor has. For instance, if the coefficient for economic growth is high, it implies a strong impact on the market index.

In the exercise, we aim to estimate the matrix \(\boldsymbol{\beta}\), which is composed of such coefficients. Calculating \(\hat{\boldsymbol{\beta}}\) as shown, gives us a vector of estimates for our regression coefficients. This vector, based on observed data, helps to make informed predictions about our outcome variable, informed by the multivariate normal distribution assumption.
Unbiased Estimator
An estimator is a rule or calculation that allows us to estimate the value of an unknown parameter in a statistical model. An unbiased estimator is one that, on average, hits the true value of the parameter being estimated across numerous samples from the population. In practice, this means that the estimates it produces do not systematically overestimate or underestimate the true value.

It's like playing darts: an unbiased estimator will have an even distribution of dart throws around the bullseye, not clustered too far to the left, right, above, or below it. The closer the darts land to the bullseye, and the more evenly dispersed they are, the more unbiased and precise the estimator is.

In our exercise, the estimator \(\hat{\boldsymbol{\beta}}\) for the regression coefficient matrix \(\boldsymbol{\beta}\) is determined to be unbiased. This implies that with repeated sampling and estimation, the average calculated \(\hat{\boldsymbol{\beta}}\) would equal the true \(\boldsymbol{\beta}\). Therefore, using this estimator gives us a reliable foundation for inferential statistics within the context of our multivariate normal model.

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Most popular questions from this chapter

With the background of the two-way classification with \(c>1\) observations per cell, show that the maximum likelihood estimators of the parameters are $$ \begin{aligned} \hat{\alpha}_{i} &=\bar{X}_{i . .}-\bar{X}_{\ldots} \\ \hat{\beta}_{j} &=\bar{X}_{. j .}-\bar{X}_{\ldots} \\ \hat{\gamma}_{i j} &=\bar{X}_{i j .}-\bar{X}_{i . .}-\bar{X}_{. j .}+\bar{X}_{\ldots} \\ \hat{\mu} &=\bar{X}_{\ldots} \end{aligned} $$ Show that these are unbiased estimators of the respective parameters. Compute the variance of each estimator.

Let \(\mathbf{A}=\left[a_{i j}\right]\) be a real symmetric matrix. Prove that \(\sum_{i} \sum_{j} a_{i j}^{2}\) is equal to the sum of the squares of the eigenvalues of \(\mathbf{A}\). Hint: If \(\boldsymbol{\Gamma}\) is an orthogonal matrix, show that \(\sum_{j} \sum_{i} a_{i j}^{2}=\operatorname{tr}\left(\mathbf{A}^{2}\right)=\operatorname{tr}\left(\mathbf{\Gamma}^{\prime} \mathbf{A}^{2} \mathbf{\Gamma}\right)=\) \(\operatorname{tr}\left[\left(\boldsymbol{\Gamma}^{\prime} \mathbf{A} \boldsymbol{\Gamma}\right)\left(\boldsymbol{\Gamma}^{\prime} \mathbf{A} \boldsymbol{\Gamma}\right)\right]\)

We wish to compare compressive strengths of concrete corresponding to \(a=3\) different drying methods (treatments). Concrete is mixed in batches that are just large enough to produce three cylinders. Although care is taken to achieve uniformity, we expect some variability among the \(b=5\) batches used to obtain the following compressive strengths. (There is little reason to suspect interaction, and hence only one observation is taken in each cell.) $$ \begin{array}{cccccc} \hline & \multicolumn{3}{c} {\text { Batch }} & & & \\ \cline { 2 - 6 } \text { Treatment } & B_{1} & B_{2} & B_{3} & B_{4} & B_{5} \\\ \hline A_{1} & 52 & 47 & 44 & 51 & 42 \\ A_{2} & 60 & 55 & 49 & 52 & 43 \\ A_{3} & 56 & 48 & 45 & 44 & 38 \\ \hline \end{array} $$ (a) Use the \(5 \%\) significance level and test \(H_{A}: \alpha_{1}=\alpha_{2}=\alpha_{3}=0\) against all alternatives. (b) Use the \(5 \%\) significance level and test \(H_{B}: \beta_{1}=\beta_{2}=\beta_{3}=\beta_{4}=\beta_{5}=0\) against all alternatives.

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 $$ \hat{d}=\frac{\sum_{j=2}^{b} \bar{X}_{\cdot j} /(b-1)-\bar{X}_{.1}}{b} $$ (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 $$ \sum_{i=1}^{a} \sum_{j=1}^{b}\left(X_{i j}-\bar{X}_{. .}\right)^{2}=Q_{3}+Q_{6}+Q_{7} $$ (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.

Given the following observations in a two-way classification with \(a=3\), \(b=4\), and \(c=2\), compute the \(F\) -statistics used to test that all interactions are equal to zero \(\left(\gamma_{i j}=0\right)\), all column means are equal \(\left(\beta_{j}=0\right)\), and all row means are equal \(\left(\alpha_{i}=0\right)\), respectively. $$ \begin{array}{ccccc} \hline \text { Row/Column } & 1 & 2 & 3 & 4 \\ \hline 1 & 3.1 & 4.2 & 2.7 & 4.9 \\ & 2.9 & 4.9 & 3.2 & 4.5 \\ 2 & 2.7 & 2.9 & 1.8 & 3.0 \\ & 2.9 & 2.3 & 2.4 & 3.7 \\ 3 & 4.0 & 4.6 & 3.0 & 3.9 \\ & 4.4 & 5.0 & 2.5 & 4.2 \\ \hline \end{array} $$
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