# Connection between Hazard Ratios, Survival, and Probability

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

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

• Thank you so much! One small follow up - is there an easy way to calculate the cumulative hazard at $x_1=0$ – Edv Beq Feb 15 at 17:04
• 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? – dandar Feb 15 at 20:12
• Yes - $\lambda_0$ is unknown isn't it? Is there a way to calculate that with odds ratios, or bayes rule, etc? – Edv Beq Feb 15 at 20:15
• 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. – dandar Feb 16 at 0:48
• 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 – dandar Feb 16 at 0:58