Proof. (i) \(\dfrac {d}{d\eta }\log (1 + e^{\eta }) = \dfrac {e^{\eta }}{1 + e^{\eta }} = \pi \), so \(\partial \ell /\partial \eta _i = y_i - m_i\pi _i = y_i - \mu _i\), and since \(\partial \eta _i/\partial \beta _j = x_{ij}\), the chain rule gives \(\partial \ell /\partial \beta _j = \sum _i x_{ij}(y_i - \mu _i)\).

(ii) \(d\pi /d\eta = e^{\eta }/(1 + e^{\eta })^2 = \pi (1 - \pi )\), so \(\partial \mu _i/\partial \beta _l = m_i\pi _i(1 - \pi _i)x_{il} = w_ix_{il}\), and \(\partial ^2\ell /\partial \beta _j\partial \beta _l = -\sum _i x_{ij}w_ix_{il}\), the \((j, l)\) element of \(-\mathbf {X}^\top \mathbf {WX}\).

(iii) For \(\mathbf {v} \ne \mathbf {0}\), \(\mathbf {v}^\top \mathbf {X}^\top \mathbf {WXv} = \sum _i w_i(\mathbf {x}_i^\top \mathbf {v})^2 > 0\), because every \(w_i > 0\) and \(\mathbf {Xv} \ne \mathbf {0}\). So the second-derivative matrix is negative definite everywhere, and \(\ell \) is strictly concave: along any line it is a function with negative second derivative, so a point with zero gradient is the unique maximum. □

Proof. Newton’s method maximises the second-order Taylor approximation of \(\ell \) at \(\boldsymbol \beta ^{(t)}\). By Theorem 11.2.1 this gives \[ \boldsymbol \beta ^{(t+1)} = \boldsymbol \beta ^{(t)} + (\mathbf {X}^\top \mathbf {WX})^{-1}\mathbf {U} . \] Since \(\mathbf {X}\boldsymbol \beta ^{(t)} = \boldsymbol \eta \) and \(\mathbf {y} - \boldsymbol \mu = \mathbf {W}(\mathbf {z} - \boldsymbol \eta )\), \begin {align*} \boldsymbol \beta ^{(t+1)} &= (\mathbf {X}^\top \mathbf {WX})^{-1}\bigl [\mathbf {X}^\top \mathbf {WX}\boldsymbol \beta ^{(t)} + \mathbf {X}^\top \mathbf {W}(\mathbf {z} - \boldsymbol \eta )\bigr ] \\ &= (\mathbf {X}^\top \mathbf {WX})^{-1}\mathbf {X}^\top \mathbf {Wz} . \end {align*}

This is the weighted least squares estimate of Theorem 6.4.1 with response \(\mathbf {z}\). □

Solution. First step. At \(\boldsymbol \beta = \mathbf {0}\), every \(\pi _i = \frac 12\), \(\mu _i = 10\) and \(w_i = 20 \times \frac 14 = 5\), so \(z_i = 0 + (y_i - 10)/5 = -1.6, -0.8, 0, 0.8, 1.6\). The weights are equal, so the weighted fit is the ordinary straight-line fit of \(z\) on \(x\): with \(\bar x = 3\) and \(\sum (x_i - 3)^2 = 10\), \begin {align*} \beta _1^{(1)} &= \frac {\sum (x_i - 3)z_i}{10} = \frac {3.2 + 0.8 + 0 + 0.8 + 3.2}{10} = 0.8 , \\ \beta _0^{(1)} &= \bar z - 0.8\bar x = -2.4 . \end {align*}

Further steps. Recomputing \(\pi _i\), \(w_i\) and \(z_i\) at each new estimate gives:

The log-likelihood rises at each step and the estimates settle after four steps: \[ \operatorname {logit}\hat \pi = -3.036 + 1.012x . \] The fitted probabilities are \(0.117, 0.267, 0.500, 0.733, 0.883\), so the fitted numbers killed are \(2.33, 5.33, 10.00, 14.67, 17.67\). They add to \(50\), the observed total, as Theorem 11.2.1 requires. The odds ratio for one unit of dose is \(e^{1.012} = 2.75\): each extra unit multiplies the odds of death by \(2.75\). Figure 23, panel 3, shows the fit.

Proof. The columns of \(\mathbf {X}\) are \(\mathbf {1}\) and \(\mathbf {z}\), so the likelihood equations say the fitted successes equal the observed ones in total and in group \(1\), hence in each group: \(m_0\hat \pi _0 = y_0\) and \(m_1\hat \pi _1 = y_1\). Taking logits, \(\hat \beta _0 = \operatorname {logit}\hat \pi _0\) and \(\hat \beta _0 + \hat \beta _1 = \operatorname {logit}\hat \pi _1\), which give the formulas. For the variance, with \(w_g = m_g\hat \pi _g(1 - \hat \pi _g)\), \begin {align*} \mathbf {X}^\top \mathbf {WX} &= \begin {pmatrix} w_0 + w_1 & w_1 \\ w_1 & w_1 \end {pmatrix} , \\ \bigl [(\mathbf {X}^\top \mathbf {WX})^{-1}\bigr ]_{22} &= \frac {w_0 + w_1}{w_0w_1} = \frac {1}{w_0} + \frac {1}{w_1} , \end {align*}

and \(\dfrac {1}{w_g} = \dfrac {1}{m_g\hat \pi _g(1 - \hat \pi _g)} = \dfrac {1}{m_g\hat \pi _g} + \dfrac {1}{m_g(1 - \hat \pi _g)} = \dfrac {1}{y_g} + \dfrac {1}{m_g - y_g}\). □