How to find limit of hazard function

[rawand]

lim(∫λu du)

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

 

[mathman]

You need to clarify. T, t, λ.

 

[rawand]

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

please i have test
thanks

 

[HallsofIvy]

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

What you first posted, [itex]\lim_{T\to\infty}\int_0^T \lambda u du[/itex] is fairly easy.
[tex]\lambda\int_0^T udu= \lambda T^2/2[/tex].
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 [itex]\lambda[/itex] is.

 

[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)