How to find limit of hazard function

1. Mar 1, 2013

rawand

lim(∫λu du)

when limit T tend to infinity and ∫ between 0 to t

2. Mar 1, 2013

mathman

You need to clarify. T, t, λ.

3. Mar 2, 2013

rawand

FR=λ(t)
λ(t)>= 0
whay
lim∫λ(u)du→∞
not:lim(t → ∞)
∫ btween(0,t)

plz i have test
thanks

4. Mar 2, 2013

HallsofIvy

That makes it even more non-understandable. At first you had $\int \lambda u du$ but then you say "$FR= \lambda(t)$. You hadn't mentioned "FR" before and now it looks like $\lambda$ is a function of t (by which I guess you mean "T", the upper limit in the original integral.

What you first posted, $\lim_{T\to\infty}\int_0^T \lambda u du$ is fairly easy.
$$\lambda\int_0^T udu= \lambda T^2/2$$.
That limit, as T goes to infinity, does not exist. (Or the limit is "infinity" which just says it does not exist for a particular reason- it grows without bound.)

The second one you posted depends upon exactly what function of T $\lambda$ is.

5. Mar 2, 2013

chiro

Hey rawand and welcome to the forums.

Just out of curiosity, does T refer to the resolution of the hazard function? (Higher t means more values between 0 and t)