Math Problems Vault

Problem 6c.1

Using the notation of Example 6.2, prove that the least squares estimates of $\alpha_{k}$ and the common slope $\beta$ (under the null hypothesis of parallelism) are given by

\tilde{\alpha }_{k} = \bar{Y}_{k}-\tilde{\beta } \bar{x}_{k}

and

\tilde{\beta } = \frac{\sum _{k} \sum _{i}\left(Y_{k i}-\bar{Y}_{k .}\right)\left(x_{k i}-\bar{x}_{k .}\right)}{\sum _{k} \sum _{i}\left(x_{k i}-\bar{x}_{k .}\right)^{2}}

Prove that

\mathrm{RSS}_{H_{3}} = \sum _{k} \sum _{i}\left(Y_{k i}-\bar{Y}_{k} \cdot \right)^{2}-\tilde{\beta }^{2} \sum _{k} \sum _{i}\left(x_{k i}-\bar{x}_{k} \cdot \right)^{2} .

If $\widehat{\beta }_{k}$ is the least squares estimate of $\beta_{k}$ under the general model, prove that

\mathrm{RSS}_{H_{1}}-\mathrm{RSS} = \sum _{k} \widehat{\beta }_{k}^{2} \sum _{i}\left(x_{k i}-\bar{x}_{k} .\right)^{2}-\tilde{\beta }^{2} \sum _{k} \sum _{i}\left(x_{k i}-\bar{x}_{k} .\right)^{2}

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Problem 6c.2

In Example 6.3 find the least squares estimates of $\alpha$ and $\beta$ when $H_{2}$ is true.

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Problem 6c.3

(a)(b)

In Example 6.4 derive the following results.

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(a)

Show that the least squares estimates of $\alpha$ and $\beta_{k}$ , under $H_{3}$ , are given by

\begin{aligned} & \left(N-\frac{x_{1.}^{2}}{\sum _{i} x_{1 i}^{2}}-\cdots -\frac{x_{K .}^{2}}{\sum _{i} x_{K i}^{2}}\right) \alpha ^{T} \\ & \quad \quad = \left(Y . .-\frac{x_{1 \cdot } \cdot \sum _{i} Y_{1 i} x_{1 i}}{\sum _{i} x_{1 i}^{2}}-\cdots -\frac{x_{K \cdot } \cdot \sum _{i} Y_{K i} x_{K i}}{\sum _{i} x_{K i}^{2}}\right) \end{aligned}

and

\beta _{k}^{T} = \frac{\sum _{i}\left(Y_{k i}-\alpha ^{T}\right) x_{k i}}{\sum _{i} x_{k i}^{2}} \quad (k = 1,2, \ldots , K)

(b)

When the values of $x$ are the same for each line so that $n_{k} = n$ and $x_{k i} = x_{i}(k = 1,2, \ldots , K)$ , prove that

\begin{gathered} \alpha ^{T} = \bar{Y} . .-\frac{\bar{x} \sum \sum Y_{k i}\left(x_{i}-\bar{x}\right)}{K \sum \left(x_{i}-\bar{x}\right)^{2}} = \bar{Y} . .-\bar{x} \frac{\sum _{k} \widehat{\beta }_{k}}{K}, \\ \operatorname {Var}\left[\alpha ^{T}\right] = \frac{\sigma ^{2} \sum x_{i}^{2}}{n K \sum \left(x_{i}-\bar{x}\right)^{2}} \end{gathered}

and

\operatorname {RSS}_{H_{3}} = \sum _{k} \sum _{i} Y_{k i}^{2}-\frac{\left(\sum \sum Y_{k i} x_{i}\right)^{2}}{\sum x_{i}^{2}}-\left(\alpha ^{T}\right)^{2} K n \frac{\sum \left(x_{i}-\bar{x}\right)^{2}}{\sum x_{i}^{2}}

Hint: $\bar{Y}$ . and $\widehat{\beta }_{k}$ are uncorrelated.

Problem 6c.4

(a)(b)(c)

The examples in Section 6.4 are all special cases of the following problem.

Consider the model

G: \mathbb {E}\left[Y\right] = \mathbf{X}_{1} \mathbf{\beta }_{1}+\mathbf{X}_{2} \mathbf{\beta }_{2} = \mathbf{X} \mathbf{\beta }

where $\mathbf{X}_{i}$ is $n \times p_{i}$ of rank $p_{i}(i = 1,2)$ . We want to test the hypothesis $H$ that all the elements of $\mathbf{\beta }_{\mathbf{2}}$ are equal (to $\beta$ , i.e., $\mathbf{\beta }_{2} = \mathbf{1}_{p_{2}} \beta$ ).

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(a)

If $\tilde{\beta }$ is the least squares estimate of $\beta$ under $H$ , show that

\tilde{\beta } = \frac{\mathbf{1}_{p_{2}}^{T} \mathbf{X}_{2}^{T}\left(\mathbf{I}_{n}-\mathbf{P}_{1}\right) \mathbf{Y}}{\mathbf{1}_{p_{2}}^{T} \mathbf{X}_{2}^{T}\left(\mathbf{I}_{n}-\mathbf{P}_{1}\right) \mathbf{X}_{2} \mathbf{1}_{p_{2}}}

where $\mathbf{P}_{1} = \mathbf{X}_{1}\left(\mathbf{X}_{1}^{T} \mathbf{X}_{1}\right)^{-1} \mathbf{X}_{1}^{T}$ .

(b)

If $\widehat{\mathbf{\beta }}_{\mathbf{2}}$ is the least squares estimate of $\mathbf{\beta }_{\mathbf{2}}$ under $G$ , prove that

\mathrm{RSS}_{H}-\operatorname {RSS}_{G} = \left(\widehat{\mathbf{\beta }}_{2}-\tilde{\beta } \mathbf{1}_{p_{2}}\right)^{T} \mathbf{X}_{2}^{T}\left(\mathbf{I}_{n}-\mathbf{P}_{1}\right) \mathbf{Y}

(c)

If $\widehat{\mathbf{Y}}_{G}$ is the fitted regression for $G$ , prove that

\widehat{\mathbf{Y}}_{G} = \mathbf{P}_{1} \mathbf{Y}+\left(\mathbf{I}_{n}-\mathbf{P}_{1}\right) \mathbf{X}_{2} \widehat{\mathbf{\beta }}_{2}

and

\operatorname {RSS}_{H}-\operatorname {RSS}_{G} = \left(\mathbf{1}_{p_{2}} \tilde{\beta }-\widehat{\mathbf{\beta }}_{2}\right)^{T} \mathbf{X}_{2}^{T}\left(\mathbf{I}_{n}-\mathbf{P}_{1}\right) \mathbf{X}_{2}\left(\mathbf{1}_{p_{2}} \tilde{\beta }-\widehat{\mathbf{\beta }}_{2}\right)

Hint: For (a) and (b) apply Theorem 3.6 with $\mathbf{Z}$ equal to $\mathbf{X}_{2}$ and $\mathbf{1}_{p_{2}}$ , respectively. For part (c) note that $\operatorname {RSS}_{G}-\operatorname {RSS}_{H} = \left\| \widehat{\mathbf{Y}}_{G}-\widehat{\mathbf{Y}}_{H}\right\|^{2}$ .