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I'm now learning the very basics of survival analysis.

Firstly, the probability function is introduced:

$$f(t)=\frac{dF(t)}{dt}=-\frac{dS(t)}{dt}$$

Where $S(t)=1-F(t)$.

We are then introduced to the hazard function:

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

Subsequently, we are told that combining the previous two expressions yield:

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

How do I justify this last step, shouldn't it be something like:

$$h(t)=\frac{-\frac{dS(t)}{dt}}{S(t)}$$

??

Jyrki Lahtonen
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Magnus
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  • Indeed. But these expressions are equivalent ! (review the chain rule). –  Feb 19 '18 at 11:48
  • This had nothing whatsoeve to do with [tag:division-algebras]. Do you pick your tags randomly rather than study what they mean? – Jyrki Lahtonen Feb 21 '18 at 21:38

2 Answers2

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Simply evaluate the derivative to obtain: $$h(t) = -\frac{d}{dt}\log(S(t)) = -\frac{1}{S(t)}\frac{dS}{dt} = \frac{f(t)}{S(t)}$$

bames
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  • What confuses me is the log I think, how did "that" find its way into our equation? – Magnus Feb 19 '18 at 12:26
  • @Magnus I suppose that whoever wrote that wants to use the integral of the hazard function somewhere, so it was convenient to express $h(t)$ as a derivative. To see where the $\log$ comes from, just integrate your expression for $h$ (which was $-\frac{1}{S(t)} \frac{dS}{dt}$). You remember that $\int \frac{1}{u}\frac{du}{dt};dt = \log(u)$, right? – bames Feb 19 '18 at 12:32
  • .....no that rule is new to me. Does it have a name? – Magnus Feb 19 '18 at 12:37
  • @Magnus I don't believe it has a name. See this. Or this, if you are interested in where it came from. – bames Feb 19 '18 at 12:40
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From your first equation

$$ \frac{{\rm d}S}{{\rm d}t} = -\frac{{\rm d}F}{{\rm d}t} = -f(t) \tag{1} $$

So that

$$ h(t) = -\frac{{\rm d}}{{\rm d}t} \log S(t) = -\frac{1}{S(t)}\frac{{\rm d} S}{{\rm d} t} = -\frac{1}{S(t)}[-f(t)] $$

caverac
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