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Let \(Z\) be a random vector with 4 components and covariance matrix \(\sigma^{2} I .\) Let \(U=Z_{1}+Z_{2}+Z_{3}+Z_{4}\) and \(V=\left(Z_{1}+Z_{2}\right)-\left(Z_{3}+Z_{4}\right) .\) Use matrix methods to find \(\operatorname{Cov}(U, V)\)

Short Answer

Expert verified
The covariance \( \operatorname{Cov}(U, V) = 0 \).

Step by step solution

01

Recognize the components of the random vector

Identify the components of the random vector \( Z \). We are given that \( Z = (Z_1, Z_2, Z_3, Z_4) \) is a random vector with a covariance matrix \( \sigma^2 I \). This means \( Z_1, Z_2, Z_3, \) and \( Z_4 \) are uncorrelated and each has variance \( \sigma^2 \).
02

Define the linear combinations for U and V

U is defined as \( U = Z_1 + Z_2 + Z_3 + Z_4 \), and V is defined as \( V = (Z_1 + Z_2) - (Z_3 + Z_4) \). These are linear combinations of the components of \( Z \).
03

Express U and V in matrix form

The random variables \( U \) and \( V \) can be represented in matrix form through a transformation applied on \( Z \). We write: \[ \begin{pmatrix} U \ V \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 & 1 \ 1 & 1 & -1 & -1 \end{pmatrix} \begin{pmatrix} Z_1 \ Z_2 \ Z_3 \ Z_4 \end{pmatrix}.\] Let \( A = \begin{pmatrix} 1 & 1 & 1 & 1 \ 1 & 1 & -1 & -1 \end{pmatrix} \).
04

Compute the covariance matrix of U and V

The covariance matrix of \( (U, V) \) is given by \( \text{Cov}(U, V) = A \text{Cov}(Z) A^T \). With \( \text{Cov}(Z) = \sigma^2 I \), where \( I \) is the identity matrix, it becomes: \[ \text{Cov}(U, V) = \sigma^2 \begin{pmatrix} 1 & 1 & 1 & 1 \ 1 & 1 & -1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 1 \ 1 & 1 \ 1 & -1 \ 1 & -1 \end{pmatrix}.\]
05

Calculate the matrix product

Perform the matrix multiplication: \[ \sigma^2 \begin{pmatrix} 1 & 1 & 1 & 1 \ 1 & 1 & -1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 1 \ 1 & 1 \ 1 & -1 \ 1 & -1 \end{pmatrix} = \sigma^2 \begin{pmatrix} 4 & 0 \ 0 & 4 \end{pmatrix}.\]
06

Interpret the covariance matrix

The resulting matrix is \( \sigma^2 \begin{pmatrix} 4 & 0 \ 0 & 4 \end{pmatrix} \). The off-diagonal element is zero, indicating that \( U \) and \( V \) are uncorrelated. Thus, \( \text{Cov}(U, V) = 0 \).

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Key Concepts

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

Random Vectors
A random vector is a collection of random variables often treated as a single entity because they are defined in a joint probability space. Think of it as an array of random values that can have correlations between one another. For example, in our exercise, the random vector \( Z \) contains four components: \( (Z_1, Z_2, Z_3, Z_4) \). Each of these components is random, meaning that they can vary and follow a specific probability distribution.
One important characteristic of the random vector \( Z \) is its covariance matrix, \( \sigma^2 I \). This indicates that each component \( Z_i \) has the same variance \( \sigma^2 \) and they are uncorrelated with one another. Knowing this makes further analysis, like calculating covariances, easier because we deal with simpler mathematical properties.
Linear Combinations
Linear combinations are fundamental in understanding random vectors and transformations. They involve adding and subtracting vectors with specific coefficients. In our problem, \( U \) and \( V \) are linear combinations of the components of \( Z \). Specifically, \( U = Z_1 + Z_2 + Z_3 + Z_4 \) and \( V = (Z_1 + Z_2) - (Z_3 + Z_4) \).
By creating linear combinations, we can simplify complex random vectors into more manageable forms. This method allows us to express these combinations in matrix form, facilitating their manipulation in mathematical equations. Understanding these combinations is crucial as they serve as the building blocks for matrix operations used in deriving properties like covariance.
Matrix Multiplication
Matrix multiplication is a core mathematical operation that allows combining matrices to transform vectors or express linear transformations. It is akin to blending two sets of numbers to produce a new set. For our random vector exercise, matrix multiplication was essential in determining the covariance of \( U \) and \( V \).
To express \( U \) and \( V \) in matrix form, we formed a matrix \( A \) representing their linear combinations and multiplied it with the covariance matrix of \( Z \). The formula used was \( \text{Cov}(U, V) = A \text{Cov}(Z) A^T \). Here, \( A \) is multiplied by the identity matrix \( \sigma^2 I \), followed by \( A^T \), the transpose of \( A \). This multiplication process simplified to give us a new covariance matrix where \( U \) and \( V \) were uncorrelated.
Transformation Matrix
A transformation matrix is a tool that applies a specific linear transformation to a vector, changing its space or configuration. In our scenario, the transformation matrix \( A = \begin{pmatrix} 1 & 1 & 1 & 1 \ 1 & 1 & -1 & -1 \end{pmatrix} \) was used to convert the components of the random vector \( Z \) into the variables \( U \) and \( V \).
This matrix enabled us to express \( U \) and \( V \) as linear combinations of the original \( Z \) components. By applying matrix multiplication, we transformed the properties of \( Z \) into the properties of \( U \) and \( V \), notably, their variance and covariance. The transformation matrix, therefore, played a vital role by allowing us to capture important statistical properties between \( U \) and \( V \) at a new level of abstraction.

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Most popular questions from this chapter

The following data come from the calibration of a proving ring, a device for measuring force (Hockersmith and Ku 1969). a. Plot load versus deflection. Does the plot look linear? b. Fit deflection as a linear function of load, and plot the residuals versus load. Do the residuals show any systematic lack of fit? c. Fit deflection as a quadratic function of load, and estimate the coefficients and their standard errors. Plot the residuals. Does the fit look reasonable? $$\begin{array}{cccc} \hline \multicolumn{5}{c} {\text { Deflection }} \\ \hline \text { Load } & \text { Run 1 } & \text { Run 2 } & \text { Run 3 } \\\ \hline 10,000 & 68.32 & 68.35 & 68.30 \\ 20,000 & 136.78 & 136.68 & 136.80 \\ 30,000 & 204.98 & 205.02 & 204.98 \\ 40,000 & 273.85 & 273.85 & 273.80 \\ 50,000 & 342.70 & 342.63 & 342.63 \\ 60,000 & 411.30 & 411.35 & 411.28 \\ 70,000 & 480.65 & 480.60 & 480.63 \\ 80,000 & 549.85 & 549.85 & 549.83 \\ 90,000 & 619.00 & 619.02 & 619.10 \\ 100,000 & 688.70 & 688.62 & 688.58 \\ \hline \end{array}$$

Show that the least squares estimates of the slope and intercept of a line may be expressed as $$ \beta_{0}=\bar{y}-\beta_{1} \bar{x} $$ and $$ \hat{\beta}_{1}=\frac{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)\left(y_{i}-\bar{y}\right)}{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}} $$

Suppose that the relation of family income to consumption is linear. Of those families in the 90 th percentile of income, what proportion would you expect to be at or above the 90 th percentile of consumption: (a) exactly \(50 \%,\) (b) less than \(50 \%,\) (c) more than \(50 \% ?\) Justify your answers.

(Weighted Least Squares) Suppose that in the model \(y_{i}=\beta_{0}+\beta_{1} x_{i}+e_{i},\) the errors have mean zero and are independent, but \(\operatorname{Var}\left(e_{i}\right)=\rho_{i}^{2} \sigma^{2},\) where the \(\rho_{i}\) are known constants, so the errors do not have equal variance. This situation arises when the \(y_{i}\) are averages of several observations at \(x_{i}\); in this case, if \(y_{i}\) is an average of \(n_{i}\) independent observations, \(\rho_{i}^{2}=1 / n_{i}\) (why?). Because the variances are not equal, the theory developed in this chapter does not apply; intuitively, it seems that the observations with large variability should influence the estimates of \(\beta_{0}\) and \(\beta_{1}\) less than the observations with small variability. The problem may be transformed as follows: $$ \rho_{i}^{-1} y_{i}=\rho_{i}^{-1} \beta_{0}+\rho_{i}^{-1} \beta_{1} x_{i}+\rho_{i}^{-1} e_{i} $$ or $$ z_{i}=u_{i} \beta_{0}+v_{i} \beta_{1}+\delta_{i} $$ where $$ u_{i}=\rho_{i}^{-1} \quad v_{i}=\rho_{i}^{-1} x_{i} \quad \delta_{i}=\rho_{i}^{-1} e_{i} $$ a. Show that the new model satisfies the assumptions of the standard statistical model. b. Find the least squares estimates of \(\beta_{0}\) and \(\beta_{1}\) c. Show that performing a least squares analysis on the new model, as was done in part (b), is equivalent to minimizing $$ \sum_{i=1}^{n}\left(y_{i}-\beta_{0}-\beta_{1} x_{i}\right)^{2} \rho_{i}^{-2} $$ This is a weighted least squares criterion; the observations with large variances are weighted less. d. Find the variances of the estimates of part (b).

Suppose that the independent variables in a least squares problem are replaced by rescaled variables \(u_{i j}=k_{j} x_{i j}\) (for example, centimeters are converted to meters.) Show that \(Y\) does not change. Does \(\hat{\beta}\) change? (Hint: Express the new design matrix in terms of the old one.)

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