0:7 0:2 0:1 0:3 0:5 0:2 0:1 0:4 0:5 P = 0:56 0:28 0:16 0:38 0:39 0:23

Transkript

1 UTKAST ENGLISH VERSION EKSAMEN I: MOT100A STOKASTISKE PROSESSER VARIGHET: 4 TIMER DATO: 16. februar 2006 TILLATTE HJELPEMIDLER: Kalkulator; Tabeller og formler i statistikk (Tapir forlag): Rottman: Matematisk formelsamling. OPPGAVESETTET BESTAR AV 4 OPPGAVER PA 3 SIDER Problem 1 Using certain criteria the share market has what could be called a bad day (state 0), an average day (state 1) or a good day (state 2). Let X n be the state at day n. The process fx n g is assumed to be a Markov chain with the following transition matrix, and with P = P 2 = 0 0 0:7 0:2 0:1 0:3 0:5 0:2 0:1 0:4 0:5 1 C A 1 0:56 0:28 0:16 0:38 0:39 0:23 C A : 0:24 0:42 0:34 a) Find P (X 1 = 2jX 0 = 0), P (X 4 = 1jX 3 = 0; X 2 = 1) and P (X 2 = 1jX 0 = 1). b) It is informed that P (X 0 = 0) = 0:2, P (X 0 = 1) = 0:5 and P (X 0 = 2) = 0:3. Find P (X 2 = 2) and P (X 2 = X 1 = X 0 = 2). c) Find the equilibrium equations for this system and solve them. If the system is started in X 0 = 0, what is the probability P (X n = 0jX 0 = 0) when n becomes large? Problem 2 Each day a share goes up with probability 0.6 and down with probability 0.4. The change one day is independent of the change on other days.

2 a) Let N be the number of days until the share goes down for the rst time, where N = 1; 2; : : :. Use rst step analysis to explain that and E(N) = 0: :6(1 + E(N)) E(N 2 ) = 0: :6(E(1 + N) 2 ) Use this to show that E(N) = 2:5 and var(n) = 3:75. b) When the share goes up, the amount it rises with, is exponentially distributed with mean 0.5kr. Let X i be the amount it rises with the i-the time it goes up. Let N be as in a). What is the interpretation of N X1 Y = X i ; i=1 where Y is set to zero if N = 1? Find the mean and variance of Y. Problem 3 Three precision tool machines need to be looked after constantly for them to work perfectly. Therefore they must be taken out and adjusted frequently. Two specialists take care of the adjustments. When a machine is up again, it works satisfactorily for a period of time that is exponentially distributed with a mean of 8 hours. The adjustments of the machines take quite some time and is exponentially distributed with a mean of 2 hours. Only one specialist can work at a machine at any time. Let the states of this system be the number of machines that are in use, such that the possible states are 0, 1, 2 or 3. a) Set up the balance equations and solve them. b) Find the expected number of machines that are in use and the expected number of specialists that are working in the equilibrium state. Further, nd the proportion of time where at least one machine is up. If you do not manage to solve a), show in principle how you would have done this point if you knew the equilibrium probabilities P 0, P 1, P 2 and P 3. Problem 4 Assume that customers arrive at an exclusive store according to a Poisson process with 3 customers per hour. Let N(t) be the number of customers that arrive within a time period of length t measured in hours. Moreover, let S n be the waiting time to the arrival of customer number n.

3 a) Find P (N(1) = 2), P (S 1 > 0:5), P (S 1 > 1:0jS 1 > 0:5) and E(S 10 )jn(2) = 8). b) In fact the customers arrive according to an inhomogeneous Poisson process. The store opens at 9 o'clock, and between 9 and 10 on the average 2 customers arrive. The average number of customers per hour then increases linearly from 2 per hour at 10 to 4 per hour at 14 o'clock. Then the number of customers per hour decreases linearly from 4 per hour at 14 to 2 per hour at 16 o'clock. In the hour between 16 and 17 there is again a constant arrival rate of 2 per hour. Find an expression for the intensity parameter (t) as a function of time. Here the parameter (t) = 2 for 0 t 1 and for 7 t 8 at a time scale which begins 9 o'clock in the morning. Find the integrated intensity parameter Z 8 m(8) = 0 (t)dt for a day of 8 hours. Use this to nd the expected number of customers that come to the store during one day and nd the corresponding standard deviation. c) One day the store manager has to go to an important meeting. He does not know when this will end, but he has told his assistant that his arrival time in the store is uniformly distributed between 10 and 11 this day. What is the expected number of customers that arrive in his absence? At last nd an expression fro the probability that 4 customers will be coming during his absence. Explicit answer with numbers is not required here.

4 Utvidet fasit (forbehold om feil!) Oppgave 1 1a) P (X 1 = 2jX 0 = 0) = 0:1, P (X 4 = 1jX 3 = 0; X 2 = 1) = P (X 4 = 1jX 3 = 0) = 0:2 P (X 2 = 1jX 0 = 1) = 0:39. 1b P (X 2 = 2) = 0:2 0:16 + 0:5 0:23 + 0:3 0:34 = 0:249: P (X 2 = 2; X 1 = 2; X 0 = 2) = P (X 0 = 2)P (X 1 = 2jX 0 = 2)P (X 2 = 2jX 1 = 2) Likevektslikninger: = 0:3 0:5 0:5 = 0:075: 0: : :1 2 = 0 0: : :4 2 = = 1. som gir lsning 0 = 17 40, 1 = 14 40, 2 = P (X n = 0jX n = 0)! a) Oppgave 2 Det er sannsynlighet 0.4 for nedgang i frste steg og 0.6 sannsynlighet for oppgang, og i det siste tilfellet ma en begynne pa nytt etter allerede a ha gatt ett steg. Lsning av den frste likningen med hensyn pa E(N), gir E(N) = 2:5. Tilsvarende fra den andre likningen, E(N 2 ) = 10. Det flger at var(n) = 10 (2:5) 2 = 3:75. 2b) Tolkning av P N 1 i=1 X i er samlet belp som aksjen er gatt opp med fr frste nedgang. E(Y ) = E(N 1)E(X i ) = 1:5 0:5 = 0:75. Siden varians i den eksponensielle fordeling er forventningen kvadrert: var(y ) = var(x i )E(N 1)+(E(X i )) 2 var(n 1) = (0:5) 2 1:5+(0:5) 2 3:75 = 1:3125: 3a) Oppgave 3 La = 1=8 og = 1=2 vre avgangsrate og inngangsrate henholdsvis. Da blir likevektslikningene: P 1 = 2P 0 2P 2 + 2P 0 = P 1 + 2P 1 3P 3 + 2P 1 = 2P 2 + P 2 P 2 = 3P 3

5 Av dette fas P 1 = 8P 0, P 2 = 4P 1 = 32P 0, P 3 = 4 3 P 2 = P 0, som innsatt i P 0 + P 1 + P 2 + P 3 = 1 gir 3b) P 0 = ; P 1 = ; P 2 = ; P 3 = : La M vre antall maskiner i bruk og N antall spesialister som er sysselsatt: E(M) = = = 2:39 E(N) = = = 0:60: Sannsynlighet for at minst en maskin er oppe er 1 P 0 = a) Oppgave 4 P (N(1) = 2) = 0:224, P (S 1 > 0:5) = e 3 0:5 = 0:22, P (S 1 > 1jS 1 > 0:5) = P (S 1 > 0:5) = 0:22, E(S 10 jn(2) = 8) = 2 + E(S 2 ) = = time. 4b) Vi har: (t) = 2; 0 t 1, (t) = (t 1); 1 t 5, (t) = 4 (t 5); 5 t 7, (t) = 2; 7 t 8. Det flger at Z 5 m(8) = (2 + 1 Z 7 2 (t 1))dt + (4 (t 5))dt + 2 = 22 5 La N vre antall kunder. Siden forventning er lik varians i Poisson fordelingen blir E(N) = 22 og SD(N) = p 22 = 4:69. 4c) La N vre antall ankomne kunder. E(NjT = t) = E(N(t)) = m(t). Det flger at m(t) = 2 + Z t 1 ( (u 1))du = 2t (t 1)2 : Z 2 E(N) = 1 (2t + 1 Z 2 4 (t 1)2 f T (t)dt = 1 2t (t 1)2 )dt = time:

6 P (N = 4jT = t) = P (N(t) = 4) = (m(t))4 e 4! med m(t) = 2t (t 1)2, og vi har m(t) Z 2 (m(t)) 4 e m(t) Z 2 (m(t)) 4 e m(t) P (N = 4) = f T (t)dt = dt: 1 4! 1 4!