TMA 4265 Stochastic Processes

Transcription

1 TMA 4265 Stochastic Processes Norges teknisk-naturvitenskapelige universitet Institutt for matematiske fag Solution - Exercise 9 Exercises from the text book 5.29 Kidney transplant T A exp( A ) T B exp( B ) T 1 Time when new kidney arrives exp(λ) T 2 Waiting time between arrival of first and second kidney exp(λ) Rules: i) First kidney goes to A (to B if A i dead) ii) Second kidney goes to B (if B is still alive) (as arrival of kidneys is Poisson process) a) Pr(A gets new kidney) Pr(T 1 < T A ) λ λ + A b) Pr(B gets new kidney) Pr(T B > T 1 )P(T A < T 1 ) + P(T B > T 1 + T 2 )P(T A > T 1 ) Pr(T B > T 1 )P(T A < T 1 ) + P(T B > T 1 + T 2 T B > T 1 )P(T B > T 1 )P(T A > T 1 ) Pr(T B > T 1 )P(T A < T 1 ) + P(T B > T 2 )P(T B > T 1 )P(T A > T 1 ) ( ) λ B + λ A λ 2 A + λ + λ B + λ A + λ 21. oktober 28 Side 1

2 5.42 N(t) Poisson(λt) S n Time of n th event Gamma(λ, n) E[S n ] n λ a) E[S 4 ] 4 λ b) E[S 4 N(1) 2] E[T 1 + T 2 + T 3 + T 4 N(1) 2] E[1 + T 3 + T 4 ] λ We know that T 3 has not yet occured, and as T 3 exp(λ), this is like restarting with T 3. c) Independent increments give E[N(4) N(2) N(1) 3] E[N(4) N(2)] E[N(4 2)] E[N(2)] 2λ 5.6 Customers arrive at a bank at a Poisson rate λ. Suppose two customers arrived during the first hour, that is, N(1) 2. We need to know that P(N(t + s) N(s) n) e λt (λt) n n!. What is the probability that: a) Both arrived during the first 2 minutes? 21. oktober 28 Side 2

3 We need to find: P(N( 1 3 ) 2 N(1) 2) P(N(1 3 ) 2 N(1) 2) P(N(1) 2) b) At least one arrived during the first 2 minutes? TMA 4265 Stochastic Processes, P(N(1 3 ) 2) P(N(2 3 ) ) P(N(1) 2) (λ 1 e λ1 3 3 )2 2! e λ2 3 (λ2 3 )! e λ λ 2 2! ) 2 λ 2 ( e λ 1 3 e λ λ 2 ( ) We have two possibilities. Either both arrived during the first 2 minutes, or only one arrived during the first 2 minutes and the second during the 4 last minutes. The first possibility was calculated in a). The second is: P(N( 1 3 ) 1 N(1) 2) P(N(1 3 ) 1 N(1) 2) P(N(1) 2) P(N(1 3 ) 1) P(N(2 3 ) 1) P(N(1) 2) (λ 1 e λ1 3 3 )1 1! e λ2 3 (λ2 3 )1 1! e λ λ 2 2! 2 9 e λ λ e λ λ The probability that at least one arrived during the first 2 minutes is then N number of errors in text Poisson(λ) P i probability that proofreader i finds the error, i 1,2 X 1 number of errors found by proofreader 1, but not by 2 X 2 number of errors found by proofreader 2, but not by 1 X 3 number of errors found by both X 4 number of errors not found a) X 1,...,X 4 have marginals like shown in b). Similarly as in exercise 5.44c), they are independent and have simultaneous distribution P r {X 1 x 1,...,X 4 x 4 } P r {X 1 x 1 }... P r {X 4 x 4 } 21. oktober 28 Side 3

4 b) We have X 1 N n Bin(n,p) p p 1 (1 p 2 ) X 1 Poisson(λ p) E[X 1 ] λp 1 (1 p 2 ) X 2 Poisson(λ p 2 (1 p 1 ))) E[X 2 ] λp 2 (1 p 1 ). E[X 3 ] λp 1 p 2 (1) X 4 Poisson(λ (1 p 1 )(1 p 2 )) E[X 4 ] λ(1 p 1 )(1 p 2 ) (2) This gives E[X 1] E[X 3 ] 1 p 2 p 2 og E[X 2] E[X 3 ] 1 p 1 p 1 c) Estimator: Put in X 1,...,X 3 for E[X 1 ],...,E[X 3 ] in the answers from b), and get ˆp 1 X 3 X 2 +X 3 and ˆp 2 X 3 X 1 +X 3 From (1) we obtain ˆλ X 3 pˆ 1 ˆp 2... X 1 + X 2 + X 3 + X 1 X 2 X 3 d) (2) gives: ˆX4 ˆλ(1 ˆp 1 )(1 ˆp 2 )... X 1 X 2 X Single-server station. V i time between successive arrivals F(iid) S i service time G(iid) X n number of customers in system immediately before n th arrival Y n number of customers in system immediately after n th departure We have X n X n D n and Y n { Yn A n Y n 1 1 A n Y n 1 ( ) where 21. oktober 28 Side 4

5 D n number of customers being serviced during n th arrival time V n A n number of customers arriving during n th departure time S n Demand: {X n } Markov chain X n X n 1 independent of {X k } k n 2 a) If V i F exp(λ), that is, forgetful, then A n is independent of all events before Y n 1, that is, independent of {Y k } k n 2. (A n is, however, dependent of S n ). From ( ) we know that Y n Y n 1 is independent of {Y k } k n 2 {Y n } is a Markov chain. (If F was not forgetful, A n would have been dependent of how much time before Y n 1 the last arrival appeared; denote this time by T. If (Y n k,...,y n 1 ) (r + k,...,r), that is, if no one arrives during the last k departure times, there is a large probability that T is large. Therefore A n is dependent of {Y k } k n 2 {Y n } NOT Markov!!! independent {Y k } k 2 T independent A n ) b) If S i G exp(), that is, forgetful, D n is independent of everything that happened prior to X n 1, that is, independent of {X k } k n 2 From ( ) we know that X n X n 1 is independent of {X k } k n 2 {X n } is a Markov chain. Using a simular argument as in a), we obtain that {X n } is not a Markov chain if G is not forgetful. c) From a) we know that when F exp(λ), then {Y n } is a Markov chain, and ( ) gives that P r {Y n k Y n 1 j} P r {Y n A n k Y n 1 j} P r {A n k j + 1 Y n 1 j} P r {A n k j + 1} (A n is independent of Y n 1, j 1) The low of total probability gives that P r {A n k } P r {A n k S n s} g sn (s)ds (λs) k k! e λs g sn (s)ds The transition probabilities for {Y n } are P Yn (j,k) { (λs) k j+1 (k j+1)! e λs g(s)ds j 1,k j 1 (λs) k k! e λs g(s)ds j,k j 21. oktober 28 Side 5

6 Similarly we have that when G exp(), {X n } is a Markov chain, and ( ) gives P r {X n k X n 1 j} P r {X n D n k X n 1 j} If we condition on the arrival time V n, we get that P r {D n j k + 1 X n 1 j} P r {D n k V n v,x n 1 j} Therefore P Xn (j,k) P j,j k+1 (v) f vn (v)dv (v) k k! e v k j (v) i ik i! e v k j + 1 ellers P j,k (v) 6.1 Given that we have n males and m females, P r{male i meets female j turing time period h } λh + o(h) The number of offspring these two produce is then a Poisson process with rate λ, and the waiting time before the first birth is T ij exp(λ) The waiting time before the first birth in the entire population is T min i,j exp(n m λ) (There are n m combinations of males and females.) [i)] In state {n,m} we have v 1 {n,m} E[Time in state {n,m} before moving on] E[T] 1 nmλ v {n,m} nmλ The probability of giving birth to a male and a female is equal, hence P {n,m}, {n+1,m} P {n,m}, {n,m+1} oktober 28 Side 6

7 Alternatively: P r{male i does NOT meet female j during h} Pr{No male meets a female during h} 1 λh + o(h) (1 λh + o(h)) nm 1 nmλh + o(h) P ii (h) We have that 1 P ii (h) nmλh + o(h) ν i lim lim nmλ h h h h 6.2 See solution in the book. 6.3 Analyze by the number of machines not functioning, and get BUT we can not analyze this any further as λ 1 is dependent of which machine breakes down first. 21. oktober 28 Side 7

8 We can alternatively consider the states (,) - Both machines functioning (1,) - Machine 1 not functioning, M2 functioning. X t (2,) - Machine 2 not functioning, M1 functioning (1,2) - M1 being repaird, M2 not functioning. (2,1) - M2 being repaird, M1 not functioning. We get P (,) (1,) (2,) (1,2) (2,1) (, ) (1,) (2,) (1, 2) 1, v (2, 1) See solution in the book. Exercises from exams Eksamen, mai 3 oppg.2 a) P{N(h) 1} P{N A (h)+n B (h) 1} P{(N A (h) 1 N B (h) ) (N A (h) N B (h) 1)} P{N A (h) 1 N B (h) } + P{N A (h) N B (h) 1} (disjoint events) P{N A (h) 1}P{N B (h) }+P{N A (h) }P{N B (h) 1} (as N A (h) and N B (h) are independent) (λ A h + o(h))(1 λ B h + o(h)) + (1 λ A h + o(h))(λ B h + o(h)) λ A h λ A λ B h 2 + o(h) + λ B h λ A λ B h 2 + o(h) (λ A + λ B )h + o(h) (as h 2 is a o(h)-function). 21. oktober 28 Side 8

9 Showing that P{N(h) 2} o(h) is easiest by first considering P{N(h) }: P{N() } P{N A (h) N B (h) } P{N A (h) }P{N B (h) } (1 λ A h + o(h))(1 λ B h + o(h)) 1 λ B h λ A h + λ A λ B h 2 + o(h) 1 (λ A + λ B )h + o(h) (siden h 2 er en o(h)-funksjon). This gives P{N(h) 2} 1 P{N(h) } P{N(h) 1} 1 (1 (λ A + λ B )h + o(h)) ((λ A + λ B )h + o(h)) o(h). b) Use that N(t) is a Poisson process with intensity λ λ A + λ B 3: P{N(1) 2} P{N(1) } + P{N(1) 1} + P{N(1) 2} λ! e λ + λ1 1! e λ + λ2 2! e λ e λ (1 + λ λ2 ).423. Use definition of conditional probability to get P{N A (1) 1 N(1) 1} P{N A(1) 1,N(1) 1} P{N(1) 1} P{N A(1) 1,N B (1) } P{N(1) 1} P{N A(1) 1}P{N B (1) } P{N(1) 1} (as N A (1) and N B (1) are independent) λ 1 A 1! e λ A λ B! e λ B λ 1 1! e λ Use that N(t) N A (t) + N B (t) to get λ A λ A + λ B 1 3 (as λ λ A + λ B ). P{N(1) 2 N A (.5) 1} P{N B (.5) + (N(1) N(.5)) 1 N A (.5) 1} P{N B (.5)+(N(1) N(.5)) 1} (as N A (.5) is independent of N B (.5) and N(1) N(.5)) P{N B (.5),N(1) N(.5) } + P{N B (.5) 1,N(1) N(.5) } +P{N B (.5),N(1) N(.5) 1} e λ B.5 e λ.5 + λ B.5 e λ B.5 e λ.5 + e λ B.5 λ.5 e λ.5 e (λ+λ B).5 (1 +.5 λ B +.5 λ) oktober 28 Side 9