11.11 Hazard Function

• The hazard function is the risk of having an event given you survived up to time t.

$$h(t) = \lim_{\Delta t \to 0} \frac{ Pr(t+ \Delta t \geq T > t | T > t)}{\Delta t}$$

Why do we care about the hazard function? It turns out that a key approach for modeling survival data as a function of covariates (i.e., regressors) relies heavily on the hazard function.

11.11.1 How is the hazard rate related to the survival probability?

Define the event (‘death’) rate:

$$f(t) = \lim_{\Delta t \to 0}\frac{ Pr(t+ \Delta t \geq T > t)}{\Delta t}$$

• $f(t)\Delta t$ is the probability of the event occurring near t, not conditioned like $h$.

• the probability of an event near (i.e. within $\Delta t$) time $t$ should be equal to the probability of surviving until time $t$ multiplied by the probability of an event near time $t$ given that you have survived until time $t$. That is:

$$f(t) = h(t)S(t)$$

The book gives a more rigorous derivation using Baye’s rule, but this an intuitive way to get there.

Now consider that if $S(t)$ is the survival probability, then $1-S(t)$ is the probability of the event occurring by time t. And intuitively we would expect that $1-S(t) = \int_{0}^{t}f(u)du$, i.e. the probability of the event occuring by time t is just the sum of the probabilities of all the times the event could have occurred.

Verification:

Note the following is basically Excercise 8. Skip it if you have not done this yet!

$$\begin{align} \frac{d S}{d t} &= lim_{\Delta t \to 0} \frac{S(t+dt) - S(t)}{\Delta t} \\ & = lim_{\Delta t \to 0} \frac{Pr(T>t+\Delta t) - Pr(T > t)}{\Delta t}\\ & = lim_{\Delta t \to 0} \frac{Pr(T>t+\Delta t) - (Pr(T > t + \Delta t) + Pr(t+ \Delta t \geq T > t))}{\Delta t} \\ & = lim_{\Delta t \to 0} \frac{- Pr(t+ \Delta t \geq T > t)}{\Delta t} \\ & = - f(t) \end{align}$$

• So we have finally:

$$\frac{d S}{d t} = - S(t)h(t)\\ \text{or}\\ \frac{d \text{ ln} S}{d t} = -h(t)$$

• log of the survival probability is the negative of the ‘cumulative’ hazard ($\int{h(t) dt}$)

$$\ln S = -\int{h(t) dt}$$

Note that this construction is similar to that of the Kaplan-Meier curve $\ln\hat{S}(d_k) = \sum_{j=1}^{k} \ln (\frac{r_j- q_j}{r_j}) \approx \sum_{j=1}^{k} (-q_j/r_j)$ where $q_i/r_j$ is the hazard at time $q_i$

Example: consider a constant hazard (e.g. radioactive decay) $h=\lambda$ (i.e. at any moment, the chance of an event is constant given no event up to that moment). In that case the survival is exponential, as we expect, $S(t) = \exp(-\lambda t)$ and $f(t) = \lambda \exp(-\lambda t)$ .