Deriving the Distribution of Total Time in a Poisson Process Series

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SUMMARY

The discussion focuses on deriving the distribution of total time in a Poisson process series when traveling from point A to C via B. The time taken for each segment is modeled using exponential distributions: Pab(t1) = k1exp(-k1t1) for A to B and Pbc(t2) = k2exp(-k2t2) for B to C. The total time t is expressed as t = t1 + t2, leading to the proposed distribution Pac(t). The correct formulation for Pac(t) is confirmed as Pac(t) = k1k2exp(-k1t1-k2t2), aligning with the properties of independent Poisson processes. The mean of this distribution can be derived 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(t1) = k2exp(-k2t2)
2So 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 the distribution 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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