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Example 6.13 (A two-way multivariate analysis of variance of plastic film data) The optimum conditions for extruding plastic film have been examined using a technique called Evolutionary Operation. (See [9].) In the course of the study that was done, three responses \( - {X}_{1} = \) tear resistance, \( {X}_{2} = \) gl...
The matrices of the appropriate sum of squares and cross products were calculated (see the SAS statistical software output in Panel \( {6.1}^{6} \) ), leading to the following MANOVA table:\n\n<table><thead><tr><th>Source of variation</th><th></th><th>SSP</th><th></th><th>d.f.</th></tr></thead><tr><td rowspan=\
No
To test for parallelism \( \left( {{H}_{01} : \mathbf{C}{\mathbf{\mu }}_{1} = \mathbf{C}{\mathbf{\mu }}_{2}}\right) \), we compute
\[ {\mathbf{{CS}}}_{\text{pooled }}{\mathbf{C}}^{\prime } = \left\lbrack \begin{array}{rrrr} - 1 & 1 & 0 & 0 \\ 0 & - 1 & 1 & 0 \\ 0 & 0 & - 1 & 1 \end{array}\right\rbrack {\mathbf{S}}_{\text{pooled }}\left\lbrack \begin{array}{rrr} - 1 & 0 & 0 \\ 1 & - 1 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \end{array}\right\rbrack \]\n\[ ...
Yes
Example 6.16 (Comparing multivariate and univariate tests for the differences in means) Suppose we collect measurements on two variables \( {X}_{1} \) and \( {X}_{2} \) for ten randomly selected experimental units from each of two groups. The hypothetical data are noted here and displayed as scatter plots and marginal ...
Let \( {\mathbf{\mu }}_{1}^{\prime } = \left\lbrack {{\mu }_{11},{\mu }_{12}}\right\rbrack \) be the population mean vector for the first group, and let \( {\mathbf{\mu }}_{2}^{\prime } = \left\lbrack {{\mu }_{21},{\mu }_{22}}\right\rbrack \) be the population mean vector for the second group. Using the \( {x}_{1} \) o...
Yes
Example 7.1 (Fitting a straight-line regression model) Determine the linear regression model for fitting a straight line\n\n\[ \n\\text{ Mean response } = E\\left( Y\\right) = {\\beta }_{0} + {\\beta }_{1}{z}_{1} \n\]\n\nto the data\n\n![10262ff6-7dc4-4708-a29b-6e887f1ef388_367_0.jpg](images/10262ff6-7dc4-4708-a29b-6e8...
The data for this model are contained in the observed response vector \( \\mathbf{y} \) and the design matrix \( \\mathbf{Z} \), where\n\n\[ \n\\mathbf{y} = \\left\\lbrack \\begin{array}{l} 1 \\\\ 4 \\\\ 3 \\\\ 8 \\\\ 9 \\end{array}\\right\\rbrack ,\\;\\mathbf{Z} = \\left\\lbrack \\begin{array}{ll} 1 & 0 \\\\ 1 & 1 \\\...
Yes
Example 7.2 (The design matrix for one-way ANOVA as a regression model) Determine the design matrix if the linear regression model is applied to the one-way ANOVA situation in Example 6.6.
We create so-called dummy variables to handle the three population means: \( {\mu }_{1} = \mu + {\tau }_{1},{\mu }_{2} = \mu + {\tau }_{2} \), and \( {\mu }_{3} = \mu + {\tau }_{3} \). We set\n\n\[ \n{z}_{1} = \left\{ {\begin{array}{ll} 1 & \text{ if the observation is } \\ & \text{ from population }1 \\ 0 & \text{ oth...
Yes
Example 7.3 (Calculating the least squares estimates, the residuals, and the residual sum of squares) Calculate the least square estimates \( \widehat{\mathbf{\beta }} \), the residuals \( \widehat{\mathbf{\varepsilon }} \), and the residual sum of squares for a straight-line model\n\n\[ \n{Y}_{j} = {\beta }_{0} + {\be...
Consequently,\n\n\[ \n\widehat{\mathbf{\beta }} = \left\lbrack \begin{matrix} {\widehat{\beta }}_{0} \\ {\widehat{\beta }}_{1} \end{matrix}\right\rbrack = {\left( {\mathbf{Z}}^{\prime }\mathbf{Z}\right) }^{-1}{\mathbf{Z}}^{\prime }\mathbf{y} = \left\lbrack \begin{array}{rr} {.6} & - {.2} \\ - {.2} & {.1} \end{array}\ri...
Yes
Example 7.4 (Fitting a regression model to real-estate data) The assessment data in Table 7.1 were gathered from 20 homes in a Milwaukee, Wisconsin, neighborhood. Fit the regression model\n\n\[ \n{Y}_{j} = {\beta }_{0} + {\beta }_{1}{z}_{j1} + {\beta }_{2}{z}_{j2} + {\varepsilon }_{j} \n\]\n\nwhere \( {z}_{1} = \) tota...
A computer calculation yields\n\n\[ \n{\left( {\mathbf{Z}}^{\prime }\mathbf{Z}\right) }^{-1} = \left\lbrack \begin{array}{rrr} {5.1523} & & \\ {.2544} & {.0512} & \\ - {.1463} & - {.0172} & {.0067} \end{array}\right\rbrack \n\]\n\nand\n\n\[ \n\widehat{\mathbf{\beta }} = {\left( {\mathbf{Z}}^{\prime }\mathbf{Z}\right) }...
Yes
Construct a \( {95}\% \) confidence interval for the mean CPU time, \( E\left( {{Y}_{0} \mid {\mathbf{z}}_{0}}\right) = \) \( {\beta }_{0} + {\beta }_{1}{z}_{01} + {\beta }_{2}{z}_{02} \) at \( {\mathbf{z}}_{0}^{\prime } = \left\lbrack {1,{130},{7.5}}\right\rbrack \) .
A computer program provides the estimated regression function\n\n\[ \widehat{y} = {8.42} + {1.08}{z}_{1} + {.42}{z}_{2} \]\n\n\[ {\left( {\mathbf{Z}}^{\prime }\mathbf{Z}\right) }^{-1} = \left\lbrack \begin{array}{rrr} {8.17969} & & \\ - {.06411} & {.00052} & \\ {.08831} & - {.00107} & {.01440} \end{array}\right\rbrack ...
Yes
To illustrate the calculations of \( \widehat{\mathbf{\beta }},\widehat{\mathbf{Y}} \), and \( \widehat{\mathbf{\varepsilon }} \), we fit a straight-line regression model (see Panel 7.2), \n\n\[ \n{Y}_{j1} = {\beta }_{01} + {\beta }_{11}{z}_{j1} + {\varepsilon }_{j1} \n\] \n\n\[ \n{Y}_{j2} = {\beta }_{02} + {\beta }_{1...
The design matrix \( \mathbf{Z} \) remains unchanged from the single-response problem. We find that \n\n\[ \n{\mathbf{Z}}^{\prime } = \left\lbrack \begin{array}{lllll} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 & 4 \end{array}\right\rbrack \;{\left( {\mathbf{Z}}^{\prime }\mathbf{Z}\right) }^{-1} = \left\lbrack \begin{array}{rr...
Yes
Example 7.9 (Testing the importance of additional predictors with a multivariate response) The service in three locations of a large restaurant chain was rated according to two measures of quality by male and female patrons. The first service-quality index was introduced in Example 7.5. Suppose we consider a regression...
\[ \left( \begin{matrix} \text{ residual sum of squares } \\ \text{ and cross products } \end{matrix}\right) = n\widehat{\mathbf{\sum }} = \left\lbrack \begin{array}{ll} {2977.39} & {1021.72} \\ {1021.72} & {2050.95} \end{array}\right\rbrack \] \[ \left( \begin{matrix} \text{ extra sum of squares } \\ \text{ and cross ...
Yes
Example 7.11 (Determining the best linear predictor, its mean square error, and the multiple correlation coefficient) Given the mean vector and covariance matrix of \( Y \) , \( {Z}_{1},{Z}_{2} \)\n\n\[ \mathbf{\mu } = \left\lbrack \begin{matrix} {\mathbf{\mu }}_{Y} \\ {\mathbf{\mu }}_{\mathbf{Z}} \\ {\mathbf{\mu }}_{\...
First,\n\n\[ \mathbf{\beta } = {\mathbf{\sum }}_{\mathbf{{ZZ}}}^{-1}{\mathbf{\sigma }}_{\mathbf{Z}Y} = {\left\lbrack \begin{array}{ll} 7 & 3 \\ 3 & 2 \end{array}\right\rbrack }^{-1}\left\lbrack \begin{array}{r} 1 \\ - 1 \end{array}\right\rbrack = \left\lbrack \begin{array}{rr} {.4} & - {.6} \\ - {.6} & {1.4} \end{array...
Yes
For the computer data of Example 7.6, the \( n = 7 \) observations on \( Y \) (CPU time), \( {Z}_{1} \) (orders), and \( {Z}_{2} \) (add-delete items) give the sample mean vector and sample covariance matrix:\n\n\[ \n\widehat{\mathbf{\mu }} = \left\lbrack \begin{matrix} \bar{y} \\ \overline{\mathbf{Z}} \\ \overline{\ma...
Result 7.13 gives the maximum likelihood estimates\n\n\[ \n\widehat{\mathbf{\beta }} = {\mathbf{S}}_{\mathbf{{ZZ}}}^{-1}{\mathbf{s}}_{\mathbf{Z}Y} = \left\lbrack \begin{array}{rr} {.003128} & - {.006422} \\ - {.006422} & {.086404} \end{array}\right\rbrack \left\lbrack \begin{matrix} {418.763} \\ {35.983} \end{matrix}\r...
Yes
Example 7.13 (Maximum likelihood estimates of the regression functions-two responses) We return to the computer data given in Examples 7.6 and 7.10. For \( {Y}_{1} = \) CPU time, \( {Y}_{2} = \) disk I/O capacity, \( {Z}_{1} = \) orders, and \( {Z}_{2} = \) add-delete items, we have Assuming normality, we find that the...
\[ {\widehat{\mathbf{\beta }}}_{0} + \widehat{\mathbf{\beta }}\mathbf{z} = \overline{\mathbf{y}} + {\mathbf{S}}_{\mathbf{{YZ}}}{\mathbf{S}}_{\mathbf{{ZZ}}}^{-1}\left( {\mathbf{z} - \overline{\mathbf{z}}}\right) \] \[ = \left\lbrack \begin{array}{l} {150.44} \\ {327.79} \end{array}\right\rbrack + \left\lbrack \begin{arr...
Yes
Example 7.14 (Calculating a partial correlation) From the computer data in Example 7.13,
\[ {\mathbf{S}}_{\mathbf{{YY}}} - {\mathbf{S}}_{\mathbf{{YZ}}}{\mathbf{S}}_{\mathbf{{ZZ}}}^{-1}{\mathbf{S}}_{\mathbf{{ZY}}} = \left\lbrack \begin{array}{ll} {1.043} & {1.042} \\ {1.042} & {2.572} \end{array}\right\rbrack \] \n\nTherefore, \n\n\[ {r}_{{Y}_{1}{Y}_{2} \cdot \mathbf{Z}} = \frac{{s}_{{Y}_{1}{Y}_{2} \cdot \m...
Yes
Example 7.15 (Two approaches yield the same linear predictor) The computer data with the single response \( {Y}_{1} = \) CPU time were analyzed in Example 7.6 using the classical linear regression model. The same data were analyzed again in Example 7.12, assuming that the variables \( {Y}_{1},{Z}_{1} \), and \( {Z}_{2}...
\[ \widehat{y} = {8.42} + {1.08}{z}_{1} + {.42}{z}_{2} \]
Yes
Example 7.16 (Incorporating time dependent errors in a regression model) Power companies must have enough natural gas to heat all of their customers' homes and businesses, particularly during the coldest days of the year. A major component of the planning process is a forecasting exercise based on a model relating the ...
Initially, we developed a regression model relating gas sendout to degree heating days, wind speed and a weekend dummy variable. Other variables likely to have some affect on natural gas consumption, like percent cloud cover, are subsumed in the error term. After several attempted fits, we decided to include not only t...
Yes
Example 8.1 (Calculating the population principal components) Suppose the random variables \( {X}_{1},{X}_{2} \) and \( {X}_{3} \) have the covariance matrix\n\n\[ \n\mathbf{\sum } = \left\lbrack \begin{array}{rrr} 1 & - 2 & 0 \\ - 2 & 5 & 0 \\ 0 & 0 & 2 \end{array}\right\rbrack \n\]\n\nIt may be verified that the eige...
Equation (8-5) can be demonstrated from first principles. For example,\n\n\[ \n\operatorname{Var}\left( {Y}_{1}\right) = \operatorname{Var}\left( {{.383}{X}_{1} - {.924}{X}_{2}}\right) \n\]\n\n\[ \n= {\left( {.383}\right) }^{2}\operatorname{Var}\left( {X}_{1}\right) + {\left( -{.924}\right) }^{2}\operatorname{Var}\left...
Yes
Can the sample variation be summarized by one or two principal components?
We find the following:\n\nCoefficients for the Principal Components (Correlation Coefficients in Parentheses)\n\n<table><thead><tr><th>Variable</th><th>\( {\widehat{\mathbf{e}}}_{1}\left( {r}_{{\widehat{y}}_{1},{x}_{k}}\right) \)</th><th>\( {\widehat{\mathbf{e}}}_{2}\left( {r}_{{\widehat{y}}_{2},{x}_{k}}\right) \)</th>...
Yes
Example 8.4 (Summarizing sample variability with one sample principal component) In a study of size and shape relationships for painted turtles, Jolicoeur and Mosi-mann [11] measured carapace length, width, and height. Their data, reproduced in Exercise 6.18, Table 6.9, suggest an analysis in terms of logarithms. (Joli...
The natural logarithms of the dimensions of 24 male turtles have sample mean vector \( {\overline{\mathbf{x}}}^{\prime } = \left\lbrack {{4.725},{4.478},{3.703}}\right\rbrack \) and covariance matrix\n\n\[ \mathbf{S} = {10}^{-3}\left\lbrack \begin{matrix} {11.072} & {8.019} & {8.160} \\ {8.019} & {6.417} & {6.005} \\ {...
Yes
Example 8.5 (Sample principal components from standardized data) The weekly rates of return for five stocks (JP Morgan, Citibank, Wells Fargo, Royal Dutch Shell, and ExxonMobil) listed on the New York Stock Exchange were determined for the period January 2004 through December 2005. The weekly rates of return are define...
Let \( {x}_{1},{x}_{2},\ldots ,{x}_{5} \) denote observed weekly rates of return for JP Morgan, Citibank, Wells Fargo, Royal Dutch Shell, and ExxonMobil, respectively. Then\n\n\[ \n{\overline{\mathbf{x}}}^{\prime } = \left\lbrack {{.0011},{.0007},{.0016},{.0040},{.0040}}\right\rbrack \n\]\n\nand\n\n\[ \n\mathbf{R} = \l...
Yes
Example 8.7 (Plotting the principal components for the turtle data) We illustrate the plotting of principal components for the data on male turtles discussed in Example 8.4. The three sample principal components are
\[ {\widehat{y}}_{1} = {.683}\left( {{x}_{1} - {4.725}}\right) + {.510}\left( {{x}_{2} - {4.478}}\right) + {.523}\left( {{x}_{3} - {3.703}}\right) \] \[ {\widehat{y}}_{2} = - {.159}\left( {{x}_{1} - {4.725}}\right) - {.594}\left( {{x}_{2} - {4.478}}\right) + {.788}\left( {{x}_{3} - {3.703}}\right) \] \[ {\widehat{y}}_{...
Yes
Example 8.8 (Constructing a confidence interval for \( {\lambda }_{1} \) ) We shall obtain a \( {95}\% \) confidence interval for \( {\lambda }_{1} \), the variance of the first population principal component, using the stock price data listed in Table 8.4 in the Exercises.\n\nAssume that the stock rates of return repr...
Therefore, with \( {95}\% \) confidence,\n\n\[ \frac{.0014}{\left( 1 + {1.96}\sqrt{\frac{2}{103}}\right) } \leq {\lambda }_{1} \leq \frac{.0014}{\left( 1 - {1.96}\sqrt{\frac{2}{103}}\right) }\text{ or }{.0011} \leq {\lambda }_{1} \leq {.0019} \]
Yes
Example 8.10 (An ellipse format chart based on the first two principal components) Refer to the police department overtime data given in Table 5.8. Table 8.1 contains the five normalized eigenvectors and eigenvalues of the sample covariance matrix \( \mathbf{S} \). The first two sample components explain \( {82}\% \) o...
Although \( n = {16} \) is not large, we use \( {\chi }_{2}^{2}\left( {.05}\right) = {5.99} \), and the ellipse becomes \[ \frac{{\widehat{y}}_{1}^{2}}{{\widehat{\lambda }}_{1}} + \frac{{\widehat{y}}_{2}^{2}}{{\widehat{\lambda }}_{2}} = {5.99} \]
Yes
Example 8.11 (A \( {T}^{2} \) -chart for the unexplained [orthogonal] overtime hours) Consider the quality control analysis of the police department overtime hours in Example 8.10. The first part of the quality monitoring procedure, the quality ellipse based on the first two principal components, was shown in Figure 8....
Since \( p = 5 \), this chart is based on \( 5 - 2 = 3 \) dimensions, and the upper control limit is \( {\chi }_{3}^{2}\left( {.05}\right) = {7.81} \) . Using the eigenvalues and the values of the principal components, given in Example 8.10, we plot the time sequence of values\n\n\[ \n{T}_{j}^{2} = \frac{{\widehat{y}}_...
Yes
Example 8.12 (Control ellipse for future principal components) In Example 8.10, we determined that case 11 was out of control. We drop this point and recalculate the eigenvalues and eigenvectors based on the covariance of the remaining 15 observations. The results are shown in Table 8.3.
Figure 8.9 gives the \( {99}\% \) prediction (8-36) ellipse for future pairs of values for the new first two principal components of overtime. The 15 stable pairs of principal components are also shown.
Yes
Example 9.1 (Verifying the relation \( \sum = {\mathrm{{LL}}}^{\prime } + \Psi \) for two factors) Consider the covariance matrix\n\n\[ \mathbf{\sum } = \left\lbrack \begin{array}{rrrr} {19} & {30} & 2 & {12} \\ {30} & {57} & 5 & {23} \\ 2 & 5 & {38} & {47} \\ {12} & {23} & {47} & {68} \end{array}\right\rbrack \]\n\n--...
The equality\n\n\[ \left\lbrack \begin{array}{rrrr} {19} & {30} & 2 & {12} \\ {30} & {57} & 5 & {23} \\ 2 & 5 & {38} & {47} \\ {12} & {23} & {47} & {68} \end{array}\right\rbrack = \left\lbrack \begin{array}{rr} 4 & 1 \\ 7 & 2 \\ - 1 & 6 \\ 1 & 8 \end{array}\right\rbrack \left\lbrack \begin{array}{rrrr} 4 & 7 & - 1 & 1 ...
Yes
Example 9.2 (Nonexistence of a proper solution) Let \( p = 3 \) and \( m = 1 \), and suppose the random variables \( {X}_{1},{X}_{2} \), and \( {X}_{3} \) have the positive definite covariance matrix\n\n\[ \n\mathbf{\sum } = \left\lbrack \begin{array}{lll} 1 & {.9} & {.7} \\ {.9} & 1 & {.4} \\ {.7} & {.4} & 1 \end{arra...
The pair of equations\n\n\[ \n{.70} = {\ell }_{11}{\ell }_{31} \n\]\n\n\[ \n{.40} = {\ell }_{21}{\ell }_{31} \n\]\n\nimplies that\n\n\[ \n{\ell }_{21} = \left( \frac{.40}{.70}\right) {\ell }_{11} \n\]\n\nSubstituting this result for \( {\ell }_{21} \) in the equation\n\n\[ \n{.90} = {\ell }_{11}{\ell }_{21} \n\]\n\nyie...
Yes
Example 9.5 (Factor analysis of stock-price data using the maximum likelihood method) The stock-price data of Examples 8.5 and 9.4 were reanalyzed assuming an \( m = 2 \) factor model and using the maximum likelihood method. The estimated factor loadings, communalities, specific variances, and proportion of total (stan...
So, for example, \[ {\widehat{h}}_{1}^{2} = {\left( {.115}\right) }^{2} + {\left( {.765}\right) }^{2} = {.58} \] \( {}^{3} \) The maximum likelihood solution leads to a Heywood case. For this example, the solution of the likelihood equations give estimated loadings such that a specific variance is negative. The softwar...
Yes
Example 9.6 (Factor analysis of Olympic decathlon data) Linden [11] originally conducted a factor analytic study of Olympic decathlon results for all 160 complete starts from the end of World War II until the mid-seventies. Following his approach we examine the \( n = {280} \) complete starts from 1960 through 2004 . T...
From a principal component factor analysis perspective, the first four eigenvalues, \( {4.21},{1.39},{1.06},{.92} \), of \( \mathbf{R} \) suggest a factor solution with \( m = 3 \) or \( m = 4 \) . A subsequent interpretation, much like Linden's original analysis, reinforces the choice \( m = 4 \) .\n\nIn this case, th...
Yes
Example 9.8 (A first look at factor rotation) Lawley and Maxwell [10] present the sample correlation matrix of examination scores in \( p = 6 \) subject areas for \( n = {220} \) male students. The correlation matrix is\n\n\[ \mathbf{R} = \left\lbrack \begin{matrix} \text{ Gaelic } & \text{ English } & \text{ History }...
All the variables have positive loadings on the first factor. Lawley and Maxwell suggest that this factor reflects the overall response of the students to instruction and might be labeled a general intelligence factor. Half the loadings are positive and half are negative on the second factor. A factor with this pattern...
No
Example 9.9 (Rotated loadings for the consumer-preference data) Let us return to the marketing data discussed in Example 9.3. The original factor loadings (obtained by the principal component method), the communalities, and the (varimax) rotated factor loadings are shown in Table 9.7. (See the SAS statistical software ...
It is clear that variables 2, 4, and 5 define factor 1 (high loadings on factor 1, small or negligible loadings on factor 2), while variables 1 and 3 define factor 2 (high loadings on factor 2, small or negligible loadings on factor 1). Variable 4 is most closely aligned with factor 1 , although it has aspects of the t...
No
Example 9.10 (Rotated loadings for the stock-price data) Table 9.8 shows the initial and rotated maximum likelihood estimates of the factor loadings for the stock-price data of Examples 8.5 and 9.5. An \( m = 2 \) factor model is assumed. The estimated specific variances and cumulative proportions of the total (standar...
The rotated loadings indicate that the bank stocks (JP Morgan, Citibank, and Wells Fargo) load highly on the first factor, while the oil stocks (Royal Dutch Shell and ExxonMobil) load highly on the second factor. (Although the rotated loadings obtained from the principal component solution are not displayed, the same p...
No
Example 9.11 (Rotated loadings for the Olympic decathlon data) The estimated factor loadings and specific variances for the Olympic decathlon data were presented in Example 9.6. These quantities were derived for an \( m = 4 \) factor model, using both principal component and maximum likelihood solution methods. The int...
The varimax rotated loadings for the \( m = 4 \) factor solutions are displayed in Table 9.9, along with the specific variances. Apart from the estimated loadings, rotation will affect only the distribution of the proportions of the total sample variance explained by each factor. The cumulative proportion of the total ...
Yes
Example 9.12 (Computing factor scores) We shall illustrate the computation of factor scores by the least squares and regression methods using the stock-price data discussed in Example 9.10. A maximum likelihood solution from \( \mathbf{R} \) gave the estimated rotated loadings and specific variances\n\n\[ \n{\widehat{\...
Weighted least squares (9-50): \( {}^{6} \)\n\n\[ \n\widehat{\mathbf{f}} = {\left( {\widehat{\mathbf{L}}}_{\mathbf{z}}^{*\prime }{\widehat{\mathbf{\Psi }}}_{\mathbf{z}}^{-1}{\widehat{\mathbf{L}}}_{\mathbf{z}}^{ * }\right) }^{-1}{\widehat{\mathbf{L}}}_{\mathbf{z}}^{*\prime }{\widehat{\mathbf{\Psi }}}_{\mathbf{z}}^{-1}\m...
Yes
Example 9.13 (Creating simple summary scores from factor analysis groupings) The principal component factor analysis of the stock price data in Example 9.4 produced the estimated loadings\n\n\[ \n\widetilde{\mathbf{L}} = \left\lbrack \begin{array}{rr} {.732} & - {.437} \\ {.831} & - {.280} \\ {.726} & - {.374} \\ {.605...
\[ \n{\widehat{f}}_{1} = {x}_{1} + {x}_{2} + {x}_{3} + {x}_{4} + {x}_{5} \n\]\n\n\[ \n{\widehat{f}}_{2} = {x}_{4} + {x}_{5} - {x}_{1} \n\]\n\nas a summary. In practice, we would standardize these new variables.\n\nIf, instead of \( \widetilde{\mathbf{L}} \), we start with the varimax rotated loadings \( {\widetilde{\ma...
Yes
Example 9.14 (Factor analysis of chicken-bone data) We present the results of several factor analyses on bone and skull measurements of white leghorn fowl. The original data were taken from Dunn [5]. Factor analysis of Dunn's data was originally considered by Wright [15], who started his analysis from a different corre...
The full data set consists of \( n = {276} \) measurements on bone dimensions:\n\n\[ \text{Head:}\;\left\{ \begin{array}{l} {X}_{1} = \text{ skull length } \\ {X}_{2} = \text{ skull breadth } \end{array}\right. \]\n\n\[ \text{Leg:}\;\left\{ \begin{array}{l} {X}_{3} = \text{ femur length } \\ {X}_{4} = \text{ tibia leng...
Yes
Compute the correlations between the first pair of canonical variates and their component variables for the situation considered in Example 10.1.
The variables in Example 10.1 are already standardized, so equation (10-15) is applicable. For the standardized variables,\n\n\[ \n{\mathbf{\rho }}_{11} = \left\lbrack \begin{array}{rr} {1.0} & {.4} \\ {.4} & {1.0} \end{array}\right\rbrack \;{\mathbf{\rho }}_{22} = \left\lbrack \begin{array}{rr} {1.0} & {.2} \\ {.2} & ...
Yes
Example 10.4 (Canonical correlation analysis of the chicken-bone data) In Example 9.14, data consisting of bone and skull measurements of white leghorn fowl were described. From this example, the chicken-bone measurements for\n\n\[ \n\\text{Head}\\left( {\\mathbf{X}}^{\\left( 1\\right) }\\right) : \\;\\left\\{ \\begin{...
\[ \n\\begin{array}{ll} \\widehat{{\\rho }_{1}^{ * }} = {.631} & \\widehat{{U}_{1}} = {.781}{z}_{1}^{\\left( 1\\right) } + {.345}{z}_{2}^{\\left( 1\\right) } \\\\ \\widehat{{V}_{1}} = {.060}{z}_{1}^{\\left( 2\\right) } + {.944}{z}_{2}^{\\left( 2\\right) } & \\end{array}\n\]\n\nand\n\n\[ \n\\begin{array}{l} {\\widehat{v...
Yes
Example 10.6 (Calculating matrices of errors of approximation) In Example 10.4, we obtained the canonical correlations between the two head and the two leg variables for white leghorn fowl. Starting with the sample correlation matrix \[ \mathbf{R} = \left\lbrack \begin{array}{ll} {\mathbf{R}}_{11} & {\mathbf{R}}_{12} \...
We first calculate (see Panel 10.1) \[ {\widehat{\mathbf{A}}}_{\mathbf{z}}^{-1} = {\left\lbrack \begin{array}{rr} {.781} & {.345} \\ - {.856} & {1.106} \end{array}\right\rbrack }^{-1} = \left\lbrack \begin{array}{rr} {.9548} & - {.2974} \\ {.7388} & {.6739} \end{array}\right\rbrack \] \[ {\widehat{\mathbf{B}}}_{\mathbf...
Yes
Example 10.7 (Calculating proportions of sample variance explained by canonical variates) Consider the job characteristic-job satisfaction data discussed in Example 10.5. Using the table of sample correlation coefficients presented in that example, we find that
\[ {R}_{{\mathbf{z}}^{\left( 1\right) } \mid {\widehat{U}}_{1}}^{2} = \frac{1}{5}\mathop{\sum }\limits_{{k = 1}}^{5}{r}_{{\widehat{U}}_{1},{z}_{k}^{\left( 1\right) }}^{2} = \frac{1}{5}\left\lbrack {{\left( {.83}\right) }^{2} + {\left( {.74}\right) }^{2} + \cdots + {\left( {.85}\right) }^{2}}\right\rbrack = {.58} \] \[ ...
Yes
Example 11.2 (Classifying a new observation into one of the two populations) A researcher has enough data available to estimate the density functions \( {f}_{1}\left( \mathbf{x}\right) \) and \( {f}_{2}\left( \mathbf{x}\right) \) associated with populations \( {\pi }_{1} \) and \( {\pi }_{2} \), respectively. Suppose \...
Given the prior probabilities and costs of misclassification, we can use (11-6) to derive the classification regions \( {R}_{1} \) and \( {R}_{2} \) . Specifically, we have\n\n\[ \n{R}_{1} : \frac{{f}_{1}\left( \mathbf{x}\right) }{{f}_{2}\left( \mathbf{x}\right) } \geq \left( \frac{10}{5}\right) \left( \frac{.2}{.8}\ri...
Yes
To construct a procedure for detecting potential hemophilia A carriers, blood samples were assayed for two groups of women and measurements on the two variables, \n\n\[ \n{X}_{1} = {\log }_{10}\left( \text{AHF activity}\right) \n\]\n\n\[ \n{X}_{2} = {\log }_{10}\text{(AHF-like antigen)} \n\]\n\nrecorded. The first grou...
The investigators provide the information\n\n\[ \n{\overline{\mathbf{x}}}_{1} = \left\lbrack \begin{array}{l} - {.0065} \\ - {.0390} \end{array}\right\rbrack ,\;{\overline{\mathbf{x}}}_{2} = \left\lbrack \begin{array}{r} - {.2483} \\ {.0262} \end{array}\right\rbrack \n\]\n\nand\n\n\[ \n{\mathbf{S}}_{\text{pooled }}^{-1...
Yes
Consider the detection of hemophilia A carriers introduced in Example 11.3. Recall that the equal costs and equal priors linear discriminant function was\n\n\[ \n\widehat{y} = {\widehat{\mathbf{a}}}^{\prime }\mathbf{x} = {\left( {\overline{\mathbf{x}}}_{1} - {\overline{\mathbf{x}}}_{2}\right) }^{\prime }{\mathbf{S}}_{\...
Fisher's solution to the separation problem can also be used to classify new observations.\n\nAllocate \( {\mathbf{x}}_{0} \) to \( {\pi }_{1} \) if\n\n\[ \n{\widehat{y}}_{0} = {\left( {\overline{\mathbf{x}}}_{1} - {\overline{\mathbf{x}}}_{2}\right) }^{\prime }{\mathbf{S}}_{\text{pooled }}^{-1}{\mathbf{x}}_{0} \n\]\n\n...
Yes
Example 11.9 (Classifying a new observation into one of three known populations) Let us assign an observation \( {\mathbf{x}}_{0} \) to one of the \( g = 3 \) populations \( {\pi }_{1},{\pi }_{2} \), or \( {\pi }_{3} \), given the following hypothetical prior probabilities, misclassification costs, and density values:
The values of \( \mathop{\sum }\limits_{\substack{{i = 1} \\ {i \neq k} }}^{3}{p}_{i}{f}_{i}\left( {\mathbf{x}}_{0}\right) c\left( {k \mid i}\right) \) [see (11-38)] are\n\n\[ k = 1 : \;{p}_{2}{f}_{2}\left( {\mathbf{x}}_{0}\right) c\left( {1 \mid 2}\right) + {p}_{3}{f}_{3}\left( {\mathbf{x}}_{0}\right) c\left( {1 \mid ...
Yes
Example 11.10 (Calculating sample discriminant scores, assuming a common covariance matrix) Let us calculate the linear discriminant scores based on data from \( g = 3 \) populations assumed to be bivariate normal with a common covariance matrix.
Given that \( {p}_{1} = {p}_{2} = {.25} \) and \( {p}_{3} = {.50} \), let us classify the observation \( {\mathbf{x}}_{0}^{\prime } = \left\lbrack {{x}_{01},{x}_{02}}\right\rbrack = \left\lbrack \begin{array}{ll} - 2 & - 1 \end{array}\right\rbrack \) according to (11-52). From (11-50),\n\n\[{\mathbf{S}}_{\text{pooled }...
Yes
Example 11.15 (Plotting sports data in two-dimensional discriminant space) Investigators interested in sports psychology administered the Minnesota Multiphasic Personality Inventory (MMPI) to 670 letter winners at the University of Wisconsin in Madison. The sports involved and the coefficients in the two discriminant f...
A plot of the group means using the first two discriminant scores is shown in Figure 11.13. Here the separation on the basis of the MMPI scores is not good, although a test for the equality of means is significant at the \( 5\% \) level. (This is due to the large sample sizes.) While the discriminant coefficients sugge...
Yes
Example 11.16 (Classifying a new observation with Fisher's discriminants) Let us use the Fisher discriminants\n\n\[ \n{\widehat{y}}_{1} = {\widehat{\mathbf{a}}}_{1}^{\prime }\mathbf{x} = {.386}{x}_{1} + {.495}{x}_{2} \]\n\n\[ \n{\widehat{y}}_{2} = {\widehat{\mathbf{a}}}_{2}^{\prime }\mathbf{x} = {.938}{x}_{1} - {.112}{...
Inserting \( {\mathbf{x}}_{0}^{\prime } = \left\lbrack {{x}_{01},{x}_{02}}\right\rbrack = \left\lbrack \begin{array}{ll} 1 & 3 \end{array}\right\rbrack \), we have\n\n\[ \n{\widehat{y}}_{1} = {.386}{x}_{01} + {.495}{x}_{02} = {.386}\left( 1\right) + {.495}\left( 3\right) = {1.87} \]\n\n\[ \n{\widehat{y}}_{2} = {.938}{x...
Yes
Example 11.17 (Logistic regression with the salmon data) We introduced the salmon data in Example 11.8 (see Table 11.2). In Example 11.8, we ignored the gender of the salmon when considering the problem of classifying salmon as Alaskan or Canadian based on growth ring measurements. Perhaps better classification is poss...
Examining the significance of individual terms under the heading Analysis of Maximum Likelihood Estimates, we see that the Wald test suggests gender is not significant ( \( p \) -value \( = {.7356} \) ). On the other hand, freshwater growth and marine are significant covariates. Gender can be dropped from the model. It...
Yes
Example 12.3 (Clustering using single linkage) To illustrate the single linkage algorithm, we consider the hypothetical distances between pairs of five objects as follows:\n\n\[ \mathbf{D} = \left\{ {d}_{ik}\right\} = \begin{array}{l} 1 \\ 2 \\ 3 \\ 4 \\ 5 \end{array}\left\lbrack \begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 0 &...
objects 5 and 3 are merged to form the cluster (35). To implement the next level of clustering, we need the distances between the cluster (35) and the remaining objects, 1, 2, and 4 . The nearest neighbor distances are\n\n\[ {d}_{\left( {35}\right) 1} = \min \left\{ {{d}_{31},{d}_{51}}\right\} = \min \{ 3,{11}\} = 3 \]...
Yes
Example 12.4 (Single linkage clustering of 11 languages) Consider the array of concordances in Table 12.3 representing the closeness between the numbers 1-10 in 11 languages. To develop a matrix of distances, we subtract the concordances from the perfect agreement figure of 10 that each language has with itself. The su...
We first search for the minimum distance between pairs of languages (clusters). The minimum distance, 1, occurs between Danish and Norwegian, Italian and French, and Italian and Spanish. Numbering the languages in the order in which they appear across the top of the array, we have\n\n\[ \n{d}_{32} = 1;\;{d}_{86} = 1;\;...
Yes
Example 12.5 (Clustering using complete linkage) Let us return to the distance matrix introduced in Example 12.3:
At the first stage, objects 3 and 5 are merged, since they are most similar. This gives the cluster (35). At stage 2, we compute\n\n\[ {d}_{\left( {35}\right) 1} = \max \left\{ {{d}_{31},{d}_{51}}\right\} = \max \{ 3,{11}\} = {11} \]\n\n\[ {d}_{\left( {35}\right) 2} = \max \left\{ {{d}_{32},{d}_{52}}\right\} = {10} \]\...
Yes
Example 12.6 (Complete linkage clustering of 11 languages) In Example 12.4, we presented a distance matrix for numbers in 11 languages. The complete linkage clustering algorithm applied to this distance matrix produces the dendrogram shown in Figure 12.7.
Comparing Figures 12.7 and 12.4, we see that both hierarchical methods yield the English-Norwegian-Danish and the French-Italian-Spanish language groups. Polish is merged with French-Italian-Spanish at an intermediate level. In addition, both methods merge Hungarian and Finnish only at the penultimate stage. However, t...
Yes
Example 12.10 (Clustering pure malt scotch whiskies) Virtually all the world's pure malt Scotch whiskies are produced in Scotland. In one study (see [22]), 68 binary variables were created measuring characteristics of Scotch whiskey that can be broadly classified as color, nose, body, palate, and finish. For example, t...
Weighted similarity coefficients \( \left\{ {s}_{ik}\right\} \) were created from binary variables representing the presence or absence of characteristics. The resulting \
No
Example 12.13 (A model based clustering of the iris data) Consider the Iris data in Table 11.5. Using MCLUST and specifically the me function, we first fit the \( p = 4 \) dimensional normal mixture model restricting the covariance matrices to satisfy \( {\mathbf{\sum }}_{k} = {\eta }_{k}\mathbf{I}, k = 1,2,3 \) .
Using the BIC criterion, the software chooses \( K = 3 \) clusters with estimated centers\n\n\[ \n{\mathbf{\mu }}_{1} = \left\lbrack \begin{array}{l} {5.01} \\ {3.43} \\ {1.46} \\ {0.25} \end{array}\right\rbrack ,\;{\mathbf{\mu }}_{2} = \left\lbrack \begin{array}{l} {5.90} \\ {2.75} \\ {4.40} \\ {1.43} \end{array}\righ...
Yes
Example 12.15 (Multidimensional scaling of public utilities) Let us try to represent the 22 public utility firms discussed in Example 12.7 as points in a low-dimensional space. The measures of (dis)similarities between pairs of firms are the Euclidean distances listed in Table 12.6. Multidimensional scaling in \( q = 1...
The stress function in Figure 12.17 has no sharp elbow. The plot appears to level out at \
No
Example 12.17 (Correspondence analysis of archaeological data) Table 12.8 contains the frequencies (counts) of \( J = 4 \) different types of pottery (called potsherds) found at \( I = 7 \) archaeological sites in an area of the American Southwest. If we divide the frequencies in each row (archaeological site) by the c...
The archaeological site profile for different types of pottery (columns) are shown in a bar graph in Figure 12.21(b). The site profiles are constructed using the column totals. The bars in the figure appear to be quite different from one another. This suggests that the various types of pottery are not distributed over ...
Yes
Example 12.19 (A biplot of universities and their characteristics) Table 12.9 gives the data on some universities for certain variables used to compare or rank major universities. These variables include \( {X}_{1} = \) average SAT score of new freshmen, \( {X}_{2} = \) percentage of new freshmen in top \( {10}\% \) of...
Because two of the variables, SAT and Expenses, are on a much different scale from that of the other variables, we standardize the data and base our biplot on the matrix of standardized observations \( {\mathbf{z}}_{i} \) . The biplot is given in Figure 12.24 on page 730.
Yes
Using the coordinates of the points in Figures 12.26 and 12.27, we obtain the initial sum of squared distances for fit:\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{{10}}{\left( {\mathbf{x}}_{j} - {\mathbf{y}}_{j}\right) }^{\prime }\left( {{\mathbf{x}}_{j} - {\mathbf{y}}_{j}}\right) = {8.547} \]
A computer calculation gives\n\n\[ \mathbf{U} = \left\lbrack \begin{array}{rr} - {.9833} & - {.1821} \\ - {.1821} & {.9833} \end{array}\right\rbrack \;\mathbf{V} = \left\lbrack \begin{array}{rr} - {1.0000} & - {.0001} \\ - {.0001} & {1.0000} \end{array}\right\rbrack \]\n\n\[ \Lambda = \left\lbrack \begin{array}{rr} {43...
Yes
Theorem 1. Let \( X \) be a random vector with finite mean \( \mu \) and finite covariance \( \sum \) . Then\n\n\[ \mathbb{E}\left\lbrack {{X}^{\prime }{\Lambda X}}\right\rbrack = \operatorname{tr}\left( {\Lambda \sum }\right) + {\mu }^{\prime }{\Lambda \mu } \]
Proof. Note: we use properties of the trace of a matrix that will be discussed in the next lecture.\n\n\[ \mathbb{E}\left\lbrack {{X}^{\prime }{\Lambda X}}\right\rbrack = \operatorname{tr}\left( {\mathbb{E}\left\lbrack {{X}^{\prime }{\Lambda X}}\right\rbrack }\right) = \mathbb{E}\left\lbrack {\operatorname{tr}\left( {{...
No
Theorem 2. (Expected Euclidean distance). Suppose \( X, Y \) are in-\n\ndependent, identically distributed random variables with mean \( \mu \) and covariance \( \sum \) . Then\n\n\[ \mathbb{E}\left\lbrack {\parallel X - Y{\parallel }_{2}^{2}}\right\rbrack = \mathop{\sum }\limits_{j}{\sigma }_{jj} \]
Proof.\n\n\( \mathbb{E}\left\lbrack {\left| \right| X - Y{\left| \right| }_{2}^{2}}\right\rbrack = \mathbb{E}\left\lbrack {{\left( X - Y\right) }^{\prime }I\left( {X - Y}\right) }\right\rbrack \)\n\n\[ = \operatorname{tr}\left( {I\sum }\right) + {\left( \mathbb{E}\left\lbrack X - Y\right\rbrack \right) }^{\prime }I\lef...
Yes
Lemma 2.3.1. If \( Z \) is an \( m \times n \) random matrix, \( D \) is an \( l \times m \) real matrix, \( E \) is an \( n \times q \) real matrix, and \( F \) is an \( l \times q \) real matrix, then\n\n\[ \mathcal{E}\left( {{DZE} + F}\right) = D\left( {\mathcal{E}Z}\right) E + F. \]
Proof. The element in the \( i \) th row and \( j \) th column of \( \mathcal{E}\left( {\mathbf{{DZE}} + \mathbf{F}}\right) \) is\n\n\[ \mathcal{E}\left( {\mathop{\sum }\limits_{{h, g}}{d}_{ih}{Z}_{hg}{e}_{gj} + {f}_{ij}}\right) = \mathop{\sum }\limits_{{h, g}}{d}_{ih}\left( {\mathcal{E}{Z}_{hg}}\right) {e}_{gj} + {f}_...
Yes
Lemma 2.3.2. If \( Y = {DX} + f \), where \( X \) is a random vector, then\n\n(32)\n\n\[ \mathcal{E}Y = D\mathcal{E}X + f \]\n\n(33)\n\n\[ \mathcal{C}\left( Y\right) = D\mathcal{C}\left( X\right) {D}^{\prime }.\]
Proof. The first assertion follows directly from Lemma 2.3.1, and the second from\n\n(34)\n\n\[ \mathcal{C}\left( Y\right) = \mathcal{E}\left( {Y - \mathcal{E}Y}\right) {\left( Y - \mathcal{E}Y\right) }^{\prime }\]\n\n\[ = \mathcal{E}\left\lbrack {{DX} + f - \left( {D\mathcal{E}X + f}\right) }\right\rbrack {\left\lbrac...
Yes
Theorem 2.3.2. The correlation coefficient \( \rho \) of any bivariate distribution is invariant with respect to transformations \( {X}_{i}^{ * } = {b}_{i}{X}_{i} + {c}_{i},{b}_{i} > 0, i = 1,2 \) . Every function of the parameters of a bivariate normal distribution that is invariant with respect to such transformation...
Proof. The variance of \( {X}_{i}^{ * } \) is \( {b}_{i}^{2}{\sigma }_{i}^{2}, i = 1,2 \), and the covariance of \( {X}_{1}^{ * } \) and \( {X}_{2}^{ * } \) is \( {b}_{1}{b}_{2}{\sigma }_{1}{\sigma }_{2}\rho \) by Lemma 2.3.2. Insertion of these values into the definition of the correlation between \( {X}_{1}^{ * } \) ...
Yes
Theorem 2.4.1. Let \( X \) (with \( p \) components) be distributed according to \( N\left( {\mathbf{\mu },\mathbf{\sum }}\right) \). Then\n\n\[ Y = {CX} \]\n\nis distributed according to \( N\left( {{C\mu },{C\sum }{C}^{\prime }}\right) \) for \( C \) nonsingular.
Proof. The density of \( Y \) is obtained from the density of \( X, n\left( {x \mid \mu ,\sum }\right) \), by replacing \( x \) by\n\n\[ x = {C}^{-1}y \]\n\nand multiplying by the Jacobian of the transformation (2),\n\n\[ {\;\operatorname{mod}\;\left| {C}^{-1}\right| } = \frac{1}{\;\operatorname{mod}\;\left| C\right| }...
Yes
Theorem 2.4.2. If \( {X}_{1},\ldots ,{X}_{p} \) have a joint normal distribution, a necessary and sufficient condition for one subset of the random variables and the subset consisting of the remaining variables to be independent is that each covariance of a variable from one set and a variable from the other set is 0.
The necessity follows from the fact that if \( {X}_{t} \) is from one set and \( {X}_{j} \) from the other, then for any density (see Section 2.2.3)\n\n(21)\n\n\[ \n{\sigma }_{ij} = \mathcal{E}\left( {{X}_{i} - {\mu }_{i}}\right) \left( {{X}_{j} - {\mu }_{j}}\right) \]\n\n\[ = {\int }_{-\infty }^{\infty }\cdots {\int }...
Yes
Theorem 2.4.5. If \( X \) is distributed according to \( N\left( {\mu ,\sum }\right) \), then \( Z = {DX} \) is distributed according to \( N\left( {{D\mu },{D\sum }{D}^{\prime }}\right) \) .
This theorem includes the cases where \( X \) may have a nonsingular or a singular distribution and \( \mathbf{D} \) may be nonsingular or of rank less than \( q \) . Since \( X \) can be represented by (34), where \( Y \) has a nonsingular distribution\n\n\( N\left( {\mathbf{v},\mathbf{T}}\right) \), we can write\n\n(...
Yes
Theorem 2.5.1. Let the components of \( X \) be divided into two groups composing the subvectors \( {X}^{\left( 1\right) } \) and \( {X}^{\left( 2\right) } \) . Suppose the mean \( \mathbf{\mu } \) is similarly divided into \( {\mathbf{\mu }}^{\left( 1\right) } \) and \( {\mathbf{\mu }}^{\left( 2\right) } \), and suppo...
As an example of the above considerations let us consider the bivariate normal distribution and find the conditional distribution of \( {X}_{1} \) given \( {X}_{2} = {x}_{2} \) . In this case \( {\mathbf{\mu }}^{\left( 1\right) } = {\mu }_{1},{\mathbf{\mu }}^{\left( 2\right) } = {\mu }_{2},{\mathbf{\sum }}_{11} = {\sig...
Yes
Theorem 2.5.2. The components of \( {X}^{\left( 1 \cdot 2\right) } \) are uncorrelated with the components of \( {\mathbf{X}}^{\left( 2\right) } \) .
Proof. The vector \( {\mathbf{X}}^{\left( 1 \cdot 2\right) } \) is \( {\mathbf{Y}}^{\left( 1\right) } - \mathcal{E}{\mathbf{Y}}^{\left( 1\right) } \) in (25) of Section 2.4.\n\nLet \( {\mathbf{\sigma }}_{\left( i\right) }^{\prime } \) be the \( i \) th row of \( {\mathbf{\sum }}_{12} \), and \( {\mathbf{\beta }}_{\left...
No
Theorem 2.5.3. For every vector \( \mathbf{\alpha } \)\n\n\[ \mathcal{V}\left( {X}_{i}^{\left( 1 \cdot 2\right) }\right) \leq \mathcal{V}\left( {{X}_{i} - {\alpha }^{\prime }{X}^{\left( 2\right) }}\right) \]
Proof. By Theorem 2.5.2\n\n\[ \mathcal{V}\left( {{X}_{i} - {\alpha }^{\prime }{X}^{\left( 2\right) }}\right) \]\n\n\[ = \mathcal{E}{\left\lbrack {X}_{i} - {\mu }_{i} - {\alpha }^{\prime }\left( {X}^{\left( 2\right) } - {\mu }^{\left( 2\right) }\right) \right\rbrack }^{2} \]\n\n\[ = \mathcal{E}{\left\lbrack {X}_{i}^{\le...
Yes
Theorem 2.5.4. For every vector \( \alpha \)\n\n\[ \operatorname{Corr}\left( {{X}_{i},{\mathbf{\beta }}_{\left( i\right) }^{\prime }{X}^{\left( 2\right) }}\right) \geq \operatorname{Corr}\left( {{X}_{i},{\mathbf{\alpha }}^{\prime }{X}^{\left( 2\right) }}\right) . \]
Proof. Since the correlation between two variables is unchanged when either or both is multiplied by a positive constant, we can assume that\n\n\( \mathcal{E}{\left\lbrack {\mathbf{\alpha }}^{\prime }\left( {\mathbf{X}}^{\left( 2\right) } - {\mathbf{\mu }}^{\left( 2\right) }\right) \right\rbrack }^{2} = \mathcal{E}{\le...
Yes
Lemma 2.6.1. Let \( {X}^{\prime } = \left( {{X}^{\left( 1\right) }{}^{\prime }{X}^{\left( 2\right) }{}^{\prime }}\right) \) . If \( {X}^{\left( 1\right) } \) and \( {X}^{\left( 2\right) } \) are independent and \( g\left( x\right) = {g}^{\left( 1\right) }\left( {x}^{\left( 1\right) }\right) {g}^{\left( 2\right) }\left(...
Proof. If \( g\left( x\right) \) is real-valued and \( X \) has a density,\n\n(5)\n\n\[ \mathcal{E}g\left( X\right) = {\int }_{-\infty }^{x}\cdots {\int }_{-\infty }^{\infty }g\left( x\right) f\left( x\right) d{x}_{1}\cdots d{x}_{p} \]\n\n\[ = {\int }_{-\infty }^{\infty }\cdots {\int }_{-\infty }^{\infty }{g}^{\left( 1...
Yes
Lemma 2.6.2. If the components of \( X \) are mutually independent,
\[ \mathcal{E}{e}^{i{t}^{\prime }X} = \mathop{\prod }\limits_{{j = 1}}^{p}\mathcal{E}{e}^{i{t}_{j}{X}_{j}} \]
Yes
Theorem 2.6.1. The characteristic function of \( X \) distributed according to \( N\left( {\mathbf{\mu },\sum }\right) \) is\n\n\[ \phi \left( t\right) = \mathcal{E}{e}^{i{t}^{\prime }X} = {e}^{i{t}^{\prime }\mu - \frac{1}{2}{t}^{\prime }{\sum t}} \] \n\nfor every real vector \( t \) .
Proof. From Corollary A.1.6 of the Appendix we know there is a nonsingular matrix \( C \) such that\n\n\[ {C}^{\prime }{\sum }^{-1}C = I\text{.} \] \n\nThus\n\n\[ {\sum }^{-1} = {C}^{\prime - 1}{C}^{-1} = {\left( C{C}^{\prime }\right) }^{-1}. \] \n\nLet\n\n\[ X - \mu = {CY} \] \n\nThen \( \mathbf{Y} \) is distributed a...
Yes
Theorem 2.6.3. If the random vector \( X \) has the density \( f\left( x\right) \) and the characteristic function \( \phi \left( t\right) \), then
\[ f\left( x\right) = \frac{1}{{\left( 2\pi \right) }^{p}}{\int }_{-\infty }^{\infty }\cdots {\int }_{-\infty }^{\infty }{e}^{-t{t}^{\prime }x}\phi \left( t\right) d{t}_{1}\cdots d{t}_{p}. \]
Yes
Theorem 2.7.1. If \( Y \) has the density \( g\left( {{y}^{\prime }y}\right) \), then \( Z = {OY} \), where \( {O}^{\prime }O = I \) , has the density \( g\left( {{z}^{\prime }z}\right) \) .
Proof. The transformation \( z = {Oy} \) has Jacobian 1.
No
Corollary 2.7.1. If \( Y \) is spherically contoured with stochastic representation \( \mathbf{Y} \triangleq R\mathbf{U} \) with \( {R}^{2} = {\mathbf{Y}}^{\prime }\mathbf{Y} \), then \( \mathbf{U} \) is spherically contoured.
Proof. If \( Z = {OY} \) and hence \( Z\overset{d}{ = }Y \), and \( Z \) has the stochastic representation \( Z = {SV} \), where \( {S}^{2} = {Z}^{\prime }Z \), then \( S = R \) and \( V = {OU} = U \) .
No
Theorem 2.7.2. If \( X \) has the density (2) and \( \mathcal{E}{R}^{2} < \infty \)
\[ \mathcal{E}X = \mu = \nu ,\;\mathcal{C}\left( X\right) = \mathcal{E}\left( {X - \mu }\right) {\left( X - \mu \right) }^{\prime } = \sum = \left( {1/p}\right) \mathcal{E}{R}^{2}\Lambda . \]
No
Lemma 3.2.1. Let \( {x}_{1},\ldots ,{x}_{N} \) be \( N \) (p-component) vectors, and let \( \bar{x} \) be defined by (3). Then for any vector \( b \)\n\n(5)\n\n\[ \mathop{\sum }\limits_{{\alpha = 1}}^{N}\left( {{x}_{\alpha } - b}\right) {\left( {x}_{\alpha } - b\right) }^{\prime } = \mathop{\sum }\limits_{{\alpha = 1}}...
Proof\n\n(6)\n\n\[ \mathop{\sum }\limits_{{\alpha = 1}}^{N}\left( {{x}_{\alpha } - b}\right) {\left( {x}_{\alpha } - b\right) }^{\prime } = \mathop{\sum }\limits_{{\alpha = 1}}^{N}\left\lbrack {\left( {{x}_{\alpha } - \bar{x}}\right) + \left( {\bar{x} - b}\right) }\right\rbrack {\left\lbrack \left( {x}_{\alpha } - \bar...
Yes
Lemma 3.2.2. If \( D \) is positive definite of order \( p \), the maximum of \[ f\left( G\right) = N\left| \log \right| G \mid - \operatorname{tr}{G}^{-1}D \] with respect to positive definite matrices \( G \) exists, occurs at \( G = \left( {1/N}\right) D \), and has the value \[ f\left\lbrack {\left( {1/N}\right) D}...
Proof. Let \( D = E{E}^{\prime } \) and \( {E}^{\prime }{G}^{-1}E = H \) . Ther \( G = E{H}^{-1}{E}^{\prime } \), and \( \left| G\right| = \left| E\right| \) \( \cdot \left| {\mathbf{H}}^{-1}\right| \cdot \left| {\mathbf{E}}^{\prime }\right| = \left| {\mathbf{H}}^{-1}\right| \cdot \left| {\mathbf{E}{\mathbf{E}}^{\prime...
Yes
Theorem 3.2.1. If \( {x}_{1},\ldots ,{x}_{N} \) constitute a sample from \( N\left( {\mathbf{\mu },\mathbf{\sum }}\right) \) with \( p < N \) , the maximum likelihood estimators of \( \mathbf{\mu } \) and \( \mathbf{\sum } \) are \( \widehat{\mathbf{\mu }} = \bar{x} = \left( {1/N}\right) \mathop{\sum }\limits_{{\alpha ...
\[ \mathop{\sum }\limits_{{\alpha = 1}}^{N}\left( {{x}_{\alpha } - \bar{x}}\right) {\left( {x}_{\alpha } - \bar{x}\right) }^{\prime } = \mathop{\sum }\limits_{{\alpha = 1}}^{N}{x}_{\alpha }{x}_{\alpha }^{\prime } - N\bar{x}{\bar{x}}^{\prime }. \]
Yes
Lemma 3.2.3. Let \( f\left( \theta \right) \) be a real-valued function defined on a set \( S \), and let \( \phi \) be a single-valued function, with a single-valued inverse, on \( S \) to a set \( {S}^{ * } \) ; that is, to each \( \theta \in S \) there corresponds a unique \( {\theta }^{ * } \in {S}^{ * } \), and, c...
Proof. By hypothesis \( f\left( {\theta }_{0}\right) \geq f\left( \theta \right) \) for all \( \theta \in S \) . Then for any \( {\theta }^{ * } \in {S}^{ * } \)\n\n\[ g\left( {\theta }^{ * }\right) = f\left\lbrack {{\phi }^{-1}\left( {\theta }^{ * }\right) }\right\rbrack = f\left( \theta \right) \leq f\left( {\theta }...
Yes
Corollary 3.2.2. If \( {x}_{1},\ldots ,{x}_{N} \) constitutes a sample from \( N\left( {\mathbf{\mu },\mathbf{\sum }}\right) \), where \( {\sigma }_{ij} = {\sigma }_{i}{\sigma }_{j}{\rho }_{ij}\left( {{\rho }_{ij} = 1}\right) \), then the maximum likelihood estimator of \( \mathbf{\mu } \) is \( \widehat{\mathbf{\mu }}...
\[ {\widehat{\rho }}_{ij} = \frac{\mathop{\sum }\limits_{{\alpha = 1}}^{N}\left( {{x}_{i\alpha } - {\bar{x}}_{i}}\right) \left( {{x}_{j\alpha } - {\bar{x}}_{j}}\right) }{\sqrt{\mathop{\sum }\limits_{{\alpha = 1}}^{N}{\left( {x}_{i\alpha } - {\bar{x}}_{i}\right) }^{2}}\sqrt{\mathop{\sum }\limits_{{\alpha = 1}}^{N}{\left...
Yes
Theorem 3.3.1. Suppose \( {X}_{1},\ldots ,{X}_{N} \) are independent, where \( {X}_{\alpha } \) is distributed according to \( N\left( {{\mathbf{\mu }}_{\alpha },\mathbf{\sum }}\right) \) . Let \( C = \left( {c}_{\alpha \beta }\right) \) be an \( N \times N \) orthogonal matrix. Then \( {Y}_{\alpha } = \mathop{\sum }\l...
Proof. The set of vectors \( {\mathbf{Y}}_{1},\ldots ,{\mathbf{Y}}_{N} \) have a joint normal distribution, because the entire set of components is a set of linear combinations of the components of \( {\mathbf{X}}_{1},\ldots ,{\mathbf{X}}_{N} \), which have a joint normal distribution. The expected value of \( {Y}_{\al...
Yes
Lemma 3.3.1. If \( C = \left( {c}_{\alpha \beta }\right) \) is orthogonal, then \( \mathop{\sum }\limits_{{\alpha = 1}}^{N}{x}_{\alpha }{x}_{\alpha }^{\prime } = \mathop{\sum }\limits_{{\alpha = 1}}^{N}{y}_{\alpha }{y}_{\alpha }^{\prime } \) , where \( {y}_{\alpha } = \mathop{\sum }\limits_{{\beta = 1}}^{N}{c}_{\alpha ...
Proof\n\n(3)\n\n\[ \mathop{\sum }\limits_{{\alpha = 1}}^{N}{y}_{\alpha }{y}_{\alpha }^{\prime } = \mathop{\sum }\limits_{\alpha }\mathop{\sum }\limits_{\beta }{c}_{\alpha \beta }{x}_{\beta }\mathop{\sum }\limits_{\gamma }{c}_{\alpha \gamma }{x}_{\gamma }^{\prime } \]\n\n\[ = \mathop{\sum }\limits_{{\beta ,\gamma }}\lef...
Yes
Theorem 3.3.3. If the m-component vector \( Y \) is distributed according to \( N\left( {\mathbf{v}, T}\right) \) (nonsingular), then \( {\mathbf{Y}}^{\prime }{T}^{-1}\mathbf{Y} \) is distributed according to the noncentral \( {\chi }^{2} \) -distribution with \( m \) degrees of freedom and noncentrality parameter \( {...
Proof. Let \( C \) be a nonsingular matrix such that \( {CT}{C}^{\prime } = I \), and define \( Z = {CY} \) . Then \( Z \) is normally distributed with mean \( \mathcal{E}Z = C\mathcal{E}Y = {C\nu } = \lambda \), say, and covariance matrix \( \mathcal{E}\left( {Z - \lambda }\right) {\left( Z - \lambda \right) }^{\prime...
Yes
Theorem 3.4.1. If \( {x}_{1},\ldots ,{x}_{N} \) are observations from \( N\left( {\mathbf{\mu },\sum }\right) \), then \( \bar{x} \) and \( S \) are sufficient for \( \mathbf{\mu } \) and \( \mathbf{\sum } \) . If \( \mathbf{\mu } \) is given, \( \mathop{\sum }\limits_{{\alpha = 1}}^{N}\left( {{x}_{\alpha } - \mathbf{\...
Proof. The density of \( {X}_{1},\ldots ,{X}_{N} \) is\n\n(2) \( \mathop{\prod }\limits_{{\alpha = 1}}^{N}n\left( {{x}_{\alpha } \mid \mathbf{\mu },\mathbf{\sum }}\right) \)\n\n\[ = {\left( 2\pi \right) }^{-\frac{1}{2}{Np}}{\left| \mathbf{\sum }\right| }^{-\frac{1}{2}N}\exp \left\lbrack {-\frac{1}{2}\operatorname{tr}{\...
Yes
Theorem 3.4.3. Let the m-component vectors \( {Y}_{1},{Y}_{2},\ldots \) be independently and identically distributed with means \( \mathcal{E}{Y}_{\alpha } = v \) and covariance matrices \( \mathcal{E}\left( {{Y}_{\alpha } - \mathbf{v}}\right) {\left( {Y}_{\alpha } - \mathbf{v}\right) }^{\prime } = T \) . Then the limi...
Proof. Let\n\n(12)\n\n\[ \n{\phi }_{n}\left( {t, u}\right) = \mathcal{E}\exp \left\lbrack {{iu}{t}^{\prime }\frac{1}{\sqrt{n}}\mathop{\sum }\limits_{{\alpha = 1}}^{n}\left( {{Y}_{\alpha } - v}\right) }\right\rbrack \n\]\n\nwhere \( u \) is a scalar and \( t \) an \( m \) -component vector. For fixed \( t,{\phi }_{n}\le...
Yes
Theorem 3.4.4. Let \( A\left( n\right) = \mathop{\sum }\limits_{{\alpha = 1}}^{N}\left( {{X}_{\alpha } - {\bar{X}}_{N}}\right) {\left( {X}_{\alpha } - {\bar{X}}_{N}\right) }^{\prime } \), where \( {X}_{1},{X}_{2},\ldots \) are independently distributed according to \( N\left( {\mathbf{\mu },\mathbf{\sum }}\right) \) an...
Proof. As shown earlier, \( A\left( n\right) \) is distributed as \( A\left( n\right) = \mathop{\sum }\limits_{{\alpha = 1}}^{n}{Z}_{\alpha }{Z}_{\alpha }^{\prime } \), where \( {\mathbf{Z}}_{1},{\mathbf{Z}}_{2},\ldots \) are distributed independently according to \( N\left( {\mathbf{0},\mathbf{\sum }}\right) \) . We a...
Yes
Theorem 3.4.5. If \( {x}_{1},\ldots ,{x}_{N} \) are independently distributed, each \( {x}_{\alpha } \) according to \( N\left( {\mathbf{\mu },\mathbf{\sum }}\right) \), and if \( \mathbf{\mu } \) has an a priori distribution \( N\left( {\mathbf{\nu },\mathbf{\Phi }}\right) \), then the a posteriori distribution of \( ...
Proof. Since \( \overline{\mathbf{x}} \) is sufficient for \( \mathbf{\mu } \), we need only consider \( \overline{\mathbf{x}} \), which has the distribution of \( \mathbf{\mu } + \mathbf{\nu } \), where \( \mathbf{\nu } \) has the distribution \( N\left\lbrack {\mathbf{0},\left( {1/N}\right) \mathbf{\sum }}\right\rbra...
Yes
Corollary 3.4.1. If \( {x}_{1},\ldots ,{x}_{N} \) are independently distributed, each \( {x}_{\alpha } \) according to \( N\left( {\mathbf{\mu },\mathbf{\sum }}\right) ,\mathbf{\mu } \) has an a priori distribution \( N\left( {\mathbf{\nu },\mathbf{\Phi }}\right) \), and the loss function is \( {\left( d - \mathbf{\mu ...
The Bayes estimator of \( \mathbf{\mu } \) is a kind of weighted average of \( \overline{\mathbf{x}} \) and \( \mathbf{v} \), the prior mean of \( \mathbf{\mu } \) . If \( \left( {1/N}\right) \mathbf{\sum } \) is small compared to \( \mathbf{\Phi } \) (e.g., if \( N \) is large), \( \mathbf{\nu } \) is given little wei...
No
Theorem 3.4.6. If \( {x}_{1},\ldots ,{x}_{N} \) are independently distributed each according to \( N\left( {\mathbf{\mu },\mathbf{\sum }}\right) \) and the loss function is \( {\left( d - \mathbf{\mu }\right) }^{\prime }Q\left( {d - \mathbf{\mu }}\right) \), then \( \bar{x} \) is a minimax estimator.
Proof. This follows from a theorem in statistical decision theory that if a procedure \( {\delta }_{0} \) is extended Bayes [i.e., if for arbitrary \( \varepsilon, r\left( {\rho ,{\delta }_{0}}\right) \leq r\left( {\rho ,{\delta }_{\rho }}\right) + \varepsilon \) for suitable \( \rho \), where \( {\delta }_{\rho } \) i...
Yes
Theorem 3.5.1. With respect to the loss function (1), the risk of the estimator (2) is less than the risk of the estimator \( \mathbf{Y} \) for \( p \geq 3 \) .
We shall show that the risk of \( \mathbf{Y} \) minus the risk of (2) is positive by applying the following lemma due to Stein (1974).\n\nLemma 3.5.1. If \( f\le
No
Lemma 3.5.1. If \( f\left( x\right) \) is a function such that\n\n(3)\n\n\[ f\left( b\right) - f\left( a\right) = {\int }_{a}^{b}{f}^{\prime }\left( x\right) {dx} \]\n\nfor all \( a \) and \( b\left( {a < b}\right) \) and if\n\n(4)\n\n\[ {\int }_{-\infty }^{\infty }\left| {{f}^{\prime }\left( x\right) }\right| \frac{1}...
Proof of Lemma. We write the left-hand side of (5) as\n\n(6)\n\n\[ {\int }_{\theta }^{\infty }\left\lbrack {f\left( x\right) - f\left( \theta \right) }\right\rbrack \left( {x - \theta }\right) \frac{1}{\sqrt{2\pi }}{e}^{-\frac{1}{2}{\left( x - \theta \right) }^{2}}{dx} \]\n\n\[ + {\int }_{-\infty }^{\theta }\left\lbrac...
Yes
Lemma 3.5.2. When \( X \) is distributed according to \( N\left( {\mu, I}\right) \) ,\n\n\[{\mathcal{E}}_{\mu }\left\{ {\begin{Vmatrix}{g}^{ + }\left( \parallel X\parallel \right) X - \mu \end{Vmatrix}}^{2}\right\} \leq {\mathcal{E}}_{\mu }\left\{ {\parallel g\left( {\parallel X\parallel }\right) X - \mu {\parallel }^{...
Proof. The right-hand side of (13) minus the left-hand side is\n\n\[{\mathcal{E}}_{\mu }\left\{ {{g}^{2}\left( {\parallel \mathbf{X}\parallel }\right) \parallel \mathbf{X}{\parallel }^{2} - {\left\lbrack {g}^{ + }\left( \parallel \mathbf{X}\parallel \right) \right\rbrack }^{2}\parallel \mathbf{X}{\parallel }^{2}}\right...
Yes
Theorem 3.5.2. The estimator\n\n\[ \n{m}^{ + }\left( y\right) = {\left( 1 - \frac{p - 2}{\parallel y - v{\parallel }^{2}}\right) }^{ + }\left( {y - v}\right) + v \]\n\nhas smaller risk than \( m\left( y\right) \) defined by (2) and is minimax.
Proof. In Lemma 3.5.2, let \( g\left( u\right) = 1 - \left( {p - 2}\right) /{u}^{2} \) and \( \mathbf{X} = \mathbf{Y} - \mathbf{v} \), and replace \( \mathbf{\mu } \) by \( \mathbf{\mu } - \mathbf{\nu } \) . The second assertion in the theorem follows from Theorem 3.4.6.
Yes
Corollary 3.5.1. For \( p \geq 3 \)\n\n\[ \left\{ {I - \frac{\min \left\lbrack {p - 2,{N}^{2}{\left( \bar{x} - \mathbf{v}\right) }^{\prime }{\sum }^{-1}{Q}^{-1}{\sum }^{-1}\left( {\bar{x} - \mathbf{v}}\right) }\right\rbrack }{N{\left( \bar{x} - \mathbf{v}\right) }^{\prime }{\sum }^{-1}{Q}^{-1}{\sum }^{-1}\left( {\bar{x...
Proof. the function \( r\left( z\right) = \min \left( {p - 2, z}\right) \) is differentiable except at \( z = \)\n\n\( p - 2 \) . The function \( r\left( z\right) \) can be approximated arbitrarily closely by a differentiable function. (For example, the corner at \( z = p - 2 \) can be smoothed by a circular arc of arb...
No
Theorem 3.5.4. If \( {h}^{\left( j\right) }\left( {y}^{\left( j\right) }\right) = {\left\lbrack {h}_{1}^{\left( j\right) }\left( {y}^{\left( j\right) }\right) ,\ldots ,{h}_{j}^{\left( j\right) }\left( {y}^{\left( j\right) }\right) \right\rbrack }^{\prime } \) is a minimax estimator of \( {\mathbf{\mu }}^{*\left( j\righ...
Proof. First consider the randomized estimator defined by\n\n\[ \Pr \left\{ {{G}_{i}\left( y\right) = {h}_{i}^{\left( j\right) }\left( {y}^{\left( j\right) }\right) }\right\} = \frac{{\alpha }_{j}}{{q}_{i}^{ * }}, \]\n\nfor the \( i \) th component. Then the risk of this estimator is\n\n\[ \mathop{\sum }\limits_{{i = 1...
Yes
Theorem 3.6.1. The covariances of the mean and covariance of a sample of \( N \) from \( {\left| \mathbf{\Lambda }\right| }^{-\frac{1}{2}}g\left\lbrack {{\left( x - \nu \right) }^{\prime }{\mathbf{\Lambda }}^{-1}\left( {x - \nu }\right) }\right\rbrack \) with \( {\mathcal{E}}^{4}{R}^{4} < \infty \) are
\[ \mathcal{E}\left( {\bar{x} - \mathbf{\mu }}\right) {\left( \bar{x} - \mathbf{\mu }\right) }^{\prime } = \frac{1}{N}\sum \] \[ \mathcal{E}\left( {{s}_{ij} - {\sigma }_{ij}}\right) \left( {\bar{x} - \mu }\right) = 0 \] \[ i, j = 1\ldots, p\text{.} \] \[ \mathcal{E}\left( {{s}_{ij} - {\sigma }_{ij}}\right) \left( {{s}_...
No
Lemma 3.6.1. The second-order moments of the elements of \( S \) are\n\n(5)\n\n\( \varepsilon {s}_{ij}{s}_{kl} = {\sigma }_{ij}{\sigma }_{kl} + \frac{1}{n}\left( {{\sigma }_{ik}{\sigma }_{jl} + {\sigma }_{il}{\sigma }_{jk}}\right) + \frac{\kappa }{N}\left( {{\sigma }_{ij}{\sigma }_{kl} + {\sigma }_{ik}{\sigma }_{jl} + ...
Proof of Lemma 3.6.1. We have\n\n(6)\n\n\[ \mathcal{E}\mathop{\sum }\limits_{{\alpha ,\beta = 1}}^{N}\left( {{x}_{i\alpha } - {\mu }_{i}}\right) \left( {{x}_{j\alpha } - {\mu }_{j}}\right) \left( {{x}_{k\beta } - {\mu }_{k}}\right) \left( {{s}_{l\beta } - {\mu }_{l}}\right) \]\n\n\[ = N\mathcal{E}\left( {{x}_{i\alpha }...
Yes
Theorem 3.6.2\n\n\[ \sqrt{n}\left\lbrack \begin{matrix} \left( {\bar{x} - \mu }\right) \\ \operatorname{vec}S - \operatorname{vec}\sum \end{matrix}\right\rbrack \]\n\n\[ \overset{d}{ \rightarrow }N\left\lbrack {\left( \begin{array}{l} \mathbf{0} \\ \mathbf{0} \end{array}\right) ,\left( \begin{matrix} \mathbf{\sum } & \...
This theorem follows from the central limit theorem for independent identically distributed random vectors (with finite fourth moments). The theorem forms the basis for large-sample inference.
No