Section 3e

3.5 Maximum Likelihood Estimation, 3.6 Orthogonal Columns in the Regression Matrix

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Problem 3e.1

Prove the statement above that the minimum is given by λj=c(j=1,2,…,p−1\lambda_{j} = c(j = 1,2, \ldots , p-1).

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Problem 3e.2

It is required to fit a regression model of the form

E[Yi]=β0+β1xi+β2ϕ(xi)(i=1,2,3) \mathbb {E}\left[Y_{i}\right] = \beta _{0}+\beta _{1} x_{i}+\beta _{2} \phi \left(x_{i}\right) \quad (i = 1,2,3)

where ϕ(x)\phi (x) is a second-degree polynomial. If x1=−1,x2=0x_{1} = -1, x_{2} = 0, and x3=1x_{3} = 1, find ϕ\phi such that the design matrix X\mathbf{X} has mutually orthogonal columns.

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Problem 3e.3

Suppose that X=(x(0),x(1),…,x(p−1),x(p))=(W,x(p))\mathbf{X} = \left(\mathbf{x}^{(0)}, \mathbf{x}^{(1)}, \ldots , \mathbf{x}^{(p-1)}, \mathbf{x}^{(p)}\right) = \left(\mathbf{W}, \mathbf{x}^{(p)}\right) has linearly independent columns.

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(a)

Using A.9.5, prove that

det⁡(XTX)=det⁡(WTW)(x(p)′x(p)−x(p)′W(WTW)−1WTx(p)) \operatorname {det}\left(\mathbf{X}^{T} \mathbf{X}\right) = \operatorname {det}\left(\mathbf{W}^{T} \mathbf{W}\right)\left(\mathbf{x}^{(p) \prime } \mathbf{x}^{(p)}-\mathbf{x}^{(p) \prime } \mathbf{W}\left(\mathbf{W}^{T} \mathbf{W}\right)^{-1} \mathbf{W}^{T} \mathbf{x}^{(p)}\right)
(b)

Deduce that

det⁡(WTW)det⁡(XTX)≥1x(p)/x(p) \frac{\operatorname {det}\left(\mathbf{W}^{T} \mathbf{W}\right)}{\operatorname {det}\left(\mathbf{X}^{T} \mathbf{X}\right)} \geq \frac{1}{\mathbf{x}^{(\mathbf{p}) / \mathbf{x}^{(\mathbf{p})}}}

and hence show that Var⁡[β^p]≥σ2(x(p)′x(p))−1\operatorname {Var}\left[\widehat{\beta }_{p}\right] \geq \sigma^{2}\left(\mathbf{x}^{(p) \prime } \mathbf{x}^{(p)}\right)^{-1} with equality if and only if x(p)′x(j)=0(j=0,1,…,p−1)\mathbf{x}^{(p) \prime } \mathbf{x}^{(j)} = 0(j = 0,1, \ldots , p-1).

Problem 3e.4

What modifications in the statement of Example 3.3 proved above can be made if the term β0\beta_{0} is omitted?

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Problem 3e.5

Suppose that we wish to find the weights βi(i=1,2,…,k)\beta_{i}(i = 1,2, \ldots , k) of kk objects. One method is to weigh each object rr times and take the average; this requires a total of krk r weighings, and the variance of each average is σ2/r\sigma^{2} / r (σ2\sigma^{2} being the variance of the weighing error). Another method is to weigh the objects in combinations; some of the objects are distributed between the two pans and weights are placed in one pan to achieve equilibrium. The regression model for such a scheme is

Y=β1x1+β2x2+⋯+βkxk+ε Y = \beta _{1} x_{1}+\beta _{2} x_{2}+\cdots +\beta _{k} x_{k}+\varepsilon

where xi=0,1x_{i} = 0,1, or -1 according as the ii th object is not used, placed in the left pan or in the right pan, ε\varepsilon is the weighing error (assumed to be the same for all weighings), and YY is the weight required for equilibrium (YY is regarded as negative if placed in the left pan). After nn such weighing operations we can find the least squares estimates β^i\widehat{\beta }_{i} of the weights.

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(a)

Show that the estimates of the weights have maximum precision (i.e., minimum variance) when each entry in the design matrix X\mathbf{X} is ±1\pm 1 and the columns of X\mathbf{X} are mutually orthogonal.

(b)

If the objects are weighed individually, show that knk n weighings are required to achieve the same precision as that given by the optimal design with nn weighings.