# 统计代写|线性回归分析代写linear regression analysis代考|Rank Deficient and Over-Parameterized Mean Functions

#### Doug I. Jones

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## 统计代写|线性回归分析代写linear regression analysis代考|Rank Deficient and Over-Parameterized Mean Functions

In the last example, several combinations of the basic predictors $W T 2, W T 9$, and WT18 were studied. One might naturally ask what would happen if more than three combinations of these predictors were used in the same regression model. As long as we use linear combinations of the predictors, as opposed to nonlinear combinations or transformations of them, we cannot use more than three, the number of linearly independent quantities.

To see why this is true, consider adding $D W 9$ to the mean function including $W T 2, W T 9$ and $W T 18$. As in Chapter 3, we can learn about adding DW9 using an added-variable plot of the residuals from the regression of Soma on WT2,WT9 and $W T 18$ versus the residuals from the regression of $D W 9$ on WT2,WT9 and WT18. Since DW9 can be written as an exact linear combination of the other predictors, $D W 9=W T 9-W T 2$, the residuals from this second regression are all exactly zero. A slope coefficient for $D W 9$ is thus not defined after adjusting for the other three terms. We would say that the four terms WT2, WT9,WT18, and DW9 are linearly dependent, since one can be determined exactly from the others. The three variables WT2,WT9 and WT18 are linearly independent because one of them cannot be determined exactly by a linear combination of the others. The maximum number of linearly independent terms that could be included in a mean function is called the rank of the data matrix $\mathbf{X}$.

Model 3 in Table 4.1 gives the estimates produced in a computer package when we tried to fit using an intercept and the five terms WT2,WT9,WT18,DW9, and DW18. Most computer programs, including this one, will select the first three, and the estimated coefficients for them. For the remaining terms, this program sets the estimates to “NA,” a code for a missing value; the word aliased is sometimes used to indicate a term that is a linear combination of terms already in the mean function, and so a coefficient for it is not estimable.

## 统计代写|线性回归分析代写linear regression analysis代考|Tests

Even if the fitted model were correct and errors were normally distributed, tests and confidence statements for parameters are difficult to interpret because correlations among the terms lead to a multiplicity of possible tests. Sometimes, tests of effects adjusted for other variables are clearly desirable, such as in assessing a treatment effect after adjusting for other variables to reduce variability. At other times, the order of fitting is not clear, and the analyst must expect ambiguous results. In most situations, the only true test of significance is repeated experimentation.
4.1.6 Dropping Terms
Suppose we have a sample of $n$ rectangles from which we want to model $\log ($ Area $)$ as a function of $\log$ (Length), perhaps through the simple regression mean function
$$\mathrm{E}(\log (\text { Area }) \mid \log (\text { Length }))=\eta_0+\eta_1 \log (\text { Length })$$
From elementary geometry, we know that Area $=$ Length $\times$ Width, and so the “true” mean function for $\log ($ Area $)$ is
$$\mathrm{E}(\log (\text { Area }) \mid \log (\text { Length }), \log (\text { Width }))=\beta_0+\beta_1 \log (\text { Length })+\beta_2 \log (\text { Width })$$
with $\beta_0=0$, and $\beta_1=\beta_2=1$. The questions of interest are: (1) can the incorrect mean function specified by (4.1) provide a useful approximation to the true mean function (4.2), and if so, (2) what are the relationships between $\eta \mathrm{s}$, in (4.1) and the $\beta \mathrm{s}$ in (4.2)?

The answers to these questions comes from Appendix A.2.4. Suppose that the true mean function were
$$\mathrm{E}\left(Y \mid X_1=\mathbf{x}_1, X_2=\mathbf{x}_2\right)=\beta_0+\boldsymbol{\beta}_1^{\prime} \mathbf{x}_1+\boldsymbol{\beta}_2^{\prime} \mathbf{x}_2$$
but we want to fit a mean function with $X_1$ only. The mean function for $Y \mid X_1$ is obtained by averaging (4.3) over $X_2$,
\begin{aligned} \mathrm{E}\left(Y \mid X_1=\mathbf{x}_1\right) & =\mathrm{E}\left[\mathrm{E}\left(Y \mid X_1=\mathbf{x}_1, X_2\right) \mid X_1=\mathbf{x}_1\right] \ & =\beta_0+\boldsymbol{\beta}_1^{\prime} \mathbf{x}_1+\boldsymbol{\beta}_2^{\prime} \mathrm{E}\left(X_2 \mid X_1=\mathbf{x}_1\right) \end{aligned}
We cannot, in general, simply drop a set of terms from a correct mean function, but we need to substitute the conditional expectation of the terms dropped given the terms that remain in the mean function.

# 线性回归代写

## 统计代写|线性回归分析代写linear regression analysis代考|Tests

4.1.6 drop Terms

$$\mathrm{E}(\log (\text { Area }) \mid \log (\text { Length }))=\eta_0+\eta_1 \log (\text { Length })$$

$$\mathrm{E}(\log (\text { Area }) \mid \log (\text { Length }), \log (\text { Width }))=\beta_0+\beta_1 \log (\text { Length })+\beta_2 \log (\text { Width })$$

$$\mathrm{E}\left(Y \mid X_1=\mathbf{x}_1, X_2=\mathbf{x}_2\right)=\beta_0+\boldsymbol{\beta}_1^{\prime} \mathbf{x}_1+\boldsymbol{\beta}_2^{\prime} \mathbf{x}_2$$

\begin{aligned} \mathrm{E}\left(Y \mid X_1=\mathbf{x}_1\right) & =\mathrm{E}\left[\mathrm{E}\left(Y \mid X_1=\mathbf{x}_1, X_2\right) \mid X_1=\mathbf{x}_1\right] \ & =\beta_0+\boldsymbol{\beta}_1^{\prime} \mathbf{x}_1+\boldsymbol{\beta}_2^{\prime} \mathrm{E}\left(X_2 \mid X_1=\mathbf{x}_1\right) \end{aligned}

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