Deriving the Hazard Rate Formula for F(t) = 1-Exp -((t-γ)/n))^β

[Buchanskii]

F(t) = 1-Exp -((t-γ)/n))^β

f(t) = dF(t)/dt = Exp/n[(t-x)/n]^β-1

h(t) = f(t)/1-F(t)

h(t)= β(t-y)^β-1/n^β



The final answer: h(t)= β(t-y)β-1/n^β, Is the correct answer.

But, I can't for the life of me work out why. Have I made a mistake in the f(t) derivation.

Can anyone help?

 

[lanedance]

its a bit hard to read, but you derivative doesn't look right

so let's start with the chain rule
$$\frac{d}{dt} g(h(t))= g'(h(t))h'(t)$$

applying to our case
$$F(t) = 1-e^{-(\frac{t-γ}{n})^\beta}$$

so let
$$g(x) = 1-e^{x}$$
$$g'(x) = e^{x}$$
$$h(t) = -(\frac{t-γ}{n})^\beta$$
$$h'(t) = -\frac{d}{dt}(\frac{t-γ}{n})^\beta$$


which gives
$$f(t) = \frac{F(t)}{dt} = e^{-(\frac{t-γ}{n})^\beta}(-\frac{d}{dt}(\frac{t-γ}{n})^\beta)$$