# Poisson probability

1. Jun 1, 2014

### aaaa202

Suppose that a system is such that in a time dt, the probability that an event A occurs, given that it has not already happened, is given by:
P(t,t+dt) = w(t) * dt

The solution for the probability that A has occured at a time t is something like:

P(t) = 1 - exp(∫0tw('t)dt')

Now suppose that w(t) is changing due to some function x such that really w(t) = w(x(t)). How do I find the probability P(x) that the event A has occured as a function of x?

2. Jun 1, 2014

### HallsofIvy

Staff Emeritus
The chain rule (or substitution for integrals):
$$\int w(t')dt'= \int w(x(t'))\left(\frac{dt'}{dx}\right)\left(\frac{dx}{dt'}\right) dt'= \int w(x)\left(\frac{dt'}{dx}\right)dx= \int_{x(0)}^{x(t)} \frac{w(x)}{\frac{dx}{dt'}}dx$$

3. Jun 1, 2014

### aaaa202

But doesn't that still give me P(t) when I carry out the integration? My problem is that I have 3 different x1(t), x2(t), x3(t) and I want P(xi) for each of these so I can see when the probability in each case becomes significant. So in the above formula when I plug in x(t) in the end I still get something which is controlled by t. Where am I confusing myself?
I also find it weird that if w(x) is an increasing function then the probability P(x) takes longer to reach a significant value when dx/dt is large. Intuitively I would think that if x(t) is driven to large values faster such that w(x) gets larger faster then the probability P(x) should do the same. Where am I wrong?

Last edited: Jun 1, 2014
4. Jun 1, 2014

### Ray Vickson

As I read it, you have three "curves $x_1(t), x_2(t), x_3(t)$ and for each of them you have a (non-homogeneous) Poisson process with rate $r_i(t) = w[x_i(t)]$. That will give you three different "counting processes" $N_1(t), N_2(t), N_3(t)$ with
$$P\{ N_i(t) = n \} = \frac{m_i(t)^n \, e^{-m_i(t)}}{n!}, \, n = 0,1,2,\ldots$$
and where
$$m_i(t) = \int_0^t w[x_i(\tau)] \, d \tau , i = 1,2,3.$$
That seems to be what you are saying; if it is not, you need to clarify what you want.