Massachusetts Institute of Technology

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1 6.0/6.3: Probabilistic Systems Aalysis (Fall 00) Problem Set 8: Solutios. (a) We cosider a Markov chai with states 0,,, 3,, 5, where state i idicates that there are i shoes available at the frot door i the morig before Oscar leaves o his ru. Now we ca determie the trasitio probabilities. Assumig i shoes are at the frot door before Oscar sets out o his ru, with probability Oscar will retur to the same door from which he set out, ad thus before his ext ru there will still be i shoes at the frot door. Alteratively, with probability Oscar returs to a differet door, ad i this case, with equal probability there will be mi{i +,5} or max{i, 0} shoes at the frot door before his ext ru. These trasitio probabilities are illustrated i the followig Markov chai: (b) Whe there are either 0 or 5 shoes at the frot door, with probability Oscar will leave o his ru from the door with 0 shoes ad hece ru barefooted. To fid the log-term probability of Oscar ruig barefooted, we must fid the steady-state probabilities of beig i states 0 ad 5, π 0 ad π 5, respectively. Note that the steady-state probabilities exist because the chai is recurret ad aperiodic. Sice this is a birth-death process, we ca use the local balace equatios. We have implyig that ad similarly, As π 0 p 0 = π p 0, π = π 0 5 π i =, { P( X t+ = i X t = i, X t = x t,...x = x ) = i = m, 0 i m π 5 =... = π = π 0. it follows that π i = 6 for i = 0,,..., 5. Hece, P(Oscar rus barefooted i the log-term) = (π 0 + π 5 ) =. 6 i=0. (a) Cosider ay possible sequece of values x, x,...,x t, i for X, X,...,X t, ad ote that 0 < i < m P( X t+ = i + X t = i, X t = x t,...x = x ) = i = 0, 0 i = m Page of 8

2 6.0/6.3: Probabilistic Systems Aalysis (Fall 00) { P( X t+ = i X t = i, X t = x t,...x = x ) = 0 < i m, 0 i = 0 P( X t+ = j X t = i, X t = x t,...x = x ) = 0, i j >. As the coditioal probabilities above oly deped o i, where X t = i, it follows that X, X,... satisfy the Markov property. The associated Markov chai is illustrated below. / / / / 0 m - m / / / / (b) Note that Y, Y,... is ot a Markov chai for m >, because does ot equal P(Y t+ = d + Y t = d, Y t = d ) = P(Y t+ = d + Y t = d, Y t = d, Y t = d ) = 0, for 0 < d < m (the idea is that if Y t = d, Y t = d, ad Y t = d, the X t = d, while if Y t = d, ad Y t = d, the X t = d). If, however, we keep track of X t ad Y t, we do have a Markov chai, because for ay possible sequece of pairs of values (x, y ),..., (x t, y t ),(i, i ) for ( X, Y ),..., ( X t, Y t ),( X t, Y t ), P(( X t+, Y t+ ) = (i +, i + ) ( X t, Y t ) = (i, i ),...( X, Y ) = (x, y )) 0 < i = i < m = i = i = 0, 0 otherwise P(( X t+, Y t+ ) = (i, i ) ( X t, Y t ) = (i, i ),...( X, Y ) = (x, y )) { = 0 < i i m, 0 otherwise P(( X t+, Y t+ ) = (i, i ) ( X t, Y t ) = (i, i ),...( X, Y ) = (x, y )) { = i = i = m, 0 otherwise from which it is clear that the coditioal probabilities oly deped o (i, i ), the values of X t ad Y t, respectively. The correspodig Markov chai is illustrated below. Page of 8

3 6.0/6.3: Probabilistic Systems Aalysis (Fall 00) / / / / (0, m) (, m) (m-,m) (m, m) / / / / / / (0, m-) (, m-) (m-, m-) / / / / / (0, ) (, ) / / (0, 0) 3. (a) If m out of idividuals are ifected, the there must be m susceptible idividuals. Each oe of these idividuals will be idepedetly ifected over the course of the day with probability ρ = ( p) m. Thus the umber of ew ifectios, I, will be a biomial radom variable with parameters m ad ρ. That is, ( ) m p I (k) = ρ k ( ρ) m k k = 0,,..., m. k (b) Let the state of the SIS model be the umber of ifected idividuals. For =, the correspodig Markov chai is illustrated below. pq+(- p)(- q) (- q) p(-q) 0 q(- p) q(- q) (c) The oly recurret state is the state with 0 ifected idividuals. (d) Let the state of the SIR model be (S, I), where S is the umber of susceptible idividuals ad I is the umber of ifected idividuals. For =, the correspodig Markov chai is illustrated below. q Page 3 of 8

4 6.0/6.3: Probabilistic Systems Aalysis (Fall 00) (0,0) (, 0) (, 0) q pq (- p)q q (0,) (, ) q(- q) (0,) (- q) p(- q) (- p)(- q) (- q) If oe did ot wish to keep track of the breakdow of susceptible ad recovered idividuals whe o oe was ifected, the three states free of ifectios could be cosolidated ito a sigle state as illustrated below. (,0) q pq (- p)q q (0,) (, ) q(- q) (0,) (- q) p(- q) (- p)(- q) (e) Ay state where the umber of ifected idividuals equals 0 is a recurret state. For =, there are either oe or three recurret states, depedig o the Markov chai draw i part (d).. (a) The process is i state 3 immediately before the first trasitio. After leavig state 3 for the first time, the process caot go back to state 3 agai. Hece J, which represets the umber of trasitios up to ad icludig the trasitio o which the process leaves state 3 for the last time is a geometric radom variable with success probability equal to 0.6. The variace for J is give by: σ p 0 J = = p 9 (- q) Page of 8

5 6.0/6.3: Probabilistic Systems Aalysis (Fall 00) (b) There is a positive probability that we ever eter state ; i.e., P(K < ) <. Hece the expected value of K is. (c) The Markov chai has 3 differet recurret classes. The first recurret class cosists of states {, }, the secod recurret class cosists of state {7} ad the third recurret class cosists of states {, 5, 6}. The probability of gettig absorbed ito the first recurret class startig from the trasiet state 3 is, /0 = /0 + /0 + 3/0 6 which is the probability of trasitio to the first recurret class give there is a chage of 3 state. Similarly, probability of absorptio ito secod ad third recurret classes are 6 ad 6 respectively. Now, we solve the balace equatios withi each recurret class, which give us the probabilities coditioed o gettig absorbed from state 3 to that recurret class. The ucoditioal steady-state probabilities are foud by weighig the coditioal steady-state probabilities by the probability of absorptio to the recurret classes. The first recurret class is a birth-death process. We write the followig equatios ad solve for the coditioal probabilities, deoted by p ad p. p p = p + p = Solvig these equatios, we get p = 3, p = 3. For the secod recurret class, p 7 =. The third recurret class is also a birth-death process, we ca fid the coditioal steady-state probabilities as follows, p = p 5 p 5 = p 6 p + p 5 + p 6 = ad thus, p = 7, p 5 = 7, p 6 = 7. Usig these data, the ucoditioal steady-state probabilities for all the states are foud as follows: π = = π = = π 3 = 0 (trasiet state) 3 π 7 = = 6 π = = 7 6 π 5 = = 7 6 π 6 = = 7 6 Page 5 of 8

6 6.0/6.3: Probabilistic Systems Aalysis (Fall 00) (d) The give coditioal evet, that the process ever eters state, chages the absorptio probabilities to the recurret classes. The probability of gettig absorbed to the first recurret class is, to the secod recurret class is 3, ad to the third recurret class is 0. Hece, the steady state probabilities are give by, π = = 3 π = = 3 6 π 3 = π = π 5 = π 6 = π 7 = = For pedagogical purposes, let us actually draw what the ew Markov chai would look like, give the evet that the process ever eters state. The resultig chai is show below. Let us see how we came up with these trasitio probabilities. / /0 S S S3 / 3/0 9/0 S7 We eed to be careful whe rescalig the ew trasitio probabilities. First of all, it is clear that the probabilities withi the recurret classes {S, S} ad {S7} do t get affected. We also ote that the self loop trasitio probability of the trasiet state S3 does t get chaged either.(this would be true for ay other trasiet state) To see that the self loop probability p 3,3 does t get chaged, we coditio o the evet that we evetually eter S or S7. Let s call the ew self loop probability, q 3,3. The, q 3,3 = P(X = S3 absorbed ito or 7, X 0 = S3) = p 3, 3 P (absorbed ito or 7 X =S3, X 0 =S3) P (absorbed ito or 7 X 0 =S3) p 3, 3 (a 3, +a 3, 7 ) = (a 3 +a 3 7 ) = p 3,3 = 0,, Now, we calculate q 3,7 ad q 3,. q 3,7 = P(X = S7 absorbed ito or 7, X 0 = S3) = 3 p 3, (a 3, +a 3, 7 ) 6 + = = = 0 p 3, 7 P (absorbed ito or 7 X =S7, X 0 =S3) P (absorbed ito or 7 X 0 =S3) Page 6 of 8

7 6.0/6.3: Probabilistic Systems Aalysis (Fall 00) q 3, = P(X = S absorbed ito or 7, X 0 = S3) = p 3, 0 3,, = (a 3 +a 3 7 ) = = p 3, P (absorbed ito or 7 X =S, X 0 =S3) P (absorbed ito or 7 X 0 =S3) Now, we ca calculate the absorptio probabilities of this ew Markov chai. The probability of gettig absorbed ito the recurret class {, }, startig from S3, is 3 0. The probability of gettig absorbed ito the recurret class {7}, startig from 3 = S3, is 3 = + 9. Thus, our calculated absorptio probabilities match the probabilities we 0 0 ituited earlier. The importat thig to take away from this example is that, whe doig problems of this sort, (i.e give we do/do t eter a particular set of recurret classes), it is eccessary to rescale the trasitio probabilities of the ew chai, comig out of ALL the trasiet states. I other words, to fid each of the ew trasitio probabilities, we coditio o the give evet, that we do or do ot eter particular recurret classes. G. a) First let the p ij s be the trasitio probabilities of the Markov chai. The m k+ () = E[R k+ X 0 = ] = E[g(X 0 ) + g(x ) g(x k+ ) X 0 = ] = p i E[g(X 0 ) + g(x ) g(x k+ ) X 0 =, X = i] i= = p i E[g() + g(x ) g(x k+ ) X = i] i= = g() + p i E[g(X ) g(x k+ ) X = i] i= = g() + p i m k (i) i= ad thus i geeral m k+ (c) = g(c) + i= p cim k (i) whe c {,..., }. Note that the third equality simply uses the total expectatio theorem. b) v k+ () = V ar[r k+ X 0 = ] = V ar[g(x 0 ) + g(x ) g(x k+ ) X 0 = ] = V ar[e[g(x 0 ) + g(x ) g(x k+ ) X 0 =, X ]] + Page 7 of 8

8 6.0/6.3: Probabilistic Systems Aalysis (Fall 00) E[V ar[g(x 0 ) + g(x ) g(x k+ ) X 0 =, X ]] = V ar[g() + E[g(X ) g(x k+ ) X 0 =, X ]] + E[V ar[g() + g(x ) g(x k+ ) X 0 =, X ]] = V ar[e[g(x ) g(x k+ ) X 0 =, X ]] + E[V ar[g(x ) g(x k+ ) X 0 =, X ]] = V ar[e[g(x ) g(x k+ ) X ]] + E[V ar[g(x ) g(x k+ ) X ]] = V ar[m k (X )] + E[v k (X )] = E[(m k (X )) ] E[m k (X )] + p i v k (i) i= = p i m k (i) ( p i m k (i)) + p i v k (i) i= i= i= so i geeral v k+ (c) = i= p cim k (i) ( i= p cim k (i)) + i= p civ k (i) whe c {,..., }. Required for 6.3; optioal for 6.0 Page 8 of 8

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