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, \(i\):
Treatment A, denoted as A, has survival curve \(S_{A}(t)\)
Treatment B, denoted as B, has survival curve \(S_{B}(t)\)
Other important notations include :
Hazard rate \(h_{A}(t)\) with dependancy to time \(t\)
Cumulative Hazard function \(H_{A}(t)\), equivalent to \(\int h_{A}(u) du\)
Hazard ratio \(HR\)
multiplication “\({.}\)”
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.