SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

and the latter with rate
(
λ ′′ = 1 −
(
γ2
1 − γ 1
))
λ ′
(iii) Continue by decomposing the λ ′′ process, etc.
To prove the result for superposition of m processes, superpose them two at a time:
(i) Superpose the λ 1 and λ 2 processes to obtain a Poisson process with rate λ ′ = λ 1 +λ 2 .
(ii) Superpose the λ ′ and λ 3 processes to obtain a Poisson process with rate λ ′′ =
λ ′ + λ 3 = λ 1 + λ 2 + λ 3 .
(iii) Continue
To prove the result for the superposition of m nonstationary processes, we prove that
for the superposition process
Pr{Y τ+∆τ −Y τ = h |Y τ = k}
49
= e−[Λ(τ+∆τ)−Λ(τ)] [Λ(τ +∆τ) − Λ(τ)] h
h!
For clarity we prove only the case m =2.
Let Y i,τ be the NSPP with rate Λ i (τ), for i =1, 2. Therefore,
Pr{Y τ+∆τ −Y τ = h |Y τ = k}
h∑
= Pr{Y 1,τ+∆τ −Y 1,τ = l, Y 2,τ+∆τ −Y 2,τ = h − l |Y 1,τ + Y 2,τ = k}
l=0
=
h∑
Pr{Y 1,τ+∆τ −Y 1,τ = l} Pr{Y 2,τ+∆τ −Y 2,τ = h − l}
l=0
h∑ e −∆t 1
(∆t 1 ) l e −∆t 2
(∆t 2 ) h−l
=
(1)
l! (h − l)!
l=0
whereweusethefactthatY 1 and Y 2 are independent NSPPs, and where ∆t i =
Λ i (τ +∆τ) − Λ i (τ). Let ∆t =∆t 1 +∆t 2 =Λ(τ +∆τ) − Λ(τ). Then (1) simplifies to
e −∆t (∆t) h
h!
32. No answer provided.
33. No answer provided.
34. No answer provided.
h∑
l=0
h!
l!(h − l)!
( ∆t1
∆t
) l ( ) h−l ∆t2
= e−∆t (∆t) h
∆t
h!
· 1