I posted this question on Cross Validated - but I think it applies here too. Also, it increases the chances of the question being answered.
Link here
If this is not acceptable - administrators please delete, and anybody else please do not take points away from me for this reason.
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What is the difference between the "Hazard Rate" and the "Killing Function" of a diffusion model?
Some Definitions:
The Killing Function
The function k(t,x) is interpreted as the killing rate. Informally, this means that if, at time t, the particle is alive and is situated at the point x, then the probability that it dies in the next h units of time is approximately k(t,x)h when h is small.
\begin{gather*} Pr(\rho \leq t+h \mid \rho > t, X(t) = x)\approx k(t, x)h & (1) \end{gather*}
And,
\begin{gather*} dX(t) = \mu dt+\sigma dW\ & (2) \end{gather*}
Hazard Rate
\begin{gather*} Pr(t \leq T \leq t+h \mid T > t) \approx \lambda (t)h & (3) \end{gather*}
That is, λ(t)h represents the instantaneous chance that an individual will die in the interval (t, t + h) given that this individual is alive at age t.
Lastly, to put it in perspective here is a picture of a diffusion with arbitrary Killing Function k(x) = a + Sqrt(t/b), where a, and b are some constants.
I added the lines for later reference.
So, these results raise a lot more question.
How do I interpret "rho" in Equation (1) for example - if I am modeling a type of bird population for with X(t)?
How do I relate the Killing Function with the Hazard Rate?
Is it OK to say that if the f(t) is the density distribution of the First-Passage-Times (Refer to Fig-2), then the Hazard rate for the diffusion (2) is:
\begin{gather*} \lambda (t) = \frac{f(t)}{1-F(t)} & (4) \end{gather*}
If I do not know the killing function - but I observe the first passage time distribution as in Fig-2: Is it possible to solve for the Killing Function?
Lastly, in the definition k(t,x) is a function of both variables {x,t}. In the literature, most of the time is referred as k(x), which is really k(X(t)) since X() is a function of t. But if one was to actually apply it - as I did in Fig-1, say:
\begin{gather*} k(x) = b[(x(t)-a)^{2}]\ & (5) \end{gather*}
I would have to express it in terms of X(t):
\begin{gather*} X(t) = a+\sqrt{\frac{t}{b}} & (6) \end{gather*}
But X(t) is reserved for the diffusion model (2) so it makes it extra confusing.
Note: Assume OP (original poster) is very unintelligent; hence, be very specific, use simple words, do not leave any algebra out, and do not hesitate to curse me out if I wrote something stupid above.
Thank you so much in advance!
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