How to interpret the hazard function and its integral?

[Eclair_de_XII]

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?

 

[StoneTemplePython]

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.

 

[Eclair_de_XII]

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 $F(t)=0$.

 

[Ray Vickson]

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 $G_B(t)$

 

[Eclair_de_XII]

Thanks, I figured it out.