Nonlinear Regression, Robust Regression & Generalized Linear Models (Logistic & Poisson)
Intrinsically nonlinear regression models Y_i = f(x_i, theta) + epsilon_i (Michaelis-Menten, Gompertz, logistic growth, exponential decay); Nonlinear Least Squares (NLS) optimization: Gauss-Newton Taylor linearization, Levenberg-Marquardt damping, and asymptotic covariance sigma^2 (J^T J)^(-1); Robust regression for heavy-tailed contamination: L_1 Least Absolute Deviations (LAD), Huber M-estimation, Iteratively Reweighted Least Squares (IRLS), and Tukey's biweight; Generalized Linear Models (GLMs): exponential dispersion family, linear predictor eta = X beta, canonical link functions g(mu), and Fisher Scoring / IRLS; Binary Logistic Regression (logit link, odds ratios, Wald/deviance tests, Hosmer-Lemeshow) and Poisson Log-Linear Regression for count data and overdispersion.
§8.1 Intrinsically Nonlinear Regression Models vs Linearizable Functional Forms
| Mechanistic Model | Nonlinear Regression Function $f(x, \boldsymbol{\theta})$ | Scientific Interpretation of Parameters | Classical Linearization & Error Distortion |
|---|---|---|---|
| Michaelis–Menten Enzyme Kinetics | $f(x, \boldsymbol{\theta}) = \dfrac{\theta_1 x}{\theta_2 + x}$ | $\theta_1 = V_{\max}$ (asymptotic maximum reaction velocity); $\theta_2 = K_m$ (half-saturation substrate concentration where $f(K_m) = V_{\max}/2$). | Lineweaver–Burk Double Reciprocal: $\dfrac{1}{Y} = \dfrac{1}{\theta_1} + \dfrac{\theta_2}{\theta_1}\left(\dfrac{1}{x}\right)$. Severely distorts additive errors ($\text{Var}(1/Y_i) \approx \sigma^2 / \mu_i^4$), giving huge leverage to small noisy $x_i$! |
| Mono-Molecular / Mitscherlich Growth | $f(x, \boldsymbol{\theta}) = \theta_1 \left(1 - \theta_2 e^{-\theta_3 x}\right)$ | $\theta_1$: upper carrying capacity asymptote; $\theta_1(1-\theta_2)$: initial response at $x=0$; $\theta_3 > 0$: exponential rate constant. | Intrinsically nonlinear in $(\theta_1, \theta_2, \theta_3)$ unless the asymptote $\theta_1$ is known a priori. |
| Three-Parameter Logistic Sigmoid | $f(x, \boldsymbol{\theta}) = \dfrac{\theta_1}{1 + \exp\left[-(\theta_2 + \theta_3 x)\right]}$ | $\theta_1$: saturation ceiling; $-\theta_2/\theta_3$: inflection point (where $f = \theta_1 / 2$); $\theta_3$: steepness rate parameter. | Linearizable via $\ln\!\left(\frac{\theta_1 - Y}{Y}\right) = -\theta_2 - \theta_3 x$ only if ceiling $\theta_1$ is known and $0 < Y < \theta_1$. |
| Gompertz Asymmetric Growth | $f(x, \boldsymbol{\theta}) = \theta_1 \exp\left(-\theta_2 e^{-\theta_3 x}\right)$ | $\theta_1$: upper asymptote; inflection occurs earlier at $f = \theta_1 / e \approx 0.368 \theta_1$ (used in tumor & actuarial modeling). | Intrinsically nonlinear when fitted with additive homoscedastic errors $Y_i = f(x_i, \boldsymbol{\theta}) + \varepsilon_i$. |
§8.2 Nonlinear Least Squares (NLS), Gauss-Newton Taylor Linearization, Levenberg-Marquardt, and Asymptotic Inference
In the nonlinear regression model $Y_i = f(\mathbf{x}_i, \boldsymbol{\theta}) + \varepsilon_i$ with $\mathbb{E}[\boldsymbol{\varepsilon}] = \mathbf{0}_n$ and $\text{Cov}(\boldsymbol{\varepsilon}) = \sigma^2 \mathbf{I}_n$, the Nonlinear Least Squares (NLS) Estimator $\hat{\boldsymbol{\theta}}$ minimizes the residual sum of squares objective function $S(\boldsymbol{\theta})$:
where $\mathbf{f}(\boldsymbol{\theta}) = (f(\mathbf{x}_1, \boldsymbol{\theta}), \dots, f(\mathbf{x}_n, \boldsymbol{\theta}))^T \in \mathbb{R}^n$. Let $\mathbf{J}(\boldsymbol{\theta}) \in \mathbb{R}^{n \times p}$ denote the $n \times p$ local Jacobian (sensitivity) matrix of first partial derivatives of the expectation surface with respect to $\boldsymbol{\theta}$:
Setting the gradient of $S(\boldsymbol{\theta})$ equal to zero gives the $p$ Nonlinear Normal Equations:
Because both $\mathbf{J}(\hat{\boldsymbol{\theta}})$ and $\mathbf{f}(\hat{\boldsymbol{\theta}})$ are nonlinear functions of $\hat{\boldsymbol{\theta}}$, this system of $p$ nonlinear equations admits no closed-form analytical solution and must be solved iteratively.
1. The Gauss–Newton Taylor Linearization Algorithm
Given a current parameter iterate $\boldsymbol{\theta}^{(m)}$ at iteration $m$, we approximate the nonlinear expectation vector $\mathbf{f}(\boldsymbol{\theta})$ around $\boldsymbol{\theta}^{(m)}$ via a first-order multivariate Taylor series expansion:
Substituting this linear approximation into $\mathbf{Y} = \mathbf{f}(\boldsymbol{\theta}) + \boldsymbol{\varepsilon}$ and rearranging terms yields the Local Linearized Pseudo-Model at iteration $m$:
Notice that for fixed $\boldsymbol{\theta}^{(m)}$, estimating the parameter increment vector $\boldsymbol{\delta}^{(m)} = \boldsymbol{\theta}^{(m+1)} - \boldsymbol{\theta}^{(m)}$ is simply an Ordinary Least Squares regression of the current residual vector $\mathbf{e}^{(m)} = \mathbf{Y} - \mathbf{f}(\boldsymbol{\theta}^{(m)})$ on the current $n \times p$ Jacobian matrix $\mathbf{J}_m = \mathbf{J}(\boldsymbol{\theta}^{(m)})$! Solving the linearized normal equations $(\mathbf{J}_m^T \mathbf{J}_m) \hat{\boldsymbol{\delta}}^{(m)} = \mathbf{J}_m^T \mathbf{e}^{(m)}$ gives the Gauss–Newton Iteration:
To understand how Gauss–Newton relates to full Newton–Raphson optimization, observe that the exact $p \times p$ Hessian matrix of $S(\boldsymbol{\theta})$ is:
Because $\mathbb{E}[e_i(\boldsymbol{\theta})] \approx 0$ near the true parameter $\boldsymbol{\theta}$ (and the residuals are small when the model fits well), Gauss–Newton drops the second-derivative intrinsic curvature term $-2 \sum e_i \nabla^2 f_i$ and approximates the Hessian by the positive semi-definite matrix $2 \mathbf{J}_m^T \mathbf{J}_m$ (which is identical to Fisher's Method of Scoring under normal errors!).
2. Levenberg–Marquardt Trust-Region Damping
When the initial guess $\boldsymbol{\theta}^{(0)}$ is far from the optimum or when the Jacobian columns are nearly collinear, $\mathbf{J}_m^T \mathbf{J}_m$ can be nearly singular, causing the undamped Gauss–Newton step $\hat{\boldsymbol{\delta}}^{(m)}$ to overshoot wildly and increase $S(\boldsymbol{\theta})$. Kenneth Levenberg (1944) and Donald Marquardt (1963) resolved this instability by adding a positive ridge damping parameter $\lambda_m > 0$ (or diagonal scaling $\lambda_m \text{diag}(\mathbf{J}_m^T \mathbf{J}_m)$) to the approximate Hessian:
The Levenberg–Marquardt Algorithm interpolates smoothly between steepest-descent gradient descent (as $\lambda_m \to \infty$, $\hat{\boldsymbol{\delta}}^{(m)} \approx \frac{1}{\lambda_m} \mathbf{J}_m^T \mathbf{e}^{(m)} = -\frac{1}{2\lambda_m} \nabla S$, guaranteeing descent in a small trust region) and quadratic Gauss–Newton convergence (as $\lambda_m \to 0$ near the minimum).
3. Asymptotic Distribution and Inference at Convergence
In finite samples, the NLS estimator $\hat{\boldsymbol{\theta}}$ is slightly biased ($\mathbb{E}[\hat{\boldsymbol{\theta}}] \neq \boldsymbol{\theta}$), the residuals do not sum to zero in general ($\sum e_i \neq 0$ unless $\mathbf{1}_n \in \mathcal{C}(\mathbf{J}(\hat{\boldsymbol{\theta}}))$), and $\text{SST} \neq \text{SSR} + \text{SSE}$. However, as $n \to \infty$ under Jennrich's regularity conditions ($\frac{1}{n}\mathbf{J}(\boldsymbol{\theta})^T\mathbf{J}(\boldsymbol{\theta}) \to \mathbf{Q}(\boldsymbol{\theta}) \succ \mathbf{0}$), $\hat{\boldsymbol{\theta}}$ is consistent and asymptotically normal:
§8.3 Robust Regression for Heavy-Tailed Contamination: L_1 LAD, Huber M-Estimation, IRLS & Tukey's Biweight
| Estimator Family | Objective Loss $\rho(u)$ | Influence Score $\psi(u) = \rho'(u)$ | IRLS Weight $w(u) = \psi(u)/u$ |
|---|---|---|---|
| Ordinary Least Squares ($L_2$) | $\dfrac{1}{2} u^2$ | $u$ (unbounded linear) | $w(u) = 1$ (constant equal weight) |
| Least Absolute Deviations ($L_1$ / LAD) | $|u|$ (Laplace MLE) | $\text{sgn}(u) \in \{-1, +1\}$ | $w(u) = \dfrac{1}{|u|}$ (solved via Barrodale–Roberts simplex LP) |
| Huber Monotone M-Estimator ($k = 1.345$ for $95\%$ normal efficiency) | $\begin{cases} \frac{1}{2}u^2, & |u| \le k \\ k|u| - \frac{1}{2}k^2, & |u| > k \end{cases}$ | $\begin{cases} u, & |u| \le k \\ k \, \text{sgn}(u), & |u| > k \end{cases}$ | $\begin{cases} 1, & |u| \le k \\ \dfrac{k}{|u|}, & |u| > k \end{cases}$ (downweights tails as $1/|u|$) |
| Tukey's Redescending Biweight (Bisquare) ($c = 4.685$ for $95\%$ normal efficiency) | $\begin{cases} \frac{c^2}{6}\left[1 - \left(1 - \frac{u^2}{c^2}\right)^3\right], & |u| \le c \\ \frac{c^2}{6}, & |u| > c \end{cases}$ | $\begin{cases} u\left(1 - \frac{u^2}{c^2}\right)^2, & |u| \le c \\ 0, & |u| > c \end{cases}$ | $\begin{cases} \left(1 - \frac{u^2}{c^2}\right)^2, & |u| \le c \\ 0, & |u| > c \end{cases}$ (completely rejects $|u| > c$!) |
§8.4 Generalized Linear Models (GLMs): Exponential Dispersion Family, Canonical Links & Fisher Scoring IRLS
| Distribution | Cumulant $b(\theta)$ | Mean $\mu = b'(\theta)$ | Variance Function $V(\mu)$ | Canonical Link $\theta = g(\mu)$ |
|---|---|---|---|---|
| Normal $\mathcal{N}(\mu, \sigma^2)$ | $\dfrac{\theta^2}{2}$ | $\mu = \theta$ | $V(\mu) = 1$ ($\phi = \sigma^2$) | Identity: $g(\mu) = \mu$ |
| Bernoulli / Binomial | $\ln(1 + e^\theta)$ | $\pi = \dfrac{e^\theta}{1 + e^\theta}$ | $V(\pi) = \pi(1 - \pi)$ ($\phi = 1$) | Logit: $g(\pi) = \ln\!\left(\dfrac{\pi}{1 - \pi}\right)$ |
| Poisson $\text{Pois}(\mu)$ | $e^\theta$ | $\mu = e^\theta$ | $V(\mu) = \mu$ ($\phi = 1$) | Log: $g(\mu) = \ln(\mu)$ |
| Gamma $\text{Gamma}(\nu, \mu)$ | $-\ln(-\theta)$ | $\mu = -\dfrac{1}{\theta}$ | $V(\mu) = \mu^2$ ($\phi = \nu^{-1}$) | Negative Reciprocal: $g(\mu) = -\dfrac{1}{\mu}$ (or Log) |
§8.5 Binary Logistic Regression: Logit Link, Odds Ratios, Deviance Tests, and IRLS
When the response variable is a binary Bernoulli indicator $Y_i \in \{0, 1\}$ (or a grouped binomial count $Y_i \sim \text{Binomial}(m_i, \pi_i)$) representing the occurrence of an event such as disease presence, loan default, or customer churn, our target is the conditional success probability $\pi(\mathbf{x}_i) = \mathbb{P}(Y_i = 1 \mid \mathbf{x}_i) = \mathbb{E}[Y_i \mid \mathbf{x}_i] \in (0, 1)$.
Fitting a naive Linear Probability Model (LPM) $\pi(\mathbf{x}_i) = \mathbf{x}_i^T \boldsymbol{\beta}$ via Ordinary Least Squares suffers from three fatal statistical defects:
1. Unbounded Predictions: Because $\mathbf{x}_i^T \boldsymbol{\beta}$ is linear and unbounded, it inevitably predicts nonsensical probabilities $\hat{\pi}_i < 0$ or $\hat{\pi}_i > 1$ for extreme covariate values.
2. Inherent Heteroscedasticity: Since $Y_i \in \{0, 1\}$ is Bernoulli, its error term $\varepsilon_i = Y_i - \pi_i$ takes only two discrete values ($1 - \pi_i$ and $-\pi_i$) with variance $\text{Var}(\varepsilon_i) = \pi_i(1 - \pi_i)$, violating homoscedasticity and invalidating OLS standard errors.
3. Constant Marginal Effects: A one-unit increase in $x_j$ changes $\pi$ by $\beta_j$ identically whether $\pi = 0.50$ or $\pi = 0.99$.
1. The Logistic Sigmoid Function, Logit Link, and Odds Ratio Interpretation
Binary Logistic Regression resolves all three defects by modeling $\pi(\mathbf{x}_i)$ via the standard logistic cumulative distribution function (sigmoid curve) $\pi(\eta) = \frac{e^\eta}{1 + e^\eta} = \frac{1}{1 + e^{-\eta}} \in (0, 1)$, whose inverse is the canonical Logit (Log-Odds) Link Function:
Here, the ratio $\Omega(\mathbf{x}_i) = \frac{\pi(\mathbf{x}_i)}{1 - \pi(\mathbf{x}_i)} = \exp(\mathbf{x}_i^T \boldsymbol{\beta}) \in (0, \infty)$ is the Odds of Success at covariate vector $\mathbf{x}_i$. How do we interpret an individual slope coefficient $\beta_j$? Increase predictor $x_j$ by $1$ unit while holding all other covariates constant:
Thus, $\beta_j$ is the change in log-odds per unit increase in $x_j$, and $\exp(\beta_j)$ is the Multiplicative Odds Ratio ($\text{OR}_j$)! If $\beta_j > 0$, $e^{\beta_j} > 1$ (odds increase by $100(e^{\beta_j} - 1)\%$ per unit increase in $x_j$); if $\beta_j = 0$, $e^{\beta_j} = 1$ (odds unchanged); if $\beta_j < 0$, $0 < e^{\beta_j} < 1$ (odds decrease). Furthermore, on the probability scale, differentiating $\pi(\mathbf{x})$ with respect to $x_j$ gives the Marginal Probability Effect:
which achieves its steepest slope $\frac{1}{4}\beta_j$ at the median effective dose ($\pi = 0.50$, where $\mathbf{x}^T\boldsymbol{\beta} = 0$, so $\text{LD}_{50} = -\beta_0 / \beta_1$ in simple logistic regression) and flattens gracefully toward zero as $\pi \to 0$ or $\pi \to 1$.
2. Log-Likelihood, IRLS Working Equations, Wald Tests, and Deviance Goodness-of-Fit
For $n$ independent Bernoulli observations $Y_i \in \{0, 1\}$, the log-likelihood function is:
Setting the score vector to zero gives the Logistic Likelihood Equations $\mathbf{U}(\hat{\boldsymbol{\beta}}) = \mathbf{X}^T (\mathbf{Y} - \hat{\boldsymbol{\pi}}) = \mathbf{0}_p$. Specializing the GLM IRLS algorithm of Section 8.4 to the canonical logit link ($V(\pi_i) = \pi_i(1 - \pi_i)$, $g'(\pi_i) = \frac{1}{\pi_i(1 - \pi_i)}$, $\phi = 1$), the IRLS weight matrix and working response at iteration $m$ are:
At convergence, $\hat{\boldsymbol{\beta}} \overset{\text{approx}}{\sim} \mathcal{N}_p\left(\boldsymbol{\beta}, (\mathbf{X}^T \hat{\mathbf{W}} \mathbf{X})^{-1}\right)$. Hypothesis testing in logistic regression employs:
1. Wald Test: $z_j = \frac{\hat{\beta}_j}{\text{SE}(\hat{\beta}_j)} \overset{d}{\to} \mathcal{N}(0, 1)$ (or $W_j = z_j^2 \sim \chi_1^2$), and the $95\%$ CI for the Odds Ratio is $\exp\left(\hat{\beta}_j \pm 1.96 \, \text{SE}(\hat{\beta}_j)\right)$. *(Caution: the Hauck–Donner effect shows that for huge $|\hat{\beta}_j|$, $\hat{\pi}_i(1 - \hat{\pi}_i) \to 0$ inflates $\text{SE}(\hat{\beta}_j)$, making the Wald statistic lose power; the Likelihood Ratio / Deviance test is always preferred!).*
2. Residual Deviance $D$ and Likelihood Ratio Drop-in-Deviance Test: The Residual Deviance is twice the log-likelihood ratio between the saturated model ($\tilde{\pi}_i = Y_i / m_i$) and the fitted model:
To test a nested Reduced model $M_R$ ($p_R$ parameters) against a Full model $M_F$ ($p_F$ parameters), the Drop-in-Deviance (Likelihood Ratio) Statistic is:
For ungrouped binary data ($m_i = 1$), overall goodness of fit is evaluated via the Hosmer–Lemeshow Test, which partitions observations into $G = 10$ deciles of predicted risk $\hat{\pi}_i$ and compares observed vs. expected event counts via a $\chi_{G-2}^2 = \chi_8^2$ statistic.
§8.6 Poisson Log-Linear Regression for Count Data, Rate Offsets, Overdispersion & Quasi-Poisson / Negative Binomial Models
When the response variable $Y_i \in \{0, 1, 2, \dots\}$ represents a non-negative integer count of discrete events occurring within a fixed spatial or temporal window—such as hospital emergency admissions per day, highway traffic accidents per intersection, or defective microchips per silicon wafer—the benchmark generalized linear model is Poisson Log-Linear Regression.
1. The Poisson Log-Linear Model, Rate Ratios, and Exposure Offsets
Assume $Y_i \sim \text{Poisson}(\mu_i)$ independently with probability mass function $f(y_i; \mu_i) = \frac{e^{-\mu_i} \mu_i^{y_i}}{y_i!}$ for $y_i = 0, 1, 2, \dots$, where the mean and variance satisfy the equidispersion property $\mathbb{E}[Y_i] = \text{Var}(Y_i) = \mu_i > 0$. Using the canonical logarithmic link function $g(\mu_i) = \ln(\mu_i) = \eta_i$ guarantees that the predicted mean count $\mu_i = \exp(\mathbf{x}_i^T \boldsymbol{\beta}) > 0$ is strictly positive for all $\boldsymbol{\beta} \in \mathbb{R}^p$:
Because the covariates act multiplicatively on the expected count $\mu_i$, exponentiating an individual regression slope $\beta_j$ yields the Incidence Rate Ratio ($\text{IRR}_j$) (or Relative Risk):
That is, a one-unit increase in $x_j$ multiplies the expected count by $e^{\beta_j}$ (a $100(e^{\beta_j} - 1)\%$ percentage change).
In many epidemiological and actuarial studies, the counts $Y_i$ are observed over unequal exposure denominators $t_i > 0$ (such as person-years at risk, population size, or kilometers of highway), and we wish to model the underlying event rate $\lambda_i = \frac{\mu_i}{t_i}$. Taking logarithms of $\frac{\mu_i}{t_i} = \exp(\mathbf{x}_i^T \boldsymbol{\beta})$ gives:
Here, $\ln(t_i)$ enters the linear predictor as an Offset Variable—a known regressor whose slope coefficient is constrained to equal exactly $1.0$ without being estimated!
2. Score Equations, IRLS Weights, and Poisson Deviance
The Poisson log-likelihood is $\ell(\boldsymbol{\beta}) = \sum_{i=1}^n \left[ Y_i (\mathbf{x}_i^T\boldsymbol{\beta}) - \exp(\mathbf{x}_i^T\boldsymbol{\beta}) - \ln(Y_i!) \right]$, yielding the score equations $\mathbf{U}(\hat{\boldsymbol{\beta}}) = \mathbf{X}^T(\mathbf{Y} - \hat{\boldsymbol{\mu}}) = \mathbf{0}_p$, the IRLS weight matrix $\mathbf{W} = \text{diag}(\mu_1, \dots, \mu_n)$, working response $z_i = \eta_i + \frac{Y_i - \mu_i}{\mu_i}$, and asymptotic covariance $\widehat{\text{Cov}}(\hat{\boldsymbol{\beta}}) = (\mathbf{X}^T \hat{\mathbf{W}} \mathbf{X})^{-1}$. The Poisson Residual Deviance and Pearson Chi-Square Statistic are:
(with $0 \ln(0) \equiv 0$). Notice that whenever an intercept is included, the first score equation $\sum_{i=1}^n (Y_i - \hat{\mu}_i) = 0$ forces the sum of fitted counts to equal the sum of observed counts ($\sum \hat{\mu}_i = \sum Y_i$), simplifying the deviance to $D = 2 \sum_{i=1}^n Y_i \ln(Y_i / \hat{\mu}_i)$.
3. Overdispersion Diagnostics and Quasi-Poisson vs Negative Binomial Remedies
In real-world count data, unobserved individual heterogeneity (frailty) or event clustering almost always causes the empirical variance to exceed the conditional mean ($\text{Var}(Y_i) > \mu_i$), a phenomenon called Overdispersion. When a well-specified Poisson model holds ($\phi = 1$), both $D / (n - p)$ and $X^2 / (n - p)$ should hover near $1.0$. If the estimated dispersion parameter $\hat{\phi}$:
exceeds unity substantially, standard Poisson standard errors are severely biased downward (producing spuriously large Wald $z$-statistics). Two standard remedies are used:
1. Quasi-Poisson GLM (Wedderburn Quasi-Likelihood): Postulates $\mathbb{E}[Y_i] = \mu_i$ and $\text{Var}(Y_i) = \phi \mu_i$ ($\phi > 1$). Because $\phi$ cancels out of the score equation $\frac{1}{\phi}\mathbf{X}^T(\mathbf{Y} - \boldsymbol{\mu}) = \mathbf{0}_p$, the point estimates $\hat{\boldsymbol{\beta}}$ are identical to ordinary Poisson MLE, while all standard errors are multiplied by the overdispersion inflation factor $\sqrt{\hat{\phi}}$, and drop-in-deviance tests use $F = \frac{(D_R - D_F)/q}{\hat{\phi}} \sim F_{q, n-p}$.
2. Negative Binomial Regression (Poisson–Gamma Mixture): Assumes $Y_i \mid u_i \sim \text{Poisson}(\mu_i u_i)$ where the unobserved frailty $u_i \sim \text{Gamma}(\alpha^{-1}, \alpha^{-1})$ has mean $1$ and variance $\alpha > 0$. Integrating out $u_i$ yields a Negative Binomial marginal distribution with mean $\mathbb{E}[Y_i] = \mu_i$ and quadratic variance function $\text{Var}(Y_i) = \mu_i + \alpha \mu_i^2$, allowing full likelihood inference and testing $H_0: \alpha = 0$.
Step-by-Step Regression Analysis Solutions
Three original questions connect calculations, definitions and interpretation. Open each solution to follow the reasoning.
A biochemistry laboratory measures initial enzymatic reaction velocity $Y_i$ across $n = 4$ substrate concentrations $x_i \in \{1, 2, 4, 8\}$ (in mM) to estimate the Michaelis–Menten nonlinear regression model:
Starting from the initial parameter vector $\boldsymbol{\theta}^{(0)} = (\theta_1^{(0)}, \theta_2^{(0)})^T = (12.0, \; 2.0)^T$:
- Derive the analytical partial derivatives $J_{i1}(\boldsymbol{\theta}) = \frac{\partial f(x_i, \boldsymbol{\theta})}{\partial \theta_1}$ and $J_{i2}(\boldsymbol{\theta}) = \frac{\partial f(x_i, \boldsymbol{\theta})}{\partial \theta_2}$, and evaluate the $4 \times 1$ initial residual vector $\mathbf{e}^{(0)} = \mathbf{Y} - \mathbf{f}(\boldsymbol{\theta}^{(0)})$, the initial sum of squared residuals $S(\boldsymbol{\theta}^{(0)})$, and the $4 \times 2$ Jacobian matrix $\mathbf{J}_0 = \mathbf{J}(\boldsymbol{\theta}^{(0)})$.
- Compute $\mathbf{J}_0^T \mathbf{J}_0$, $\mathbf{J}_0^T \mathbf{e}^{(0)}$, the Gauss–Newton step increment $\hat{\boldsymbol{\delta}}^{(0)} = (\mathbf{J}_0^T \mathbf{J}_0)^{-1} \mathbf{J}_0^T \mathbf{e}^{(0)}$, and the updated parameter vector $\boldsymbol{\theta}^{(1)} = \boldsymbol{\theta}^{(0)} + \hat{\boldsymbol{\delta}}^{(0)}$.
- Verify that $S(\boldsymbol{\theta}^{(1)}) < S(\boldsymbol{\theta}^{(0)})$, and taking $\boldsymbol{\theta}^{(1)}$ as the converged NLS estimate with $\mathbf{J}_0^T \mathbf{J}_0$ as the approximate information matrix, compute the asymptotic standard errors $\text{SE}(\hat{\theta}_1)$ and $\text{SE}(\hat{\theta}_2)$.
Differentiating $f(x_i, \boldsymbol{\theta}) = \frac{\theta_1 x_i}{\theta_2 + x_i}$ with respect to $\theta_1$ and $\theta_2$ gives the Jacobian columns:
At $\boldsymbol{\theta}^{(0)} = (12.0, \; 2.0)^T$, we evaluate $f_i^{(0)} = \frac{12 x_i}{2 + x_i}$, $e_i^{(0)} = Y_i - f_i^{(0)}$, $J_{i1}^{(0)} = \frac{x_i}{2 + x_i}$, and $J_{i2}^{(0)} = -\frac{12 x_i}{(2 + x_i)^2}$ across the $n = 4$ points:
- $x_1 = 1$: $f_1^{(0)} = \frac{12}{3} = 4.000 \implies e_1^{(0)} = 4.2 - 4.0 = +0.200$, $J_{11}^{(0)} = \frac{1}{3} \approx 0.3333$, $J_{12}^{(0)} = -\frac{12}{9} \approx -1.3333$.
- $x_2 = 2$: $f_2^{(0)} = \frac{24}{4} = 6.000 \implies e_2^{(0)} = 6.5 - 6.0 = +0.500$, $J_{21}^{(0)} = \frac{2}{4} = 0.5000$, $J_{22}^{(0)} = -\frac{24}{16} = -1.5000$.
- $x_3 = 4$: $f_3^{(0)} = \frac{48}{6} = 8.000 \implies e_3^{(0)} = 8.4 - 8.0 = +0.400$, $J_{31}^{(0)} = \frac{4}{6} \approx 0.6667$, $J_{32}^{(0)} = -\frac{48}{36} \approx -1.3333$.
- $x_4 = 8$: $f_4^{(0)} = \frac{96}{10} = 9.600 \implies e_4^{(0)} = 10.2 - 9.6 = +0.600$, $J_{41}^{(0)} = \frac{8}{10} = 0.8000$, $J_{42}^{(0)} = -\frac{96}{100} = -0.9600$.
The initial sum of squared residuals is $S(\boldsymbol{\theta}^{(0)}) = (0.2)^2 + (0.5)^2 + (0.4)^2 + (0.6)^2 = 0.04 + 0.25 + 0.16 + 0.36 = 0.8100$, and:
We compute the $2 \times 2$ matrix $\mathbf{J}_0^T \mathbf{J}_0$ and the $2 \times 1$ vector $\mathbf{J}_0^T \mathbf{e}^{(0)}$:
- $(\mathbf{J}_0^T \mathbf{J}_0)_{11} = (1/3)^2 + (1/2)^2 + (2/3)^2 + (4/5)^2 = \frac{1}{9} + \frac{1}{4} + \frac{4}{9} + \frac{16}{25} = 0.5556 + 0.2500 + 0.6400 = 1.4456$
- $(\mathbf{J}_0^T \mathbf{J}_0)_{22} = (-4/3)^2 + (-3/2)^2 + (-4/3)^2 + (-0.96)^2 = \frac{16}{9} + \frac{9}{4} + \frac{16}{9} + 0.9216 = 3.5556 + 2.2500 + 0.9216 = 6.7272$
- $(\mathbf{J}_0^T \mathbf{J}_0)_{12} = (1/3)(-4/3) + (1/2)(-3/2) + (2/3)(-4/3) + (0.8)(-0.96) = -\frac{4}{9} - \frac{3}{4} - \frac{8}{9} - 0.768 = -1.3333 - 0.7500 - 0.7680 = -2.8513$
- $(\mathbf{J}_0^T \mathbf{e}^{(0)})_1 = (1/3)(0.2) + (1/2)(0.5) + (2/3)(0.4) + (0.8)(0.6) = 0.0667 + 0.2500 + 0.2667 + 0.4800 = 1.0633$
- $(\mathbf{J}_0^T \mathbf{e}^{(0)})_2 = (-4/3)(0.2) + (-3/2)(0.5) + (-4/3)(0.4) + (-0.96)(0.6) = -0.2667 - 0.7500 - 0.5333 - 0.5760 = -2.1260$
With $\det(\mathbf{J}_0^T \mathbf{J}_0) = (1.4456)(6.7272) - (-2.8513)^2 = 9.7248 - 8.1301 = 1.5947$:
Updating $\boldsymbol{\theta}^{(1)} = \boldsymbol{\theta}^{(0)} + \hat{\boldsymbol{\delta}}^{(0)}$ gives:
Evaluating $f(x_i, \boldsymbol{\theta}^{(1)}) = \frac{12.6844 x_i}{1.9740 + x_i}$ at $x_i \in \{1, 2, 4, 8\}$ gives fitted values $\mathbf{f}(\boldsymbol{\theta}^{(1)}) = (4.265, \; 6.384, \; 8.493, \; 10.174)^T$ and residuals $\mathbf{e}^{(1)} = (-0.065, \; +0.116, \; -0.093, \; +0.026)^T$, whose sum of squares is:
A single Gauss–Newton step reduced the residual sum of squares by $96.7\%$! With $n - p = 4 - 2 = 2$ error degrees of freedom, the error variance estimate is $s^2 = \frac{0.0270}{2} = 0.0135$, and the asymptotic standard errors are:
A cardiovascular epidemiology team fits a binary logistic regression model $\ln\!\left(\frac{\pi_i}{1 - \pi_i}\right) = \beta_0 + \beta_1 x_{i1} + \beta_2 x_{i2}$ across $n = 250$ patients to predict $5$-year coronary heart disease incidence $Y_i \in \{0, 1\}$ from standardized systolic blood pressure $x_{i1}$ (in SD units) and smoking status $x_{i2} \in \{0 = \text{non-smoker}, 1 = \text{smoker}\}$. Maximum likelihood estimation via Fisher Scoring IRLS yields:
The maximized log-likelihoods are: Null (intercept-only) model $\ell_0 = -142.50$; Reduced model with $x_1$ only $\ell(x_1) = -131.20$; Full model with $(x_1, x_2)$ $\ell(x_1, x_2) = -126.40$.
- Compute the point estimates and $95\%$ Wald confidence intervals for the Odds Ratios $\text{OR}_1 = e^{\beta_1}$ and $\text{OR}_2 = e^{\beta_2}$, and interpret each in clinical terms.
- For a smoking patient ($x_2 = 1$) whose systolic blood pressure is $1.5$ SD above the mean ($x_1 = 1.5$), compute the linear predictor $\hat{\eta}_0$, the predicted disease probability $\hat{\pi}_0$, the IRLS working variance weight $\hat{w}_0 = \hat{\pi}_0(1 - \hat{\pi}_0)$, and the marginal probability effect $\frac{\partial \hat{\pi}_0}{\partial x_1}$ of blood pressure at this patient's risk profile.
- Conduct both the overall Likelihood Ratio (Drop-in-Deviance) Test of $H_0: \beta_1 = \beta_2 = 0$ ($\chi_{0.05, 2}^2 = 5.991$) and the partial drop-in-deviance test vs. Wald test for smoking status $H_0: \beta_2 = 0$ ($\chi_{0.05, 1}^2 = 3.841$).
Using $\widehat{\text{OR}}_j = \exp(\hat{\beta}_j)$ and the $95\%$ log-odds interval $\hat{\beta}_j \pm 1.96 \, \text{SE}(\hat{\beta}_j)$ exponentiated to the odds-ratio scale:
1. Blood Pressure ($x_1$, per $1\text{ SD}$ increase):
Holding smoking status constant, each $1\text{ SD}$ increase in systolic blood pressure increases the odds of coronary heart disease by $82.2\%$ ($95\%$ CI: $35.8\%$ to $144.5\%$).
2. Smoking Status ($x_2 = 1$ vs. $x_2 = 0$):
At any given blood pressure level, smokers have $2.46$ times the odds of coronary heart disease compared to non-smokers.
For a patient with $x_{01} = 1.5$ and $x_{02} = 1$, the estimated log-odds linear predictor $\hat{\eta}_0$ is:
Applying the inverse-logit sigmoid function $\hat{\pi}_0 = \frac{1}{1 + e^{-\hat{\eta}_0}}$:
Because $\hat{\pi}_0 = 0.50$ sits right at the inflection point of the logistic curve, this patient achieves the maximal possible IRLS working weight and maximal marginal sensitivity:
Thus, at this risk profile, a $1\text{ SD}$ increase in blood pressure increases the absolute probability of disease by approximately $15.0$ percentage points.
1. Overall Model Likelihood Ratio Test ($H_0: \beta_1 = \beta_2 = 0$, $q = 2$ df):
The model as a whole is highly statistically significant ($p < 10^{-7}$).
2. Partial Drop-in-Deviance vs. Wald Test for Smoking ($H_0: \beta_2 = 0$, $q = 1$ df):
- Drop-in-Deviance (LRT): $\Delta D = -2 \left[ \ell(x_1) - \ell(x_1, x_2) \right] = -2 \left[ -131.20 - (-126.40) \right] = 2(4.80) = 9.60 > \chi_{0.05, 1}^2 = 3.841$.
- Wald Chi-Square: $W_2 = \left(\frac{\hat{\beta}_2}{\text{SE}(\hat{\beta}_2)}\right)^2 = \left(\frac{0.900}{0.300}\right)^2 = 3.00^2 = 9.00 > 3.841$.
Both tests confirm that smoking status adds highly significant predictive power beyond blood pressure.
This two-part problem examines Iteratively Reweighted Least Squares (IRLS) in both Robust Regression and Generalized Linear Models:
1. Part A (Huber Robust Regression IRLS): In a simple linear regression $Y_i = \beta_0 + \beta_1 x_i + \varepsilon_i$, an analyst observes $n = 5$ data points at $x_i \in \{-2, -1, 0, 1, 2\}$ with responses $\mathbf{Y} = (1.0, \; 2.0, \; 3.0, \; 4.0, \; 10.0)^T$ (where $Y_5 = 10.0$ is a severe outlier above the clean line $Y = 3 + x$).
- Compute the ordinary least squares estimates $(\hat{\beta}_{0,\text{OLS}}, \hat{\beta}_{1,\text{OLS}})$.
- Starting from the initial median/resistant fit $\hat{\beta}_0^{(0)} = 3.0, \; \hat{\beta}_1^{(0)} = 1.0$ with robust scale $\hat{\sigma} = 1.0$ and Huber tuning constant $k = 1.345$, compute the scaled residuals $u_i^{(0)}$, the Huber IRLS weights $w_i^{(0)} = \min(1, \, k/|u_i^{(0)}|)$, and perform one IRLS Weighted Least Squares update to find $(\hat{\beta}_0^{(1)}, \hat{\beta}_1^{(1)})$. Compare its resistance to the outlier against OLS!
2. Part B (Overdispersed Quasi-Poisson Regression with Exposure Offset): An epidemiologist fits a Poisson log-linear model $\ln(\mu_i) = \ln(t_i) + \beta_0 + \beta_1 x_i$ (where $t_i$ is person-years of exposure and $x_i \in \{0, 1\}$ is occupational hazard exposure) across $n = 50$ industrial plants ($p = 2$). The Poisson MLE gives $\hat{\beta}_0 = -4.605$, $\hat{\beta}_1 = +0.6931$ with naive Poisson standard error $\text{SE}_{\text{Pois}}(\hat{\beta}_1) = 0.120$, and Pearson chi-square statistic $X^2 = \sum_{i=1}^{50} (Y_i - \hat{\mu}_i)^2 / \hat{\mu}_i = 192.0$.
- Compute the Incidence Rate Ratio $\widehat{\text{IRR}}_1 = e^{\hat{\beta}_1}$, the expected count $\hat{\mu}_0$ for an exposed plant ($x_0 = 1$) with $t_0 = 500$ person-years, the overdispersion estimate $\hat{\phi}$, and the Quasi-Poisson adjusted standard error and $95\%$ confidence interval for $\text{IRR}_1$.
For $x_i \in \{-2, -1, 0, 1, 2\}$, we have $\bar{x} = 0$ and $S_{xx} = 4 + 1 + 0 + 1 + 4 = 10$. With $\mathbf{Y} = (1, 2, 3, 4, 10)^T$:
- $\bar{Y} = \frac{1 + 2 + 3 + 4 + 10}{5} = \frac{20}{5} = 4.00 \implies \hat{\beta}_{0,\text{OLS}} = 4.00$
- $S_{xy} = (-2)(1) + (-1)(2) + (0)(3) + (1)(4) + (2)(10) = -2 - 2 + 0 + 4 + 20 = 20 \implies \hat{\beta}_{1,\text{OLS}} = \frac{20}{10} = 2.00$
Notice how the single contaminated observation $Y_5 = 10.0$ (whose true clean value would be $5.0$) doubled the OLS slope from $\beta_1 = 1.00$ to $\hat{\beta}_{1,\text{OLS}} = 2.00$ and shifted the intercept from $3.00$ to $4.00$!
At the initial resistant fit $\hat{Y}_i^{(0)} = 3.0 + 1.0 x_i = (1, 2, 3, 4, 5)^T$ with $\hat{\sigma} = 1.0$ and $k = 1.345$, the initial residuals $e_i^{(0)} = Y_i - \hat{Y}_i^{(0)}$ and standardized residuals $u_i^{(0)} = e_i^{(0)} / 1.0$ are:
Since $|u_i^{(0)}| = 0 \le 1.345$ for $i = 1, 2, 3, 4$, those four clean points receive full Huber weight $w_1^{(0)} = w_2^{(0)} = w_3^{(0)} = w_4^{(0)} = 1.000$. For the outlier $i = 5$ ($|u_5^{(0)}| = 5.0 > 1.345$), Huber's weight function downweights observation $5$ to:
Now perform Weighted Least Squares with weights $\mathbf{w}^{(0)} = (1, \, 1, \, 1, \, 1, \, 0.269)^T$:
- $\sum w_i = 4.2690$, $\quad \sum w_i x_i = -2 - 1 + 0 + 1 + 0.269(2) = -1.4620 \implies \bar{x}_w = \frac{-1.4620}{4.2690} = -0.34247$
- $\sum w_i Y_i = 1 + 2 + 3 + 4 + 0.269(10) = 12.6900 \implies \bar{Y}_w = \frac{12.6900}{4.2690} = 2.97259$
- $\sum w_i x_i^2 = 4 + 1 + 0 + 1 + 0.269(4) = 7.0760 \implies S_{xx,w} = 7.0760 - 4.2690(-0.34247)^2 = 7.0760 - 0.5007 = 6.5753$
- $\sum w_i x_i Y_i = -2 - 2 + 0 + 4 + 0.269(20) = 5.3800 \implies S_{xy,w} = 5.3800 - 4.2690(-0.34247)(2.97259) = 5.3800 + 4.3459 = 9.7259$
Therefore, the Huber M-estimates are:
(And under Tukey's redescending biweight with $c = 4.685$, since $|u_5^{(0)}| = 5.0 > 4.685$, the outlier receives weight $w_5 = 0$, recovering the exact clean parameters $\hat{\beta}_0 = 3.00, \; \hat{\beta}_1 = 1.00$!).
In the Poisson log-linear rate model $\ln(\mu_i) = \ln(t_i) + \beta_0 + \beta_1 x_i$:
1. Incidence Rate Ratio ($\text{IRR}_1$) and Expected Count at $t_0 = 500, x_0 = 1$:
2. Overdispersion Parameter $\hat{\phi}$ and Quasi-Poisson Standard Error:
With $n - p = 50 - 2 = 48$ degrees of freedom and $X^2 = 192.0$:
Because the empirical count variance is $4$ times larger than the Poisson mean, the naive Poisson standard error $\text{SE}_{\text{Pois}}(\hat{\beta}_1) = 0.120$ is understated by a factor of $2$! The Quasi-Poisson standard error is:
Consequently, the $95\%$ Quasi-Poisson confidence interval for $\text{IRR}_1 = e^{\beta_1}$ is: