How to Derive the Distribution of Two Poisson Processes in Series?

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SUMMARY

The discussion focuses on deriving the distribution of the total time, t, taken to travel from point A to C via point B, where both segments follow Poisson distributions. The distributions are defined as Pab(t1) = k1exp(-k1t1) for A to B and Pbc(t2) = k2exp(-k2t2) for B to C. The correct approach to find the total distribution Pac(t) is to use the convolution of the two distributions, leading to Pac(t) = k1k2exp(-k1t1-k2t2). The mean of this distribution can be expressed in terms of k1 and k2.

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Homework Statement



Imagine you want to go from A to C via B. So you have two steps: A to B and B to C. Let's assume the time taken (t1) to go from A to B is Poisson and is given by Pab(t1) and the for B to C is t2 and the distribution is Pbc(t2). You are given:

Pab(t1) = k1exp(-k1t1) and Pbc(t2) = k2exp(-k2t1)

So the total time to go from A to C is t = t1 + t2

Derive the distribution of t, that is, find Pac(t). Also find the mean of this distribution in terms of k1 and k2.

Homework Equations



We can use the basic rules of probability,

The Attempt at a Solution



Is it as simple as this?:

Pac(t) = k1Pab(t1) + k2Pbc(t2)

or is it:

Pac(t) = k1k2exp(-k1t1-k2t2)
 
Last edited:
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I am leaning towards the second equation I wrote above, as, if you take the case that k1 = k2 = 1, then the resulting equation is that of a normal poisson with t = t1 + t2.

I only want to know if this thinking is correct, I know how to continue from there.
 

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