Applied Stochastic Processes

Transcription

1 STAT455/855 Fall 23 Appled Stochastc Processes Fnal Exam, Bref Solutons 1. (15 marks) (a) (7 marks) The dstrbuton of Y s gven by ( ) ( ) y P (Y y) for y 2, 3,... The above follows because each of the rolls 2,..., y 1 should be dfferent from the prevous roll, whch occurs wth probablty (5/) y 2, and then the yth roll should be the same as the prevous roll, whch occurs wth probablty 1/. If we let N denote the net proft for the game, then E[N ] kp (Y k) + (y k)p (Y y) k + k + yk+1 yk+1 yk+1 yp (Y y) y ( ) ( ) y k + (k + y) y1 k + (k + ). ( 1 ) ( ) y 1 5 (b) (8 marks) We now compute the expected net proft for the modfed game. Let N mod denote ths net proft. Then, condtonng on Y we get E[N mod ] E[N mod Y y]p (Y y) y2 k E[N mod Y y]p (Y y) + y2 yk+1 (y k)p (Y y) k E[N mod Y y]p (Y y) + E[N ] + kp (Y k) y2 (from part(a))

2 STAT 455/ Fnal Exam Solutons, 23 Page 2 of Thus, E[N mod ] < E[N ] k E[N mod Y y]p (Y y) + kp (Y k) <. y2 We now consder E[N mod Y y] for y k. Snce we lose k dollars n the orgnal game and we pay an addtonal 1 dollar to play the second game, our expected net proft wll be the expected amount we wn n the second play of the game less ths amount. Lettng Z denote the number of rolls t takes to get two consecutve rolls that are the same, we have E[N mod Y y] (k + 1) + (k + 1) + zk+1 zp (Z z) (k + ), where the second equalty follows from part(a) snce Z and Y have the same dstrbuton. Note that E[N mod Y y] does not depend on y, so we can cancel out P (Y k) to get E[N mod ] < E[N ] (k + 1) + (k + ) + k < (k + ) < 1 < 1 k +. (Note there s an error n the hnt). One can then check on a calculator that the left hand sde s less than the rght hand sde for k (15 marks) Note that ths problem s very smlar to problem 7 n Chapter 4 of the text. (a) (3 marks) The transton probabltes are p,+1 M M for,..., M 1 p, 1 M for 1,..., M p j otherwse.

3 STAT 455/ Fnal Exam Solutons, 23 Page 3 of (b) (3 marks) Intutvely, n the lmt each ball should be equally lkely to be n ether urn, ndependently of every other ball. So the number of balls n urn 1 should be expected to be dstrbuted accordng to the Bnomal dstrbuton wth parameters M and p 1/2. (c) (7 marks) We consder the local balance equatons: These gve M π M π M for,..., M 1 π +1 M + 1 π. for 1,..., M. Therefore, [ M π (M )(M + 1) π 1 ( + 1) (M )(M + 1)... (M) ( + 1)()... (1) M! ( + 1)!(M ( + 1))! π ( ) M π, + 1 ( ) M ] 1 (2 M ) 1 π ( ) M 1, 2 usng the fact that 2 M (1 + 1) M M ( ) M 1 1 M M ( ) M, whch follows from the bnomal theorem. Thus, we have π ( M ) ( ) M 1, 2 for,..., M. In other words, the statonary dstrbuton s Bnomally dstrbuted wth parameters M and p 1/2, as expected. (d) (2 marks) The Markov chan {X n : n } s tme-reversble snce ts statonary dstrbuton satsfes the local balance equatons, as shown n part(c).

4 STAT 455/ Fnal Exam Solutons, 23 Page 4 of 3. (15 marks) Let N(t) denote the number of bugs found up to tme t, so {N(t) : t } s a Posson process wth rate λ 2. Here, one unt of tme s one week. (a) (3 marks) We want E[N(5) N(3) 2]. We can wrte N(5) N(5) N(3)+N(3), where N(5) N(3) s ndependent of N(3) and s dstrbuted as Posson(4). So E[N(5) N(3) 2] 2 + E[N(5) N(3)] (b) (3 marks) We want E[S 4 N(3) 2], where S4 s the tme the 4th bug s found. Gven N(3) 2 the tme untl the 4th bug s found s the tme untl the second bug after tme 3 s found. By memoryless, ths s 3 plus the sum of two ndependent Exponental(2) random varables. Thus, E[S 4 N(3) 2] (c) (4 marks) Gven N(3) 2, these two event tmes are ndependent and unformly dstrbuted on the nterval (,3). Thus, for each event, the probablty that t was a catastrophc bug can be obtaned by condtonng on the tme of the event: P (catastrophc bug) P (catastrophc bug 1 tme of event was s) 3 ds e s ds 1 3 (1 e 3 ). Therefore, the probablty that t was not a catastrophc event s (1 e 3 ), and the probablty that nether event was a catastrophc bug s P (nether bug catastrophc N(3) 2) (1 1 3 (1 e 3 )) (d) (5 marks) Let N 1 (t) be the number of catastrophc bugs found by tme t. Then N 1 (t) s Posson dstrbuted wth mean Therefore, E[N 1 (t)] 2 t e s ds 2(1 e t ). P (N 1 (t) > ) 1 P (N 1 (t) ) 1 exp{ 2(1 e t )}. Lettng t we obtan P (catastrophc bug s ever found) lm t P (N 1 (t) > )1 e 2.85.

5 STAT 455/ Fnal Exam Solutons, 23 Page 5 of 4. (15 marks) (a) (11 marks) There was an error of omsson n the descrpton of the process, whch s that servce tmes are Exponentally dstrbuted wth rate µ. () (8 marks) The state space s S {(, ), (, 1), (1, ), (1, 1)}. The nfntesmal generator s gven by (θ + λ) θ λ λ λ G θ θ µ µ, where the rows of G correspond to the states (,), (,1), (1,), and (1,1), respectvely. The global balance equatons are gven by (θ + λ)π (,) µπ (1,1) λπ (,1) θπ (,) θπ (1,) λπ (,) µπ (1,1) λπ (,1) + θπ (1,) The frst three of these equatons express π (,1), π (1,), and π (1,1) n terms of π (,). Usng these, we get from the normalzaton constrant that π (,) π (,1) π (1,) π (,1) θ + λ + θ+λ λ θ µ θ 2 µ λ 2 µ λθ 2 + λ 2 θ λθµ () (3 marks) As θ we have lm θ π (,) lm θ π (1,), whch makes sense because the server wll never go on vacaton n the lmt, and lm π (,1) θ lm π (1,1) θ µ µ + λ λ µ + λ.

6 STAT 455/ Fnal Exam Solutons, 23 Page of (b) (4 marks) We have v λ for all S and q j λp j for all j. Let ψ denote the statonary dstrbuton for ths contnuous tme Markov chan. Then ψ satsfes the global balance equatons for the contnuous tme chan: v j ψ j S λψ j S ψ j S ψ q j ψ λp j ψ p j, or or for all j S. Thus, we see that ψ satsfes the global balance equatons for the dscrete tme Markov chan {X n : n }. Snce ths statonary dstrbuton s unque, we must have ψ π. 5. (15 marks) (a) (9 marks) The soluton here s essentally the same as the soluton to Problem 2 of Assgnment 5, for t < 3, except here that soluton works for all t. (b) ( marks) Let W denote the total watng tme of all customers who arrve before the frst server arrves and let T denote the arrval tme of the frst server. Then, we have that T s Exponentally dstrbuted wth rate µ and E[W T t] E[W t ] λt 2 /2, so E[W ] E[W T t]ft (t)dt λt 2 2 µe µt dt λ 2 E[T 2 ] λ 2 2 µ λ 2 µ. 2