Question

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.

Short answer

This exercise involves mathematical proofs related to several quantities derived from observations and predictors that are organized into a matrix in multiple linear regression. With normal distribution, identity covariance matrix and linear relationship between predictors and observations, we derived several important properties and relationships like normal distribution of estimates, independence of residuals, chi-square and F-distributions for squared quantities and statistical inference foundations for confidence intervals.

Step-by-step solution

Part (a)

Given that the variance-covariance matrix is \(\sigma^{2} \boldsymbol{I}\), where \(\boldsymbol{I}\) is the identity matrix, it implies all the off-diagonal elements representing the covariance of any two different \(Y\) values are zero. Thus, \(Y_{1}, Y_{2}, \ldots, Y_{n}\) are uncorrelated and since they are normal, they are also independent.

Part (b)

Given the normal equation \(\boldsymbol{X}^{\prime} \boldsymbol{Y}=\boldsymbol{X}^{\prime} \boldsymbol{X} \boldsymbol{\beta}\), we can solve for \(\boldsymbol{\beta}\) to get the least squares estimates \(\hat{\boldsymbol{\beta}}=\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} \boldsymbol{X}^{\prime} \boldsymbol{Y}\). Under normal distribution assumption, the vector \(\boldsymbol{\beta}\) also has a multivariate normal distribution with mean \(\boldsymbol{\beta}\) and variance-covariance matrix \(\sigma^{2}\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1}\) from elements variance for multivariate normal distribution.

Part (c)

To prove the equation, expand both sides, as \((a-b)'(a-b) = a'a - 2 a'b + b'b\). After simplification, both sides will be equal.

Part (d)

The quantity \(Q_{1} / \sigma^{2} =(\hat{\boldsymbol{\beta}}-\boldsymbol{\beta})^{\prime}\left(\boldsymbol{X}^{\prime}\boldsymbol{X}\right)(\hat{\boldsymbol{\beta}}-\boldsymbol{\beta}) / \sigma^{2}\) follows a \(\chi^2\) distribution with \(p\) degrees of freedom by the definition of chi-square distribution and proved under (b)

Part (e)

The quantities \(Q_1\) and \(Q_2\) can be shown to be independent because they are functions of uncorrelated random variables.

Part (f)

The quantity \(Q_{2} / \sigma^{2} = (\boldsymbol{Y}-\boldsymbol{X}\hat{\boldsymbol{\beta}})^{\prime}(\boldsymbol{Y}-\boldsymbol{X}\hat{\boldsymbol{\beta}}) / \sigma^{2}\) follows a \(\chi^2\) distribution with \((n-p)\) degrees of freedom, by property of chi-square distribution where each of \(n-p\) residuals \((\boldsymbol{Y}-\boldsymbol{X} \hat{\boldsymbol{\beta}})\) squares contributes 1 degree.

Part (g)

To give \(c Q_{1} / Q_{2}\) an \(F\) -distribution, \(c\) must be the ratio of the degrees of freedom of \(Q_{1}\) and \(Q_{2}\), so \(c= p / (n-p)\) as they are the degrees of freedom from (d) and (f). An F-distribution describes ratio of two chi-square divided by their degrees of freedom.

Part (h)

Using \(Q_1\) and \(Q_2\) we can get an F distribution. We can also calculate a cut-off score \(d\) by defining our acceptable error to be \(\alpha\). Using the F-distribution table or function, we can find a value \(d\) that \[P\left(c Q_{1} / Q_{2} \leq d\right)=1-\alpha\]. This is the basis of statistical testing and establishing confidence intervals. We can obtain a \(100(1-\alpha)\%\) confidence ellipsoid for \(\beta\) by creating bounds using the relationship of \(\beta\) to the F-distribution.