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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

$$ HR=\frac{p}{1-p} \quad (2) $$

  1. Is $r$=$HR$? I don't understand the change in notation.

  2. 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?

  3. How do I interpret $p$ in equation (2)? What is it? A simple example would help

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$S(t|x_{1})=\exp(-\Lambda(t|x_{1}))$ 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.

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  • $\begingroup$ Thank you so much! One small follow up - is there an easy way to calculate the cumulative hazard at $x_1=0$ $\endgroup$ – Edv Beq yesterday
  • $\begingroup$ The cumulative hazard at $x_{1}=0$ is $\Lambda(t|x_{1}=0)=\int_{0}^{t}\lambda_{0}(u)du$. Is this what you meant? $\endgroup$ – dandar 23 hours ago
  • $\begingroup$ Yes - $\lambda_0$ is unknown isn't it? Is there a way to calculate that with odds ratios, or bayes rule, etc? $\endgroup$ – Edv Beq 23 hours ago
  • $\begingroup$ The model used above is a Cox proportional Hazards model and uses partial likelihood (PL) to estimate the $\beta$ parameter, which essentially ignores the $\lambda_{0}(t)$ function since it does not affect inference about $\beta$. However I understand that the PL estimate $\hat{\beta}$ of $\beta$ is then used by "plugging " it in to $\lambda(t|x_{1})$ - i.e. $\hat{\lambda}(t|x_{1})=\lambda_{0}(t)\exp(\hat{\beta}_{1}x_{1})$ and in turn using $\hat{\lambda}(t|x_{1})$ to estimate $S(t|x_{1})$. Essentially I think this is a post-modelling estimation problem - i.e. you need not conduct this step. $\endgroup$ – dandar 18 hours ago
  • $\begingroup$ The above is called model-based estimate of the baseline hazard function - non-model based estimators include the Breslow and Nelson-Aalen estimators. So for example non-model based estimators can construct estimated survival curves without fitting a Cox PH model - i.e. KM survival method. The model fitting should permit easy adjustment for covariates, just as in regression it is easier to adjust for covariates than non-regression methods. Finally if infererence about $\beta$ is all that is required then estimating $\lambda_{0}(t)$ is not needed - however survival curves are often desired $\endgroup$ – dandar 18 hours ago

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