1 UNIVERSITY OF LONDON IMPERIAL COLLEGE OF SCIENCE, TECHNOLOGY AND MEDICINE Course:M3S4/M4S4 Setter: Hand Checker: Crowder Editor: McCoy External: Kent Date: December 2003 BSc and MSci EXAMINATIONS (MATHEMATICS) MAY-JUNE 2004 This paper is also taken for the relevant examination for the Associateship SOLUTIONS M3S4/M4S4 APPLIED PROBABILITY © 2000 University of London M3S4/M4S4 Page of 10. 1 2 1. SOLUTION (a) [Seen] (i) [2] pn t h pn t . 1 h o h pn 1 t . h o h o h (ii) [2] Simplifying and rearranging the above, we obtain pn t h pn t o h pn1 t pn t h h so that, in the limit as h 0 , dpn t pn 1 t pn t (1a) dt In the special case of n 0 the equation is dp0 t (1b) p0 t dt (iii) [6] Multiplying the equations in (1) by s n and summing over n 0,1, 2,... yields d d p0 t pn t s n p0 t pn t s n pn 1 t s n dt n 1 dt n 1 n 1 Then, using s, t pn t s n , this becomes n 0 1 s t (iv) [3] Integrating this equation yields log 1 s t A s so that B s exp 1 s t with A s ln B s , and using the condition given in the question that s,0 1 we obtain B s 1. (v) [2] This is the pgf of a Poisson distribution with parameter t (b) [5] [Seen similar] The number of casualties which have occurred by time t is a compound Poisson process. In the lectures we derived that the expected number of casualties in such a process is E S t t , where is the mean of the number of casualties at each accident (full marks for either remembering this or deriving it). Since the mean of a © 2000 University of London M3S4/M4S4 Page of 10. 2 3 G0 q distribution is q p , we have that the mean number of casualties by time t is qt p . © 2000 University of London M3S4/M4S4 Page of 10. 3 4 2. SOLUTION [Part seen] (i) [5] P X x P X x | T t P t dt e t t e t dt x! 0 x x x! e t x t dt 0 x x! x ! x 1 x for x 0,1, 2,... . This is a geometric distribution with parameter a G0 distribution. (ii) [5] We have a geometric distribution with p 0 0 . so that E x xp x xqp x qp qp In the lectures we called it d x p dp 0 d 1 qp p dp 1 p q 2 q Likewise, the pgf is 0 0 s s x qp x q s x p x q 1 sp s (iii) [5] From the lectures, the probability that the population will die out is given by the smallest positive root of . Now, from q 1 p we obtain 2 p q 0 so that p q 1 0 with solutions q p and 1. If q p the smaller of these is 1, and if q p the smaller is q p . In terms of and , these are, respectively, 1 and, for , (iv) [5] From the lectures, we know that the expected number born in generation n is n n , where is the mean of the offspring probability distribution. Hence, © 2000 University of London M3S4/M4S4 Page of 10. 4 5 n n . If the expected number of births up to and including the nth generation is thus 2 ... n 1 As n this converges to 1 1 1 n © 2000 University of London M3S4/M4S4 Page of 10. 5 6 3. SOLUTION (a) [Seen] (i) [2] P , i 0 , i 1 (ii) [1] A set of states, C, which it is impossible to leave: pij 0 if i C , j C (iii) [2] Two states communicate if there is a non-zero probability of moving (perhaps in more than one step) from either state to the other. A communicating class is a set of communicating states. Since, by definition, each state communicates with itself, each state does lie in a communicating class. To prove the assertion we need to show that no state lies in more than one such class. This follows from the fact that if a state s lies in two such classes, A and B, then it is possible to get from an state in A to any state in B, and vice versa, via s. Thus A and B form a single communicating class. An informal argument like this is acceptable, or a more formal one if they give it. (b) [Unseen] (i) [3] (ii) [2] Communicating classes: {0,1}, {2}, {3}. Classes {2} and {3} are closed since, once entered, they cannot be left. (iii) [2] Any distribution of the form 0,0, p,1 p will be a stationary distribution. (c) [Unseen] (i) [6] Using the equations describing the first two elements and the fourth in Q we obtain the equations 0.5 0 0.25 1 0 0.5 0 0.5 1 0.5 2 1 3 0.5 3 and combining these with i 1 yields 0 0.25, 1 0.5, 2 0.25, 3 0 . (ii) [2] The mean time between returns is 1 0 4 © 2000 University of London M3S4/M4S4 Page of 10. 6 7 4. SOLUTION (i) [2] [Seen] Traffic intensity is the ratio of the expected value of the service time distribution to that of the inter-arrival time distribution. The traffic intensity of an M / M /1 queue is . (ii) [5] [Seen] For a steady state, we have 0 px1 px1 px and 0 p1 p0 . From these, p1 p0 , p2 2 p1 , and in general px x p0 . (iii) [5] [Seen] For the px , x 0,1, 2,... to form a proper distribution, we must have p x 1 . This is only possible if 0 1 from which p 0 1 . The steady 0 state queue length distribution is thus geometric. In the lectures, we called it G0 (iv) [8] [Unseen] The mean queue length is 1 , either from memory or by deriving the mean of the geometric distribution in (iii). Thus a . By the memoryless property of the exponential distribution, the mean of the service time distribution is 1 . Hence b 1 . From these two equations we obtain a b 1 a , so that the mean inter-arrival time is b 1 a a . © 2000 University of London M3S4/M4S4 Page of 10. 7 8 5. SOLUTION (a) [8] [Unseen] The equation is in Lagrange form with f 2st g s h 2t The auxiliary equations are ds dt d . 2 st s 2t The first equation gives ds 2tdt from which c1 s t 2 . ds d from which c2 s . s so that s s t 2 . The second equation gives Hence s s t 2 Now using the condition that se s when t 0 we obtain ses s s or s es . Hence the solution is s, t se s t . 2 (b) [Seen similar] (i) [2] The bacterial population is not wiped out at time t if no drug treatment has occurred by that time. Since the rate at which doses of treatment are administered is , the time interval between treatments is distributed as M . Hence, P(no treatment by time t) = exp t , so that PY t 0 exp t . (ii) [2] Either: since no treatment has occurred by time t, the population behaves according to a simple birth process, so that, from the lectures, its size at time t has a geometric distribution G1 e t . Or, from the pgf given in the question, the distribution is a geometric distribution G1 e t . So, PY t y | Y t 0 e t 1 e t y 1 y 1,2,... (iii) [5] The unconditional probability distribution of Y t is given by PY t 0 1 PY t 0 1 e t and when y 0 , PY t y PY t y | Y t 0 PY t 0 e t 1 e t y 1 e t 1 e t The pgf of Y t is given by © 2000 University of London e t y 1 M3S4/M4S4 Page of 10. 8 9 s, t P Y t y s y y 0 1 e t e t 1 e t y 1 sy y 1 1 e t 1 se t 1 s 1 e t e t s 1 1 s 1 e t (iv) [3] Differentiating s, t with respect to s yields s, t e t se t 1 e t 2 s 1 s 1 e t 1 s 1 e t So that s, t s t e © 2000 University of London e t e s 1 e t t 1 e t e 2 t M3S4/M4S4 Page of 10. 9