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

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To derive the distribution of the total time t from A to C via B, the user considers two Poisson processes: Pab(t1) for A to B and Pbc(t2) for B to C. The proposed equations for the total distribution Pac(t) are either a simple addition of the individual distributions or a combined exponential form. The user favors the second equation, suggesting it aligns with the characteristics of a Poisson process when k1 and k2 equal 1. The discussion seeks confirmation on the correctness of this approach before proceeding further.
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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)
 
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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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