Proportional hazards model

Remarks

For two covariate vectors $x$ and $x'$, the hazard ratio $h(t \mid x) / h(t \mid x') = e^{\beta^T (x - x')}$ does not depend on $t$; this is the proportional hazards property. Equivalently, the survival functions satisfy $S(t \mid x) = S_0(t)^{e^{\beta^T x}}$, where $S_0$ is the baseline survival function.

Given right-censored data $(\tilde t_i, \delta_i, x_i)$ without tied times, $\beta$ is estimated by maximising the partial likelihood $$L(\beta) = \prod_{i \,:\, \delta_i = 1} \frac{e^{\beta^T x_i}}{\sum_{j \,:\, \tilde t_j \ge \tilde t_i} e^{\beta^T x_j}}.$$

Example: the Weibull model with shape $k > 0$ and hazard $h(t \mid x) = k t^{k-1} \lambda^{-k} e^{\beta^T x}$ is a proportional hazards model with $h_0(t) = k t^{k-1} \lambda^{-k}$.

Non-example: if a treatment has hazard $h_1(t) = 2t$ and the control has $h_0(t) = 1$, the ratio $2t$ changes with time and the curves cross at $t = 1/2$. No constant $\beta$ satisfies $h_1 = h_0 e^{\beta}$, so these two groups do not follow a proportional hazards model.