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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 \(\beta\) is the \(3 \times 1\) regression coeffient 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 \(\boldsymbol{\beta}\) and the covariance matrix is \(\sigma^2 (\boldsymbol{X}' \boldsymbol{X})^{-1}\). Additionally, after conducting matrix computations with the provided data, the \(\hat{\boldsymbol{\beta}}\) matrix is calculated.

Step by step solution

01

Formulate mean and covariance matrices

Start by understanding that in a multivariate normal distribution, the mean matrix of \(\hat{\boldsymbol{\beta}}\) is \(\boldsymbol{\beta}\) and the covariance matrix of \(\hat{\boldsymbol{\beta}}\) is \(\sigma^2 (\boldsymbol{X}' \boldsymbol{X})^{-1}\). The proof of this lies in the properties of multivariate normal distribution with regression context, and mainly involves matrix and probability rules.
02

Express the mean and covariance matrices

After the formulation of the mean and covariance matrices, the mean matrix will be the \(3 \times 1\) coefficient matrix \(\boldsymbol{\beta}\), and the covariance matrix would be \(\sigma^2 (\boldsymbol{X}' \boldsymbol{X})^{-1}\). The precise values of these matrix are not given in the problem and hence cannot be computed.
03

Calculation of \(\hat{\boldsymbol{\beta}}\)

To compute \(\hat{\boldsymbol{\beta}}\), first transpose matrix \(\boldsymbol{X}\), multiply the result by \(\boldsymbol{X}\) to obtain \(\boldsymbol{X}^T\boldsymbol{X}\) matrix, and then take the inverse of the resultant matrix. Next, take the transposed original matrix \(\boldsymbol{X}\) and multiply it by the transposed vector of given \(\boldsymbol{Y}\) values. Finally, multiply the two results to compute \(\hat{\boldsymbol{\beta}}\). Moreover, the value of \(\hat{\boldsymbol{\beta}}\) usually helps to comprehend how the linear regressors contribute to the predicted outcomes.
04

Computation of \(\hat{\boldsymbol{\beta}}\) values

For calculating the values of \(\hat{\boldsymbol{\beta}}\), follow the steps detailed in the previous step with the given data. This computation requires knowledge of matrix inversion, transposition, and multiplication, most of which can be performed with a calculator or a software package like R or Python. The resulting \(\hat{\boldsymbol{\beta}}\) values describe the influence of each regressor on the observed variable, modifying the effect for every unit increase in the corresponding predictor.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample of size \(n\) from a distribution which is \(N\left(0, \sigma^{2}\right)\). Prove that \(\sum_{1}^{n} X_{i}^{2}\) and every quadratic form, which is nonidentically zero in \(X_{1}, X_{2}, \ldots, X_{n}\), are dependent.

Suppose \(\mathbf{A}\) is a real symmetric matrix. If the eigenvalues of \(\mathbf{A}\) are only 0 's and 1 's then prove that \(\mathbf{A}\) is idempotent.

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

Let \(\mathbf{X}^{\prime}=\left[X_{1}, X_{2}\right]\) be bivariate normal with matrix of means \(\boldsymbol{\mu}^{\prime}=\left[\mu_{1}, \mu_{2}\right]\) and positive definite covariance matrix \(\mathbf{\Sigma}\). Let $$ Q_{1}=\frac{X_{1}^{2}}{\sigma_{1}^{2}\left(1-\rho^{2}\right)}-2 \rho \frac{X_{1} X_{2}}{\sigma_{1} \sigma_{2}\left(1-\rho^{2}\right)}+\frac{X_{2}^{2}}{\sigma_{2}^{2}\left(1-\rho^{2}\right)} $$ Show that \(Q_{1}\) is \(\chi^{2}(r, \theta)\) and find \(r\) and \(\theta\). When and only when does \(Q_{1}\) have a central chi-square distribution?

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a normal distribution \(N\left(\mu, \sigma^{2}\right)\). Show that $$ \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}=\sum_{i=2}^{n}\left(X_{i}-\bar{X}^{\prime}\right)^{2}+\frac{n-1}{n}\left(X_{1}-\bar{X}^{\prime}\right)^{2}, $$ where \(\bar{X}=\sum_{i=1}^{n} X_{i} / n\) and \(\bar{X}^{\prime}=\sum_{i=2}^{n} X_{i} /(n-1)\). Hint: \(\quad\) Replace \(X_{i}-\bar{X}\) by \(\left(X_{i}-\bar{X}^{\prime}\right)-\left(X_{1}-\bar{X}^{\prime}\right) / n\). Show that \(\sum_{i=2}^{n}\left(X_{i}-\bar{X}^{\prime}\right)^{2} / \sigma^{2}\) has a chi-square distribution with \(n-2\) degrees of freedom. Prove that the two terms in the right-hand member are independent. What then is the distribution of $$ \frac{[(n-1) / n]\left(X_{1}-\bar{X}^{\prime}\right)^{2}}{\sigma^{2}} ? $$

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