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By taking t = 0, we nd P {Ik = 1} = 1 /( 1 + 2 ) Since {N (t)} is a Poisson process with rate 1 + 2 , we have P {Zk > t} = exp [ ( 1 + 2 )t] Hence P {Ik = 1, Zk > t} = P {Ik = 1}P {Zk > t}, showing that P {Ik = 1 | Zk = t} = 1 /( 1 + 2 ) independently of t (b) Obviously, the process {Ni (t)} satis es the properties (A), (B) and (D) To verify property (C), note that P {one arrival of type i in (t, t + t]} = ( t)pi + o( t) = ( pi ) t + o( t) It remains to prove that the processes {N1 (t)} and {N2 (t)} are independent Fix t > 0 Then, by conditioning,

P {N1 (t) = k, N2 (t) = m}

Subscribed enhanced dialled service (D-CSI) Serving network dialled service (N-CSI)

P {N1 (t) = k, N2 (t) = m | N (t) = n}P {N (t) = n}

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= P {N1 (t) = k, N2 (t) = m | N (t) = k + m}P {N (t) = k + m} = ( t)k+m k+m k m p1 p2 e t k (k + m)! ( p1 t)k p2 t ( p2 t)m e , k! m!

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showing that P {N1 (t) = k, N2 (t) = m} = P {N1 (t) = k}P {N2 (t) = m} The remarkable result (115) states that the next arrival is of type i with probability i /( 1 + 2 ) regardless of how long it takes until the next arrival This result is characteristic for competing Poisson processes which are independent of each other As an illustration, suppose that long-term parkers and short-term parkers arrive at a parking lot according to independent Poisson processes with respective rates 1 and 2 Then the merged arrival process of parkers is a Poisson process with rate 1 + 2 and the probability that a newly arriving parker is a long-term parker equals 1 /( 1 + 2 ) Example 112 A stock problem with substitutable products A store has a leftover stock of Q1 units of product 1 and Q2 units of product 2 Both products are taken out of production Customers asking for product 1 arrive according to a Poisson process with rate 1 Independently of this process, customers asking for product 2 arrive according to a Poisson process with rate 2 Each customer asks for one unit of the concerning product The two products serve as substitute for each other, that is, a customer asking for a product that is sold out is satis ed with the other product when still in stock What is the probability distribution of the time until both products are sold out What is the probability that product 1 is sold out before product 2 To answer the rst question, observe that both products are sold out as soon as Q1 + Q2 demands have occurred The aggregated demand process is a Poisson process with rate 1 + 2 Hence the time until both products are sold out has an Erlang (Q1 + Q2 , 1 + 2 ) distribution To answer the second question, observe that product 1 is sold out before product 2 only if the rst Q1 + Q2 1 aggregated demands have no more than Q2 1 demands for product 2 Hence, by (115), the desired probability is given by.

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