How to interpret the hazard function and its integral?

In summary: G_A(t) = k⋅ln|F(t)-1|##If you would like, you can also solve for ##G_A(t)## by integrating ##G(t)##:$$G_A(t) = k⋅ln|F(t)-1|$$
  • #1
Eclair_de_XII
1,083
91

Homework Statement


"Suppose that the accident rate for one workplace ##A## is ##k## times the rate of another workplace ##B##. In other words, ##\lambda_A(t)=k⋅\lambda_B(t)##. Conclude that the probability of no accidents in workplace ##A## is the probability of no accidents in workplace ##B## to the ##k##-th power.

In other words, if ##\lambda_A(t)=k⋅\lambda_B(t)##, then ##P(X^c|A)=P(X^c|B)^k## where ##X## denotes the event of an accident."

Homework Equations


Hazard function: ##\lambda(t)=\frac{F'(t)}{1-F(t)}##
-denotes the probability of an accident happening on ##t+\delta t## given no accidents have happened by ##t##

The Attempt at a Solution


I'm having trouble interpreting this function, honestly. I mean, what do I get if I integrate ##\lambda(t)##? Is it not just the probability of an accident happening between some initial time ##t_0## and some final time ##t_f## for either workplace? What I got, through integration, was:

(Assuming ##\lambda## is constant)

##\lambda_B(t)=\frac{F'(t)}{1-F(t)}##
##\lambda_Bt=\frac{F'(t)}{1-F(t)}dt##
##\lambda_Bt=\int_{t_0}^{t_f} \frac{F'(t)}{1-F(t)}dt=P(X|B)##

##P(X|B)=-\int_{t_0}^{t_f} \frac{F'(t)}{F(t)-1}dt=-\int_{t_0}^{t_f} \frac{F'(t)}{F(t)-1}dt=-ln|F(t)-1|=ln|F(t_0)-1|-ln|F(t_f)-1|=ln|\frac{F(t_0)-1}{F(t_f)-1}|##

Then what I get for ##P(X|A)## is:

##P(X|A)=-k⋅\int_{t_0}^{t_f} \frac{F'(t)}{F(t)-1}dt=-k⋅ln|F(t)-1|=k⋅[ln|F(t_0)-1|-ln|F(t_f)-1|]=k⋅ln|\frac{F(t_0)-1}{F(t_f)-1}|=ln|\frac{F(t_0)-1}{F(t_f)-1}|^k=P(X|B)^k##

I'm confused on how ##\lambda(t)## works... Can anyone tell me how to interpret its integral? Can anyone tell me if I integrated the wrong function?
 
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  • #2
Can you start by telling us what ##F(t)## and ##F'(t)## are? I think CDF and PDF, but of what?

Given the ##\lambda##'s in here, I have a strong suspicion we're talking about Poissons and exponential inter-arrivals, but maybe this is about more general renewal processes or something completely different... To the extent its Poisson related there should be some simplifications that make this a lot easier.
 
  • #3
StoneTemplePython said:
Can you start by telling us what ##F(t)## and ##F'(t)## are?

Well, ##F(t)## is the c.d.f. describing the probability of having an accident before time ##t##. So... ##F(t)=Pr(T \leq t)## and ##F'(t)## is the derivative of that. So ##F(t)## is the area under ##F'(t)## but above ##F(t)=0##.
 
  • #4
Eclair_de_XII said:

Homework Statement


"Suppose that the accident rate for one workplace ##A## is ##k## times the rate of another workplace ##B##. In other words, ##\lambda_A(t)=k⋅\lambda_B(t)##. Conclude that the probability of no accidents in workplace ##A## is the probability of no accidents in workplace ##B## to the ##k##-th power.

In other words, if ##\lambda_A(t)=k⋅\lambda_B(t)##, then ##P(X^c|A)=P(X^c|B)^k## where ##X## denotes the event of an accident."

Homework Equations


Hazard function: ##\lambda(t)=\frac{F'(t)}{1-F(t)}##
-denotes the probability of an accident happening on ##t+\delta t## given no accidents have happened by ##t##

The Attempt at a Solution


I'm having trouble interpreting this function, honestly. I mean, what do I get if I integrate ##\lambda(t)##? Is it not just the probability of an accident happening between some initial time ##t_0## and some final time ##t_f## for either workplace? What I got, through integration, was:

(Assuming ##\lambda## is constant)

##\lambda_B(t)=\frac{F'(t)}{1-F(t)}##
##\lambda_Bt=\frac{F'(t)}{1-F(t)}dt##
##\lambda_Bt=\int_{t_0}^{t_f} \frac{F'(t)}{1-F(t)}dt=P(X|B)##

##P(X|B)=-\int_{t_0}^{t_f} \frac{F'(t)}{F(t)-1}dt=-\int_{t_0}^{t_f} \frac{F'(t)}{F(t)-1}dt=-ln|F(t)-1|=ln|F(t_0)-1|-ln|F(t_f)-1|=ln|\frac{F(t_0)-1}{F(t_f)-1}|##

Then what I get for ##P(X|A)## is:

##P(X|A)=-k⋅\int_{t_0}^{t_f} \frac{F'(t)}{F(t)-1}dt=-k⋅ln|F(t)-1|=k⋅[ln|F(t_0)-1|-ln|F(t_f)-1|]=k⋅ln|\frac{F(t_0)-1}{F(t_f)-1}|=ln|\frac{F(t_0)-1}{F(t_f)-1}|^k=P(X|B)^k##

I'm confused on how ##\lambda(t)## works... Can anyone tell me how to interpret its integral? Can anyone tell me if I integrated the wrong function?

Note that
$$\lambda(t) = -\frac{G'(t)}{G(t)},$$
where ##G(t) = P(T > t) = 1 - F(t)## (for a lifetime ##T## with hazard function ##\lambda(t)##).
Thus,
$$G(t) = \exp \left(-\int_0^t \lambda(s) \, ds \right). $$
Now just compare ##G_A(t)## with ##G_B(t)##
 
  • #5
Thanks, I figured it out.
 

1. What is the hazard function and why is it important to interpret?

The hazard function is a fundamental concept in survival analysis, which is used to model and understand the probability of an event occurring at a specific time. It is important to interpret because it helps to identify and quantify the risk of an event happening over time, which can be useful in various fields such as medicine, engineering, and social sciences.

2. How is the hazard function related to the survival function?

The hazard function and the survival function are complementary to each other. While the survival function tells us the probability of an event not occurring until a specific time, the hazard function tells us the probability of an event occurring at a specific time. In other words, the hazard function is the instantaneous risk of an event happening at a particular time, given that it has not occurred yet.

3. What does the integral of the hazard function represent?

The integral of the hazard function represents the cumulative risk of an event occurring up to a specific time. It is also known as the cumulative hazard function and is often used to estimate the survival function, which gives the probability of an event not occurring until a specific time.

4. How can the hazard function be interpreted graphically?

The hazard function can be interpreted graphically by plotting it against time. This results in a curve that shows the instantaneous risk of an event happening at different points in time. The shape of the curve can provide valuable insights into the nature of the event and its risk over time.

5. How can the hazard function be used in practical applications?

The hazard function has various practical applications, such as in predicting the failure rate of a product, estimating the time to event in survival analysis, and analyzing the risk of developing a disease. It can also be used to compare the risk of an event between different groups or populations and to identify potential risk factors that may influence the hazard function.

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