5.3.4 problem 47
a. Let’s develop the integral ∫t0h(s)ds
∫t0h(s)ds=∫t0f(s)1−F(s)ds=∫t0f(s)G(s)ds
where G(t)=1−F(t) is
the survival function.
Doing the substitution v=logG(s),
we have
dv=G'(s)G(s)ds=−F'(s)G(s)ds=−f(s)G(s)ds
With the substitution, the lower limit of integration becomes
logG(0)=log1=0, and the upper
limit becomes logG(t).
Therefore,
|
∫t0h(s)ds=∫logG(t)0−dv=−logG(t) | (5.45) |
logG(t)=−∫t0h(s)ds
G(t)=1−F(t)=exp(−∫t0h(s)ds)
F(t)=1−exp(−∫t0h(s)ds) , for all t>0
b. The PDF is the derivative of the CDF
f(t)=F'(t)=[1−exp(−∫t0h(s)ds)]'
|
f(t)=−exp(−∫t0h(s)ds)⋅(−∫t0h(s)ds)' | (5.46) |
To calculate the derivative of the integral in the relation above, let’s apply the
differential in both sides of equation (5.45)
(∫t0h(s)ds)'=(−logG(t))'=−G'(t)G(t)=(F(t)−1)'G(t)=f(t)1−F(t)=h(t)
Substituting this result in equation (5.46)
f(t)=−exp(−∫t0h(s)ds)⋅(−h(t))
f(t)=h(t)exp(−∫t0h(s)ds) , for all t>0