It was the novel, rather than the lecture, that got me to the finish line



Pictured : Lago Maggiore, Locarno, Switzerland (2024)

A mathematical proof of \(S_{A}(t) = S_{B}(t)^{HR}\), the Proportional hazards assumption.

Under the Proportional hazards assumption, any ratios of hazard functions at any one time are constant.

I provide an example assuming we are comparing the efficacy of two treatment groups denoted by, \(i\):

Other important notations include :

The proportional hazards assumption is such that the hazard functions for any individual is a fixed proportion of hazard to any other covariate.

Since we are interested in proofing this equation (1) :

\[ S_{A}(t) = S_{B}(t)^{HR} \] We denote the following as equation (2) : \[ exp^{(-H_{A}(t))} = exp^{(-H_{B}(t))} {.} \space {HR} \]

We know the following, that the Cumulative hazard function is the negative log of the Survival function :

\[ H_{i}(t) = -logS(t) \] We also know that the exponential of the negative Cumulative hazard function is the Survival function : \[ S(t) = exp(-H_{i}(t)) \] Via the law of calculus, it is true that the integral of the hazard function (density) is the cumulative hazard function with respect to time, \(H_{i}(t)\): \[ H_{i}(t) = -\int h_{i}(u) \space du \]

Therefore, rewriting (2) we obtain :

\[ \frac {exp^{(-H_{A}(t))}} {exp^{(-H_{B}(t))}} = {HR} = \frac {exp(log S(t))} {exp(log{S(t))}} \] Which can be simplified to equation (3), two minus signs in the former:

\[ exp^{-\int h_{A}(u) \space du - \space(-\int h_{B}(u) \space du \space) } = \frac {exp^{-\int h_{A}(u) \space du} } {exp^{-\int h_{B}(u) \space du }} = HR \]

which resembles the hazard function with covariates :

\[ h_{A}(t) = h_{B}(t) \space {.} \space exp^{(\vec{X\space }^{T} \vec{B})} = exp^{-\int h_{A}(u) \space du - \space(-\int h_{B}(u) \space du \space) } \] which is simplified to :

\[ \frac {h_{A}(t)} {h_{B}(t)} = exp^{(\vec{X\space }^{T} \vec{B})} = exp^{-\int h_{A}(u) \space du - \space(-\int h_{B}(u) \space du \space) } \]

Where we fit a proportional hazards model :

\[ \frac {exp^{-log(S_{A}(t))}} {exp^{-log(-S_{B}(t))}} = \frac {h_{A}(t)} {h_{B}(t)} = exp^{(\vec{X\space }^{T} \vec{B})} \]

We recognise this result with equation (3).

This concludes the proof.

References :