Section 4.6

Classical Least Squares

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Problem 4.6.1

Hooke's law says that the displacement yy of an ideal spring is proportional to the force xx that is applied — i.e., y=kxy = kx for some constant kk. Consider a spring in which kk is unknown. Various masses are attached, and the resulting displacements shown below are observed. Using these observations, determine the least squares estimate for kk.

x (lb)y (in)511.1715.4817.51022.01226.3 \begin{array}{c|c} x \text{ (lb)} & y \text{ (in)} \\ \hline 5 & 11.1 \\ 7 & 15.4 \\ 8 & 17.5 \\ 10 & 22.0 \\ 12 & 26.3 \end{array}

Figure 4.6.6Figure 4.6.6

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Problem 4.6.2

Show that the slope of the line that passes through the origin in R2\mathbb {R}^{2} and comes closest in the least squares sense to passing through the points {(x1,y1),(x2,y2),…,(xn,yn)}\left\{ (x_1, y_1), (x_2, y_2), \ldots , (x_n, y_n)\right\} is given by m=∑ixiyi/∑ixi2m = \sum_i x_i y_i / \sum_i x_i^2.

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Problem 4.6.3

A small company has been in business for three years and has recorded annual profits (in thousands of dollars) as follows.

Year123Sales743 \begin{array}{c|ccc} \text{Year} & 1 & 2 & 3 \\ \hline \text{Sales} & 7 & 4 & 3 \end{array}

Assuming that there is a linear trend in the declining profits, predict the year and the month in which the company begins to lose money.

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Problem 4.6.4

An economist hypothesizes that the change (in dollars) in the price of a loaf of bread is primarily a linear combination of the change in the price of a bushel of wheat and the change in the minimum wage. That is, if BB is the change in bread prices, WW is the change in wheat prices, and MM is the change in the minimum wage, then B=αW+βMB = \alpha W + \beta M. Suppose that for three consecutive years the change in bread prices, wheat prices, and the minimum wage are as shown below.

Year 1Year 2Year 3B+\1+\1+\1W+\1+\20\M+\10\−\1 \begin{array}{c|ccc}& \text{Year 1} & \text{Year 2} & \text{Year 3} \\ \hline B & +\1 & +\1 & +\1 \\ W & +\1 & +\2 & 0\ \\ M & +\1 & 0\ & -\1 \end{array}

Use the theory of least squares to estimate the change in the price of bread in Year 4 if wheat prices and the minimum wage each fall by $1.

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Problem 4.6.5

Suppose that a researcher hypothesizes that the weight loss of a pint of ice cream during storage is primarily a linear function of time. That is,

y=α0+α1t+ε, y = \alpha _0 + \alpha _1 t + \varepsilon ,

where y=y = the weight loss in grams, t=t = the storage time in weeks, and ε\varepsilon is a random error function whose mean value is 00. Suppose that an experiment is conducted, and the following data is obtained.

Time (t)12345678Loss (y).15.21.30.41.49.59.72.83 \begin{array}{c|cccccccc} \text{Time } (t) & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \text{Loss } (y) & .15 & .21 & .30 & .41 & .49 & .59 & .72 & .83 \end{array}
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(a)

Determine the least squares estimates for the parameters α0\alpha_0 and α1\alpha_1.

(b)

Predict the mean weight loss for a pint of ice cream that is stored for 20 weeks.

Problem 4.6.6

After studying a certain type of cancer, a researcher hypothesizes that in the short run the number (yy) of malignant cells in a particular tissue grows exponentially with time (tt). That is, y=α0eα1ty = \alpha_0 e^{\alpha_1 t}. Determine least squares estimates for the parameters α0\alpha_0 and α1\alpha_1 from the researcher's observed data given below.

t (days)12345y (cells)16274574122 \begin{array}{c|ccccc} t \text{ (days)} & 1 & 2 & 3 & 4 & 5 \\ \hline y \text{ (cells)} & 16 & 27 & 45 & 74 & 122 \end{array}
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Problem 4.6.7

Using least squares techniques, fit the following data

x−5−4−3−2−1012345y27912131414131084 \begin{array}{c|ccccccccccc} x & -5 & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline y & 2 & 7 & 9 & 12 & 13 & 14 & 14 & 13 & 10 & 8 & 4 \end{array}

with a line y=α0+α1xy = \alpha_0 + \alpha_1 x and then fit the data with a quadratic y=α0+α1x+α2x2y = \alpha_0 + \alpha_1 x + \alpha_2 x^2. Determine which of these two curves best fits the data by computing the sum of the squares of the errors in each case.

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Problem 4.6.8

Consider the time (TT) it takes for a runner to complete a marathon (26 miles and 385 yards). Many factors such as height, weight, age, previous training, etc.can influence an athlete's performance, but experience has shown that the following three factors are particularly important:

x1=Ponderal index=height (in.)(weight (lbs.))1/3,x2=Miles run the previous 8 weeks,x3=Age (years). x_1 = \text{Ponderal index} = \frac{\text{height (in.)}}{\left(\text{weight (lbs.)}\right)^{1/3}}, \qquad x_2 = \text{Miles run the previous 8 weeks}, \qquad x_3 = \text{Age (years)}.

A linear model hypothesizes that the time TT (in minutes) is given by T=α0+α1x1+α2x2+α3x3+εT = \alpha_0 + \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 + \varepsilon, where ε\varepsilon is a random function accounting for all other factors and whose mean value is assumed to be zero. On the basis of the five observations given below, estimate the expected marathon time for a 43-year-old runner of height 74 in., weight 180 lbs., who has run 450 miles during the previous eight weeks.

Tx1x2x318113.16192319313.58034221213.82073122113.14093824812.548245 \begin{array}{c|ccc} T & x_1 & x_2 & x_3 \\ \hline 181 & 13.1 & 619 & 23 \\ 193 & 13.5 & 803 & 42 \\ 212 & 13.8 & 207 & 31 \\ 221 & 13.1 & 409 & 38 \\ 248 & 12.5 & 482 & 45 \end{array}

What is your personal predicted mean marathon time?

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Problem 4.6.9

For A∈Rm×n\mathbf{A} \in \mathbb {R}^{m \times n} and b→∈Rm\overrightarrow {b} \in \mathbb {R}^{m}, prove that x→2\overrightarrow {x}_2 is a least squares solution for Ax→=b→\mathbf{A}\overrightarrow {x} = \overrightarrow {b} if and only if x→2\overrightarrow {x}_2 is part of a solution to the larger system

(Im×mAAT0n×n)(x→1x→2)=(b→0→). \begin{pmatrix} \mathbf{I}_{m \times m} & \mathbf{A} \\ \mathbf{A}^T & \mathbf{0}_{n \times n} \end{pmatrix} \begin{pmatrix} \overrightarrow {x}_1 \\ \overrightarrow {x}_2 \end{pmatrix} = \begin{pmatrix} \overrightarrow {b} \\ \overrightarrow {0} \end{pmatrix}.
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Remark.
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It is not uncommon to encounter least squares problems in which A\mathbf{A} is extremely large but very sparse (mostly zero entries). For these situations, the system above will usually contain significantly fewer nonzero entries than the system of normal equations, thereby helping to overcome the memory requirements that plague these problems. Using this system also eliminates the undesirable need to explicitly form the product ATA\mathbf{A}^T \mathbf{A} — recall from Example 4.5.1 that forming ATA\mathbf{A}^T \mathbf{A} can cause loss of significant information.

Problem 4.6.10

In many least squares applications, the underlying data matrix Am×n\mathbf{A}_{m \times n} does not have independent columns — i.e., rk⁡(A)<n\operatorname {rk}\left(\mathbf{A}\right) < n — so the corresponding system of normal equations ATAx→=ATb→\mathbf{A}^T \mathbf{A}\overrightarrow {x} = \mathbf{A}^T \overrightarrow {b} will fail to have a unique solution. This means that in an associated linear estimation problem of the form

y=α1t1+α2t2+⋯+αntn+ε y = \alpha _1 t_1 + \alpha _2 t_2 + \cdots + \alpha _n t_n + \varepsilon

there will be infinitely many least squares estimates for the parameters αi\alpha_i, and hence there will be infinitely many estimates for the mean value of yy at any given point (t1,t2,…,tn)(t_1, t_2, \ldots , t_n) — which is clearly undesirable. In order to remedy this problem, we restrict ourselves to making estimates only at those points (t1,t2,…,tn)(t_1, t_2, \ldots , t_n) that are in the row space of A\mathbf{A}. If

t→=(t1t2⋮tn)∈R(AT),and ifx→=(α^1α^2⋮α^n) \overrightarrow {t} = \begin{pmatrix} t_1 \\ t_2 \\ \vdots \\ t_n \end{pmatrix} \in R(\mathbf{A}^T), \quad \text{and if} \quad \overrightarrow {x} = \begin{pmatrix} \hat{\alpha }_1 \\ \hat{\alpha }_2 \\ \vdots \\ \hat{\alpha }_n \end{pmatrix}

is any least squares solution (i.e., ATAx→=ATb→\mathbf{A}^T \mathbf{A}\overrightarrow {x} = \mathbf{A}^T \overrightarrow {b}), prove that the estimate defined by

y^=t→Tx→=∑i=1ntiα^i \hat{y} = \overrightarrow {t}^T \overrightarrow {x} = \sum _{i=1}^{n} t_i \hat{\alpha }_i

is unique in the sense that y^\hat{y} is independent of which least squares solution x→\overrightarrow {x} is used.

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