How to find limit of hazard function

AI Thread Summary
The discussion centers on finding the limit of the hazard function expressed as lim(∫λu du) as T approaches infinity. Participants clarify the variables involved, specifically T, t, and λ, with some confusion regarding the notation and definitions. The integral ∫ from 0 to T of λu du is noted to yield λT²/2, which diverges as T approaches infinity, indicating that the limit does not exist. There is also a question about whether T relates to the resolution of the hazard function, suggesting that a higher t allows for more values between 0 and t. Overall, the conversation highlights the complexities in defining and calculating limits in the context of hazard functions.
rawand
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lim(∫λu du)

when limit T tend to infinity and ∫ between 0 to t
 
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You need to clarify. T, t, λ.
 
FR=λ(t)
λ(t)>= 0
whay
lim∫λ(u)du→∞
not:lim(t → ∞)
∫ btween(0,t)

please i have test
thanks
 
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.
 
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)
 
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