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This post was helpful in explaining the intuition for the hazard function. In it, the hazard function is explained to be

$$h(t) = \frac{f(t)}{S(t)},$$

where $f(t)$ is the PDF, $S(t)$ is the survival function. This leads to the identity

$$h(t) = -\frac{d}{dt} \log (S(t)). $$

Is there any intuition to be gained from it? The hazard function is the instantaneous rate of death at time t given that the patient survived to time t. It is surprising to me that somehow that equals the rate of change of the log-transform of the survival function.

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  • $\begingroup$ It is not only the effect of "log-transform", but also the differential operator! To me, the second form is just a math equivalence (perhaps easier to remember) to the first form, whose intuition is as yourself stated. $\endgroup$ Commented Feb 11 at 2:24
  • $\begingroup$ By the way, the expression you listed had an obvious typo -- there should not be a negative sign in any of expressions. $\endgroup$ Commented Feb 11 at 2:27

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To facilitate analysis, denote the rate-of-change of $S$ over the interval $[t,t+\Delta]$ as:

$$S'_\Delta(t) \equiv \frac{S(t+\Delta) - S(t)}{\Delta},$$

and observe that this rate-of-change becomes the derivative $S'(t)$ as $\Delta \downarrow 0$. Now, conditional on surviving up to time $t$, the probability of failure within a period $\Delta > 0$, expressed as a rate over that period, is given by:

$$\begin{align} \frac{\mathbb{P}(t < T \leqslant t+\Delta | T > t)}{\Delta} &= \frac{1}{\Delta} \frac{\mathbb{P}(t < T \leqslant t+\Delta)}{\mathbb{P}(T > t)} \\[6pt] &= \frac{1}{\Delta} \frac{S(t) - S(t+\Delta)}{S(t)} \\[6pt] &= - \frac{1}{\Delta} \frac{S(t+\Delta) - S(t)}{S(t)} \\[6pt] &= - \frac{1}{S(t)} \frac{S(t+\Delta) - S(t)}{\Delta} \\[6pt] &= - \frac{S'_\Delta(t)}{S(t)} \\[6pt] &\approx - \frac{S'(t)}{S(t)} \\[6pt] &= - \frac{d}{dt} \log(S(t)). \\[6pt] \end{align}$$

Taking $\Delta \downarrow 0$ then makes this approximation exact and you have the hazard function at time $t$. As to the intuition of this result, it mostly comes from the fact that the conditioning in the probability gives a denominator $S(t)$, which is the term that leads to the logarithmic derivative.

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    $\begingroup$ This is great. Because the conditional probability reduces to scaling by S(t), the hazard function is just the derivative of S(t) on a relative (multiplicative) scale. Taking the log of S(t) transforms it to that relative scale. $\endgroup$ Commented Feb 11 at 10:52
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An alternative way is to see the survival as an exponential function $$S(t) = e^{-\int_0^t h(\tau) d\tau}$$

this expression occurs in What distribution to use to model time before a train arrives?

Intuitively you can see it as an exponentially declining function where the rate in time is variable and the total hazard needs to be integrated.

Another intuition about this exponential term would be to consider a game with dice rolls or some other probability, where we roll a dice untill some event occurs (and we consider survival as the event not having occured yet).

The probability of survival after $x$ dice rolls with $p$ probabability survival per roll is an exponential function $$P\left(\text{survive} | \text{rolls}=x\right) = p^x$$ and we could consider the number of dice rolls as a function of time $x(t)$

$$S(t) = p^{x(t)}$$

in addition this number could be a sum of dice rolls $n_k$ that occured every $k$-th second

$$S(t) = p^{\sum_{k = 1}^t n_k}$$

Then the logaritm is

$$\log\left[S(t)\right] \propto \sum_{k = 1}^t n_k$$ and expresses the number of dice rolls. The hazard is the difference or derivative, the number of dice rolls that we had in a particular second.

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