How do I calculate the hazard function from the survival rate?

Question

I'm trying to calculate the hazard function for a type of mechanical component, given a dataset with the start and failure times of each component. In the dataset, all components eventually fail.

First, I calculated the percentage of components that survived greater than 3 months as $s = 0.458$.

Second, I calculated the hazard function, assuming that the hazard function for failure is constant over each period, like this:

\begin{align} h(t) &= (1-s)^{t} \\ &= (1-0.458)^3 \\ &= 0.15922 \end{align}

where $t$ is the length of each period (in this case, 3 months, although I'm unsure about this).

However, the notes I'm following calculate the hazard function like this:

\begin{align} h(t) &= 1-0.458^{1/3} \\ &= 0.2292 \end{align}

The lack of parentheses and the different power confuse me. Am I, my notes, or both of us incorrect?

Answer 1

It seems that you have "grouped data", where evaluations are no longer made continuously but rather take place at fixed time points. Consider the picture below. The time axis is divided into $3$ intervals, $I_1$, $I_2$, and $I_3$.

Under the assumptions that

\begin{align} \Pr[T > 1] & = p^{\tfrac{1}{3}} \\ \Pr[T > 2] & = \Pr[T > 2 \,|\, T > 1] \, \Pr[T > 1] = p^{\tfrac{1}{3}} \, p^{\tfrac{1}{3}} = p^{\tfrac{2}{3}} \\ \Pr[T > 3] & = \Pr[T > 3 \,|\, T > 2] \, \Pr[T > 2] = p^{\tfrac{1}{3}} \, p^{\tfrac{2}{3}} = p \end{align}

the hazard to have the event in $I_2$, say, is then given by

$$ h(I_2) = 1 - \Pr[T > 2 \,|\, T > 1] = 1 - p^{\tfrac{1}{3}}. $$

Answer 2

Let the positive continuous random variable $X$ denote the time of failure of a component. The survival rate $s(t)$ at time $t = T$ is related to the hazard rate $h(t)$ via $$s(T) = P\{X > T\} = \exp\left(-\int_0^T h(t)\,\mathrm dt\right)$$ where the integral is, of course, the area under the curve $h(t)$ from $0$ up to $T$. If $h(t)$ has constant value $a$ on this interval, then $$s(T) = \exp(-aT) ~ \Rightarrow ~ a = \frac{1}{T}\left(-\ln s(T)\right).$$ Your notes seem to be defining the hazard function as the probability of failing in one month, that is, $P\{X \leq 1\}$ with time measured in months. This is easily calculated as $1-s(1) = 1 -\exp(-a)$ where we know that $s(3) = \exp(-3a) = 0.458$. This gives $$1-s(1) = 1-\exp(-a) = 1 - 0.458^{1/3}.$$

As I said in my comment, the hazard rate $h(t)$ for a continuous random variable is defined as $$h(t) = \lim_{\epsilon \downarrow 0}P\{t < X \leq t+\epsilon\mid X > t\}.$$ If the survival data is integer-valued, whether because of the model chosen or the way the experiment is conducted (e.g. we only know the percentage that survived $n$ months because the experimenter only checked once each month on how many survived), then this definition simplifies to $$\hat{h}(n) = P\{X = n \mid X > n-1\}.$$ If $\hat{h}(n)$ is assumed to have the same value $p$ for all $n\geq 1$, then $X$ is a geometric random variable with parameter $p$, and given that $s(3) = P\{X > 3\} = (1-p)^3 = 0.458$, we get that $$\hat{h}(1) = p = 1 - 0.458^{1/3}.$$ If data is available about the measured values of $s(1)$, $s(3)$, etc. the assumption about the constant value of $\hat{h}(n)$ should be cross-checked.