Property 1:
S(t) = e–H(t)
or equivalently
H(t) = −ln S(t)
Proof: Since S(t) = 1 − F(t), the derivative S′(t) = − F′(t) = − f(t). Thus
Taking the integral of both sides yields
From which we get H(t) = -ln S(t). Taking the exponential of both sides yields the first result.
Observation: Some properties of the hazard function are
Observation: If the hazard function is constant then t has an exponential distribution; i.e. a constant hazard functions corresponds to an exponentially distributed lifetime. To see this, let
Then c(1−F(t)) = F′(t), which has a solution when F(t) = 1 − ce-ct. Alternatively
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