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Often in regression the mean of the random variable Y is a linear function of p -values x1,x2,,xp, say β1x1+β2x2++βpxp, where β=(β1,β2,,βp) are the regression coefficients. Suppose that n values, Y=(Y1,Y2,,Yn) are observed for the x -values in X=[xij], where X is an n×p design matrix and its ith row is associated with Yi,i=1,2,,n. Assume that Y is multivariate normal with mean Xβ and variance-covariance matrix σ2I, where I is the n×n identity matrix. (a) Note that Y1,Y2,,Yn are independent. Why? (b) Since Y should approximately equal its mean Xβ, we estimate β by solving the normal equations XY=XXβ for β. Assuming that XX is non- singular, solve the equations to get β^=(XX)1XY. Show that β^ has a multivariate normal distribution with mean β and variance-covariance matrix σ2(XX)1 (c) Show that (YXβ)(YXβ)=(β^β)(XX)(β^β)+(YXβ^)(YXβ^) say Q=Q1+Q2 for convenience. (d) Show that Q1/σ2 is χ2(p). (e) Show that Q1 and Q2 are independent. (f) Argue that Q2/σ2 is χ2(np). (g) Find c so that cQ1/Q2 has an F -distribution. (h) The fact that a value d can be found so that P(cQ1/Q2d)=1α could be used to find a 100(1α) percent confidence ellipsoid for β. Explain.

Short Answer

Expert verified
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

01

Part (a)

Given that the variance-covariance matrix is σ2I, where I is the identity matrix, it implies all the off-diagonal elements representing the covariance of any two different Y values are zero. Thus, Y1,Y2,,Yn are uncorrelated and since they are normal, they are also independent.
02

Part (b)

Given the normal equation XY=XXβ, we can solve for β to get the least squares estimates β^=(XX)1XY. Under normal distribution assumption, the vector β also has a multivariate normal distribution with mean β and variance-covariance matrix σ2(XX)1 from elements variance for multivariate normal distribution.
03

Part (c)

To prove the equation, expand both sides, as (ab)(ab)=aa2ab+bb. After simplification, both sides will be equal.
04

Part (d)

The quantity Q1/σ2=(β^β)(XX)(β^β)/σ2 follows a χ2 distribution with p degrees of freedom by the definition of chi-square distribution and proved under (b)
05

Part (e)

The quantities Q1 and Q2 can be shown to be independent because they are functions of uncorrelated random variables.
06

Part (f)

The quantity Q2/σ2=(YXβ^)(YXβ^)/σ2 follows a χ2 distribution with (np) degrees of freedom, by property of chi-square distribution where each of np residuals (YXβ^) squares contributes 1 degree.
07

Part (g)

To give cQ1/Q2 an F -distribution, c must be the ratio of the degrees of freedom of Q1 and Q2, so c=p/(np) 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.
08

Part (h)

Using Q1 and Q2 we can get an F distribution. We can also calculate a cut-off score d by defining our acceptable error to be α. Using the F-distribution table or function, we can find a value d that P(cQ1/Q2d)=1α. This is the basis of statistical testing and establishing confidence intervals. We can obtain a 100(1α)% confidence ellipsoid for β by creating bounds using the relationship of β to the F-distribution.

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Key Concepts

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

Multivariate Normal Distribution
In regression analysis, understanding the multivariate normal distribution is crucial. It is a generalization of the univariate normal distribution to more than one variable. This concept comes into play when we examine the regression model Y=Xβ+ϵ where
  • Y represents the vector of observed values,
  • X is the design matrix containing rows that are associated with these observations,
  • β is a vector of regression coefficients, and
  • ϵ is the error term.
When the random variable Y follows a multivariate normal distribution with mean Xβ and a variance-covariance matrix σ2I, it means that each element of Y is a normally distributed random variable, and they are independent of each other. This independence is due to the identity matrix I, which indicates zero covariance between different variables of Y. This feature of multivariate normal distribution helps simplify the process of estimating the regression parameters β.
Least Squares Estimation
Least squares estimation is a method used to estimate unknown parameters in a regression model. In simple terms, it minimizes the sum of squared differences between the observed and predicted values. For regression models, the normal equation used is XY=XXβ By solving this equation, we find the estimated coefficients, denoted as β^, given by β^=(XX)1XY This method is chosen because it provides the best linear unbiased estimator (BLUE) under the assumption that errors are normally distributed with constant variance. Furthermore, because the errors are assumed normally distributed, β^ itself follows a multivariate normal distribution with mean β and variance σ2(XX)1. This property is particularly useful because it allows us to make statistical inferences about the regression coefficients.
Chi-Square Distribution
The chi-square distribution is essential in regression analysis for hypothesis testing and constructing confidence intervals. It is used as a measure of the distribution of variance. In the context of regression, we consider quantities such as Q1/σ2 where Q1=(β^β)(XX)(β^β) This expression yields a χ2 distribution with p degrees of freedom, where p is the number of predictors in the regression model. This is because the term β^β is normally distributed, and its quadratic form with respect to the matrix XX produces a χ2 distribution. Another form related to the error or residual component, Q2/σ2 where Q2=(YXβ^)(YXβ^) also follows a chi-square distribution but with np degrees of freedom. Here, n stands for the number of observations. Understanding these distributions helps in evaluating the goodness-of-fit and conducting statistical tests in regression analysis.
Confidence Ellipsoid
A confidence ellipsoid is an extension of the confidence interval to multiple dimensions. In the regression context, it helps to give a bounded region where the true parameter β lies with a certain probability. By combining the chi-square distribution findings with the F-distribution, we can design a confidence ellipsoid for β. Using statistics like cQ1/Q2 which follows an F-distribution, a critical value d can be determined such that P(cQ1/Q2d)=1α where α is the significance level.The ellipsoid is defined for a 100(1α)% confidence level. Geometrically, this region is shaped like an ellipsoid, illustrating the joint confidence in estimates of the multiple regression coefficients. This is crucial in assessing how far the estimated β^ might stray from the actual β due to sampling variability, thus providing a multidimensional insight into the precision of our regression estimates.

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Most popular questions from this chapter

Let A be the real symmetric matrix of a quadratic form Q in the observations of a random sample of size n from a distribution which is N(0,σ2). Given that Q and the mean X¯ of the sample are independent, what can be said of the elements of each row (column) of A ? Hint: Are Q and X2 independent?

Let X1,X2,,Xn be a random sample from a normal distribution N(μ,σ2). Show that i=1n(XiX¯)2=i=2n(XiX¯)2+n1n(X1X¯)2, where X¯=i=1nXi/n and X¯=i=2nXi/(n1). Hint: Replace XiX¯ by (XiX¯)(X1X¯)/n. Show that i=2n(XiX¯)2/σ2 has a chi-square distribution with n2 degrees of freedom. Prove that the two terms in the right-hand member are independent. What then is the distribution of [(n1)/n](X1X¯)2σ2?

Show that j=1bi=1a(XijX¯i.)2=j=1bi=1a(XijX¯iX¯.j+X¯..)2+aj=1b(X¯.jX¯..)2.

Let X1,X2,X3,X4 denote a random sample of size 4 from a distribution which is N(0,σ2). Let Y=14aiXi, where a1,a2,a3, and a4 are real constants. If Y2 and Q=X1X2X3X4 are independent, determine a1,a2,a3, and a4.

Let Xijk,i=1,,a;j=1,,b,k=1,,c, be a random sample of size n=abc from a normal distribution N(μ,σ2). Let X¯=k=1cj=1bi=1aXijk/n and X¯ir=k=1cj=1bXijk/bc. Prove that i=1aj=1bk=1c(XijkX¯)2=i=1aj=1bk=1c(XijkX¯i..)2+bci=1a(X¯i.X¯)2 Show that i=1aj=1bk=1c(XijkX¯i..)2/σ2 has a chi-square distribution with a(bc1) degrees of freedom. Prove that the two terms in the right-hand member are independent. What, then, is the distribution of bci=1a(X¯i.X¯)2/σ2? Furthermore, let X.j.=k=1ci=1aXijk/ac and X¯ij.=k=1cXijk/c. Show that i=1aj=1bk=1c(XijkX¯)2=i=1aj=1bk=1c(XijkX¯ij.)2+bci=1a(X¯inX¯)2+acj=1b(X¯.jX¯)2+ci=1aj=1b(X¯ij.X¯i.X.j.+X) Prove that the four terms in the right-hand member, when divided by σ2, are independent chi-square variables with ab(c1),a1,b1, and (a1)(b1) degrees of freedom, respectively.

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