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Chapter 9: Problem 5

Let \(X_{1}, X_{2}, X_{3}, X_{4}\) be a random sample of size \(n=4\) from the normal distribution \(N(0,1) .\) Show that \(\sum_{i=1}^{4}\left(X_{i}-\bar{X}\right)^{2}\) equals $$ \frac{\left(X_{1}-X_{2}\right)^{2}}{2}+\frac{\left[X_{3}-\left(X_{1}+X_{2}\right) / 2\right]^{2}}{3 / 2}+\frac{\left[X_{4}-\left(X_{1}+X_{2}+X_{3}\right) / 3\right]^{2}}{4 / 3} $$ and argue that these three terms are independent, each with a chi-square distribution with 1 degree of freedom.

Short Answer

Expert verified
The given expression for the sample variance of the normal distribution can be derived. The terms in the expression are independent and follow a chi-square distribution with 1 degree of freedom which is synonymous with the properties of Chi-square distribution.

Step by step solution

01

Understanding the Sample Variance

The sample variance of a distribution is given by: \( \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}\) where \( \bar{X}\) is the sample mean and \( n\) is the number of samples. Here \( n=4\) and \( X_{i}\) are random samples from the N(0,1) distribution.
02

Substituting the Given Expression

The given expression can be rewritten as: \[ \sum_{i=1}^{4}\left(X_{i}-\bar{X}\right)^{2} = \frac{\left(X_{1}-X_{2}\right)^{2}}{2}+\frac{\left[X_{3}-\left(X_{1}+X_{2}\right)/ 2\right]^{2}}{3 / 2}+\frac{\left[X_{4}-\left(X_{1}+X_{2}+X_{3}\right) / 3\right]^{2}}{4 / 3} \]
03

Arguing the Independence and Chi-Square Distribution

Each term in the above expression represents a random variable that is a function of normally distributed random variables. As per Cochran's theorem, the formed random variables are independent if the sum of their degrees of freedom equals the total degrees of freedom. Here, each random variable has 1 degree of freedom, summing up to 3 which is equal to 4-1 (n-1: total degrees of freedom in variance calculation). Therefore, the random variables are independent. Additionally, these random variables follow a Chi-square distribution with 1 degree of freedom because they are squares of standard normal variables which is a property of chi-square distribution.

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Most popular questions from this chapter

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 \(\sum_{j} \sum_{i} a_{i j}^{2}=\operatorname{tr}\left(\mathbf{A}^{2}\right)=\operatorname{tr}\left(\mathbf{\Gamma}^{\prime} \mathbf{A}^{2} \mathbf{\Gamma}\right)=\) \(\operatorname{tr}\left[\left(\mathbf{\Gamma}^{\prime} \mathbf{A} \mathbf{\Gamma}\right)\left(\mathbf{\Gamma}^{\prime} \mathbf{A} \boldsymbol{\Gamma}\right)\right]\)

Student's scores on the mathematics portion of the ACT examination, \(x\), and on the final examination in the first-semester calculus ( 200 points possible), \(y\), are given. (a) Calculate the least squares regression line for these data. (b) Plot the points and the least squares regression line on the same graph. (c) Find point estimates for \(\alpha, \beta\), and \(\sigma^{2}\). (d) Find 95 percent confidence intervals for \(\alpha\) and \(\beta\) under the usual assumptions. $$ \begin{array}{cc|cc} \hline \mathrm{x} & \mathrm{y} & \mathrm{x} & \mathrm{y} \\ \hline 25 & 138 & 20 & 100 \\ 20 & 84 & 25 & 143 \\ 26 & 104 & 26 & 141 \\ 26 & 112 & 28 & 161 \\ 28 & 88 & 25 & 124 \\ 28 & 132 & 31 & 118 \\ 29 & 90 & 30 & 168 \\ 32 & 183 & & \\ \hline \end{array} $$

Two experiments gave the following results: $$ \begin{array}{cccccc} \hline \mathrm{n} & \bar{x} & \bar{y} & s_{x} & s_{y} & \mathrm{r} \\ \hline 100 & 10 & 20 & 5 & 8 & 0.70 \\ 200 & 12 & 22 & 6 & 10 & 0.80 \\ \hline \end{array} $$ Calculate \(r\) for the combined sample.

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

Assume that the sample \(\left(x_{1}, Y_{1}\right), \ldots,\left(x_{n}, Y_{n}\right)\) follows the linear model \((9.6 .1)\). Suppose \(Y_{0}\) is a future observation at \(x=x_{0}-\bar{x}\) and we want to determine a predictive interval for it. Assume that the model \((9.6 .1)\) holds for \(Y_{0}\); i.e., \(Y_{0}\) has a \(N\left(\alpha+\beta\left(x_{0}-\bar{x}\right), \sigma^{2}\right)\) distribution. We will use \(\hat{\eta}_{0}\) of Exercise \(9.6 .4\) as our prediction of \(Y_{0}\) (a) Obtain the distribution of \(Y_{0}-\hat{\eta}_{0}\). Use the fact that the future observation \(Y_{0}\) is independent of the sample \(\left(x_{1}, Y_{1}\right), \ldots,\left(x_{n}, Y_{n}\right)\) (b) Determine a \(t\) -statistic with numerator \(Y_{0}-\hat{\eta}_{0}\). (c) Now beginning with \(1-\alpha=P\left[-t_{\alpha / 2, n-2}
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