Q35. Logistic regression and an adjusted effect
[4 marks]
For independent patients, disease is coded 1 for disease and 0 otherwise. The variable group is coded 0 for group A and 1 for group B.
glm(disease ~ group + cholesterol, family = binomial)(a) Write down explicitly the statistical model corresponding to this code.
(b) The estimated coefficient of group is 0.70. Interpret this estimate after accounting for cholesterol.
(c) Does the estimate mean that group B has a disease probability 0.70 greater than group A? Explain.
Definitions. Let Y_i=1 indicate disease, g_i=0 for group A and g_i=1 for group B, and c_i denote cholesterol. Write p_i=\mathbb P(Y_i=1\mid g_i,c_i). The odds of probability p are p/(1-p) and its logit is \log\{p/(1-p)\}. An odds ratio compares odds, not probabilities; comparisons here hold cholesterol fixed.
(a) Model statement. Conditional on the fixed covariates, Y_i\mid g_i,c_i\overset{\mathrm{ind}}{\sim} \operatorname{Bernoulli}(p_i), \qquad p_i=\mathbb{P}(Y_i=1\mid g_i,c_i), with \boxed{ \log\left(\frac{p_i}{1-p_i}\right) =\beta_0+\beta_Gg_i+\beta_Cc_i=\eta_i. } Equivalently, exponentiating and solving for p_i gives \frac{p_i}{1-p_i}=e^{\eta_i} \quad\Longrightarrow\quad p_i=e^{\eta_i}(1-p_i) \quad\Longrightarrow\quad p_i=\frac{e^{\eta_i}}{1+e^{\eta_i}}.
(b) Interpretation of \widehat\beta_G=0.70. For two patients with the same cholesterol, \frac{\widehat{\operatorname{odds}}(Y=1\mid g=1,c)} {\widehat{\operatorname{odds}}(Y=1\mid g=0,c)} =e^{\widehat\beta_G}=e^{0.70}\approx2.01. Conclusion: Group B has approximately 2.01 times the fitted odds of disease of group A, after accounting for cholesterol. The estimate is an association; it is not automatically a causal effect.
(c) Why this is not a probability increase of 0.70. The coefficient acts on log odds, not directly on probability. If p_A is the fitted probability for group A at a given cholesterol and r=e^{0.70}, then \frac{p_B}{1-p_B}=r\frac{p_A}{1-p_A} \quad\Longrightarrow\quad \boxed{p_B=\frac{rp_A}{1-p_A+rp_A}.} The difference p_B-p_A therefore depends on p_A. For example, if p_A=0.20, then p_B\approx0.335, a difference of about 0.135, not 0.70.