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 \(f(t) = \lambda \exp(-\lambda t)\) .