According to this wiki link, the hazard ratio relates to survival according to the following equation
$$
S_1(t)=S_0(t)^r \quad (1)
$$
where $S$ is the survival and $r$ is the hazard ratio. So from the equation (1) above we can solve for the hazard ratio
$$
r=\frac{ln(S_1(t))}{ln(S_0(t))}=\frac{H_1(t)}{H_0(t)}=\frac{h_1(t)}{h_0(t)}
$$
Where $h$, and $H$ is the hazard rate and cumulative function respectively.
Furthermore, according to the wiki the hazard ratio can be calculated with the odds probability formula
$$
HR=\frac{p}{1-p} \quad (2)
$$
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Is $r$=$HR$? I don't understand the change in notation.
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The relation $S_1(t)=S_0(t)^r$ – is it valid only for Cox
Proportional Hazard or is it a generalization for the vanilla
survival analysis as well? -
How do I interpret $p$ in equation (2)? What is it? A simple example
would help
Best Answer
$S(t|x_{1})=\exp(-\Lambda(t|x_{1}))$ where $\Lambda(t|x_{1})=\int_{0}^{t}\lambda(u|x_{1})du$, here $x_{1}=1$ for group 1 and $0$ if not. For a hazard function modlled as $\lambda(t|x_{1})=\lambda_{0}(t)\exp(\beta_{1}x_{1})$ the hazard ratio $r$ is defined as
\begin{align*} r &= \frac{\lambda(t|x_{1}=1)}{\lambda(t|x_{1}=0)}\\ &=\frac{\lambda_{0}(t)\exp(\beta_{1})}{\lambda_{0}(t)\exp(0)}\\ &=\frac{\exp(\beta_{1})}{1}\\ &=\exp(\beta_{1}) \end{align*}
Thus
\begin{align*} \Lambda(t|x_{1}=1)&=\int_{0}^{t}\lambda(u|x_{1}=1)du\\ &=\int_{0}^{t}\lambda_{0}(t)rdu\\ &=r\int_{0}^{t}\lambda(u|x_{1}=0)du\\ &=r\Lambda(t|x_{1}=0) \end{align*}
Accordingly
\begin{align*} S(t|x_{1}=1)&=\exp(-\Lambda(t|x_{1}=1))\\ &=\exp(-r\Lambda(t|x_{1}=0))\\ &=\frac{1}{r\exp(\Lambda(t|x_{1}=0))}\\ &=\left[\frac{1}{\exp(\Lambda(t|x_{1}=0))}\right]^{r}\\ &=\left[\exp(-\Lambda(t|x_{1}=0))\right]^{r}\\ &=\left[S(t|x_{1}=0)\right]^{r}\\ \end{align*}
So yes $r=HR$. The deriavation of $S(t|x_{1}=1)=\left[S(t|x_{1}=0)\right]^{r}$ above depedned on the defintion of the HR and the model form $\lambda(t|x_{1})=\lambda_{0}(t)\exp(\beta_{1}x_{1})$. Since no time-varying covariates are in the exponential term the HR is constant over time - i.e. proportional hazards since $\lambda(t|x_{1}=1)=r\lambda(t|x_{1}=0)$ shows hazard of $\lambda(t|x_{1}=1)$ is proportional to hazard of $\lambda(t|x_{1}=0)$ with time constant multiplicative factor $r=\exp(\beta_{1})$. Finally noting that $Pr[T>t|x_{1}]=S(t|x_{1})$
\begin{align*} Odds(x_{1}=1)&=\frac{Pr[T>t|x_{1}=1]}{1-Pr[T>t|x_{1}=0]}\\ &=\frac{S(t|x_{1}=1)}{1-S(t|x_{1}=1)}\\ &=\frac{\exp(-r\Lambda(t|x_{1}=0))}{1-\exp(-r\Lambda(t|x_{1}=0))}\\ &=\frac{1}{\exp(r\Lambda(t|x_{1}=0))-1}\\ &=\frac{2}{\exp(r\Lambda(t|x_{1}=0))} \end{align*}
and
\begin{align*} Odds(x_{1}=0)&=\frac{1}{\exp(\Lambda(t|x_{1}=0))-1}\\ &=\frac{2}{\exp(\Lambda(t|x_{1}=0))} \end{align*}
So
\begin{align*} \frac{Odds(x_{1}=1)}{Odds(x_{1}=0)}&=\frac{\exp(\Lambda(t|x_{1}=0))}{\exp(r\Lambda(t|x_{1}=0))}\\ &=\exp[\Lambda(t|x_{1}=0)(1-r)]\\ &=\exp[-\Lambda(t|x_{1}=0)(r-1)] \end{align*}
This implies
\begin{align*} r=\frac{\Lambda(t|x_{1}=0)-\log(OR)}{\Lambda(t|x_{1}=0)} \end{align*}
Edit: I am not so sure about $r=p/1-p$ for some $p$, since if $p$ is a probability this looks like an odds rather than an odds ratio (OR). The only thing I can think of is the following: assuming in the expression $\exp[-\Lambda(t|x_{1}=0)(r-1)]$ that $\Lambda(t|x_{1}=0)(r-1)$ is "small" then using $e^{x}\approx 1-x$ then $\exp[-\Lambda(t|x_{1}=0)(r-1)]\approx 1-\Lambda(t|x_{1}=0)(r-1)$ so that
\begin{align*} OR &\approx 1-\Lambda(t|x_{1}=0)(r-1)\\ \Longleftrightarrow & r = \frac{\Lambda(t|x_{1}=0)+1-OR}{\Lambda(t|x_{1}=0)} \end{align*}
We see the above is equal to $p/1-p$ for $p=\Lambda(t|x_{1}=0)+1$ if $OR=1$. Thus if $\Lambda(t|x_{1}=0)(r-1)$ is "small" and OR=1 then the HR is just the odds of an event for group 0.