Proof. The model matrix is the \(n \times n\) Vandermonde matrix with rows \((1, x_i, x_i^2, \dots , x_i^{n-1})\). If \(\mathbf {Xc} = \mathbf {0}\), the polynomial \(c_0 + c_1x + \dots + c_{n-1}x^{n-1}\), of degree at most \(n - 1\), has the \(n\) distinct roots \(x_1, \dots , x_n\), so it is the zero polynomial and \(\mathbf {c} = \mathbf {0}\). So \(\mathbf {X}\) is non-singular, the equations \(\mathbf {X}\boldsymbol \beta = \mathbf {y}\) have a solution, and least squares finds it, with \(\mathbf {e} = \mathbf {0}\). □

Proof. By the binomial theorem, \[ z^j = d^{-j}(x - c)^j = d^{-j}\sum _{i=0}^j \binom {j}{i}(-c)^{j-i}x^i , \] so each column \(z^j\) is a combination of \(1, x, \dots , x^j\), and \(\mathbf {X}_z = \mathbf {X}_x\mathbf {T}\) with \(\mathbf {T}\) upper triangular. The diagonal entries of \(\mathbf {T}\) are \(d^{-j} \ne 0\), so \(\mathbf {T}\) is non-singular. Then the hat matrices agree: \begin {align*} \mathbf {H}_z &= \mathbf {X}_x\mathbf {T}(\mathbf {T}^\top \mathbf {X}_x^\top \mathbf {X}_x\mathbf {T})^{-1}\mathbf {T}^\top \mathbf {X}_x^\top \\ &= \mathbf {X}_x\mathbf {T}\mathbf {T}^{-1}(\mathbf {X}_x^\top \mathbf {X}_x)^{-1}(\mathbf {T}^\top )^{-1}\mathbf {T}^\top \mathbf {X}_x^\top = \mathbf {H}_x , \end {align*}

so \(\hat {\mathbf {y}} = \mathbf {Hy}\), \(\mathbf {e}\) and every sum of squares are the same for both. Matching the coefficients of \(x^k\) in \(\sum _j\gamma _jz^j = \sum _j\beta _jx^j\), only \(z^k\) contains \(x^k\), with coefficient \(d^{-k}\), so \(\gamma _kd^{-k} = \beta _k\).

For the last statement, take \(y = \beta _0 + \beta _2x^2 + \varepsilon \). In terms of \(x\), the variable \(z^2 = (x^2 - 2cx + c^2)/d^2\) contains \(x\) when \(c \ne 0\), so the column space of \((\mathbf {1}, \mathbf {z}^2)\) contains a vector with a non-zero \(x\)-component, which \((\mathbf {1}, \mathbf {x}^2)\) does not. The two models are different models. □

Proof. Pairing each \(u_i\) with the observation whose deviation is \(-u_i\), the odd powers cancel in pairs, and a deviation equal to \(0\) contributes nothing. Then \(\mathbf {1}^\top \mathbf {u} = \sum u_i = 0\), and \(\mathbf {u}^\top \mathbf {u}^2 = \sum u_i^3 = 0\). The sample covariance of \(u\) and \(u^2\) is \(\frac {1}{n}\sum u_i u_i^2 - \bar u\,\overline {u^2} = 0 - 0\), so the correlation is zero. □

Solution. Centre at \(\bar x = 4\), so \(u = x - 4\) takes the values \(-3, \dots , 3\). The needed sums are \begin {gather*} \textstyle \sum u = 0, \quad \sum u^2 = 28, \quad \sum u^3 = 0, \quad \sum u^4 = 196, \\ \textstyle \sum y = 155, \quad \sum uy = 84, \quad \sum u^2y = 536 . \end {gather*} The normal equations for \(y = \beta _0 + \beta _1u + \beta _2u^2 + \varepsilon \) are \[ \begin {pmatrix} 7 & 0 & 28 \\ 0 & 28 & 0 \\ 28 & 0 & 196 \end {pmatrix} \begin {pmatrix} \hat \beta _0 \\ \hat \beta _1 \\ \hat \beta _2 \end {pmatrix} = \begin {pmatrix} 155 \\ 84 \\ 536 \end {pmatrix} . \] The middle equation, uncoupled from the others by Theorem 10.1.3, gives \(\hat \beta _1 = 84/28 = 3\). The other two have determinant \(7(196) - 28^2 = 588\), so \begin {align*} \hat \beta _0 &= \frac {196(155) - 28(536)}{588} = \frac {15372}{588} = 26.143 , \\ \hat \beta _2 &= \frac {7(536) - 28(155)}{588} = \frac {-588}{588} = -1 . \end {align*}

The fitted model is \[ \hat y = 26.143 + 3u - u^2 = -1.857 + 11x - x^2 , \] the second form by substituting \(u = x - 4\) (Theorem 10.1.2: the fits are the same). The fitted values are \(8.14, 16.14, 22.14, 26.14, 28.14, 28.14, 26.14\), and the residuals are \[ -0.14, \quad -0.14, \quad 0.86, \quad -0.14, \quad -1.14, \quad 0.86, \quad -0.14 , \] so \(SS_{\mathrm {Res}} = 2.857\) on \(7 - 3 = 4\) degrees of freedom and \(MS_{\mathrm {Res}} = 0.714\). With \(\sum y^2 = 3771\), \(SS_T = 3771 - 155^2/7 = 338.857\).

The quadratic term. The \((3,3)\) element of \((\mathbf {X}^\top \mathbf {X})^{-1}\) is \(7/588 = 1/84\), so \[ \operatorname {se}(\hat \beta _2) = \sqrt {0.714/84} = 0.0922, \qquad t_0 = \frac {-1}{0.0922} = -10.84 , \] and \(|t_0| > t_{0.025,4} = 2.776\). The curvature is real: a straight line would not do.

The turning point. The fitted yield is greatest where \(3 - 2u = 0\), at \(u = 1.5\), or \(x = 5.5\), where \(\hat y = 26.143 + 4.5 - 2.25 = 28.39\) kg. Its standard error is found in Example 10.1.6.

Proof. This is the delta method (Lemma 6.1.1) with two variables. Let \(g(a, b) = -a/(2b)\). To first order around \((\beta _1, \beta _2)\), \[ g(\hat \beta _1, \hat \beta _2) \approx g(\beta _1, \beta _2) + g_a(\hat \beta _1 - \beta _1) + g_b(\hat \beta _2 - \beta _2) , \] with \(g_a = -1/(2\beta _2)\) and \(g_b = \beta _1/(2\beta _2^2) = -u_0/\beta _2\). The variance of the linear function on the right is \(g_a^2\operatorname {Var}(\hat \beta _1) + 2g_ag_b\operatorname {Cov}(\hat \beta _1, \hat \beta _2) + g_b^2\operatorname {Var}(\hat \beta _2)\). Since \(g_a^2 = 1/(4\beta _2^2)\), \(2g_ag_b = u_0/\beta _2^2\) and \(g_b^2 = u_0^2/\beta _2^2\), taking out the factor \(1/(4\beta _2^2)\) gives the formula. □

Solution. From \((\mathbf {X}^\top \mathbf {X})^{-1}\), \(\operatorname {Var}(\hat \beta _1) = \sigma ^2/28\), \(\operatorname {Var}(\hat \beta _2) = \sigma ^2/84\) and the covariance is zero. With \(\hat \beta _2 = -1\), \(\hat u_0 = 1.5\) and Theorem 10.1.5, \begin {align*} \widehat {\operatorname {Var}}(\hat u_0) &= \frac {MS_{\mathrm {Res}}}{4}\Bigl [\frac {1}{28} + \frac {4(1.5)^2}{84}\Bigr ] \\ &= \frac {0.714}{4}\Bigl [\frac {3}{84} + \frac {9}{84}\Bigr ] = \frac {0.714}{28} = 0.0255 , \end {align*}

so \(\operatorname {se}(\hat u_0) = 0.160\). With \(t_{0.025,4} = 2.776\) the interval for \(u_0\) is \(1.5 \pm 0.44\), and adding \(4\), \[ 5.06 \le x_0 \le 5.94 . \] The best irrigation level is between about \(51\) and \(59\) mm a week. Figure 20 shows the fit.