ST3236: Stochastic ProcessTutorial 10

TA: Mar Choong Hock

Email: g0301492@nus.edu.sg

Exercises: 11

Question 1For i = 1, … , n. Let {Xi(t) : t > 0} be independent Poisson processes, each with the sameparameter . Find the distribution of the first time that at least one event has occurred inevery process.

Question 1

Note: You can give either the c.d.f. or the p.d.f. as the answer for this case, the c.d.f. is

F(t) = 1 - {1 - exp(-t)}n

Question 2Let {X(t) : t 0} be a Poisson process of rate . Suppose it is known that X(t) = n. For n = 1, 2, … , determine the mean of the first arrival time W1 and Wn.

Note: The Wn here is conditioned on X(t) = n. This is different from tutorial 9.

Question 2Let Y1, … , Yn be IID and uniformly distributed on (0, t]. Then W1 has the same distribution as Y(1)

and Wn has the same distribution as Y(n).

Note: Yi is a random variable that represents theposition of customer i, on the time axis (measured w.r.t t = 0), after placing the customer on the axis (uniform distribution). It is not necessary the ith arrived customer.

By definition, Y(1) = min {Y1, Y2, …, Yn}, that is, it is the first arrived customer.

Question 2Then W1 has the same distribution as Y(1) and Wn has the same distribution as Y(n) (since it is the maximum of all Yi it represents the last arrived customer). The distribution of Y(1) is

Question 2

Question 2

Question 3Let {X(t) : t 0} be a Poisson process of rate . Suppose it is known that X(t) = 2.

Determine the mean W1W2, the product of the first two arrival times

Question 3Let Y1 and Y2 be IID and distributed uniformly on (0; t]. We have

E(Y1) = E(Y2) = t/2

Let Y(1) and Y(2) be the order statistics of Y1 and Y2. Then (W1, W2) have the same distribution as (Y(1), Y(2)). Because

Y(1)Y(2) = Y1Y2

ThusE(W1W2) = E(Y1Y2) = E(Y1)E(Y2) = t2/4

Question 4Customers arrive at a certain facility according to a Poisson process of rate . Suppose that it is knowthat five customers arrived in the first t hours.

(a)Determine the mean total waiting time E{W1 + … +W5 | X(t) = 5}

(b) Determine the mean total waiting time E{W1 + … +W5 +W6 | X(t) = 5}

Question 4Let Y1, …, Y5 be IID and distributed uniformly on (0, t]. We have

E(Y1) = … = E(Y5) = t/2

Let Y(1), …, Y(5) be the order statistics. Then (W1, … ,W5) have the same distribution as(Y(1), … , Y(5)). Because

Y(1) + … + Y(5) = Y1 + … + Y5

Question 4aThusE(W1 + … +W5 | X(t) = 5) = E(Y1 + … + Y5) = E(Y1) + … + E(Y5) = 5t/2

Question 4bE{W1 + … +W5 +W6|X(t) = 5}= E{W1 + … +W5 | X(t) = 5 } + E {W6 | X(t) = 5}= 5t/2 + (t +1/)

(See lecture notes for the calculation of E{W6 | X(t) = 5} )