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1 B. Sc. Examiatio by course uit 26 MTH734U: Topics i Probability & Stochastic Processes[SOLUTIONS] Duratio: 3 hours Date ad time: To Be Determied Apart from this page, you are ot permitted to read the cotets of this questio paper util istructed to do so by a ivigilator. You should attempt ALL questios. Marks awarded are show ext to the questios. Calculators are NOT permitted i this examiatio. The uauthorised use of a calculator costitutes a examiatio offece. Complete all rough workigs i the aswer book ad cross through ay work that is ot to be assessed. Possessio of uauthorised material at ay time whe uder examiatio coditios is a assessmet offece ad ca lead to expulsio from QMUL. Check ow to esure you do ot have ay otes, mobile phoes, smartwatches or uauthorised electroic devices o your perso. If you do, raise your had ad give them to a ivigilator immediately. It is also a offece to have ay writig of ay kid o your perso, icludig o your body. If you are foud to have hidde uauthorised material elsewhere, icludig toilets ad cloakrooms it shall be treated as beig foud i your possessio. Uauthorised material foud o your mobile phoe or other electroic device will be cosidered the same as beig i possessio of paper otes. A mobile phoe that causes a disruptio i the exam is also a assessmet offece. Exam papers must ot be removed from the examiatio room. Examier(s): Dr Dudley Stark, Dr Jes Starke TURN OVER

2 Page 2 MTH734U (26) Questio. (a) [Stated i lecture] (b) [Proved i lecture] We kow that Therefore, N(t) = max{ : S t}. 3 M(t) = E(N(t)) = M(t) = P(N(t) ). = {N(t) } = {S t}. 2 P(S t) = = (c) [Proved i Lecture/Similar to Coursework] F (t). 2 The p.d.f. of a Expoetial(θ) distributed radom variable is f(x) = θe θx ad the c.d.f. is F (x) = e θx. µ = µ x xf(x) dx = [ F (u)du = θ = θ = xθe θx dx = θ. x x [ ( e θu ]du e θu du = θ(θ ( e θx )) = e θx = F (x). 3 (d) [Theorem proved i Lecture for reewal process but o]y stated for delayed reewal processes] M D (t) = E(N D (t)) = E(E(N D (t) X )) = E(N D (t) X = x)g(x) dx. If x t, the give that X = x, N D (t) is distributed as + N(t x). 2 If x t, the give that X = x, N D (t) =. Therefore, M D (t) = = = G(t) + E( + N(t x))g(x) dx 2 g(x) dx + M(t x)g(x) dx M(t x)g(x) dx.

3 MTH734U (26) Page 3 Questio 2. (a) (i) [Stated i lecture] It is the semi-markov process with trasitio matrix P i,j for which the process always stays i a state exactly oe uit of time before makig a trasitio. (ii) [Stated i lecture] The equatios for the statioary distributio are π i = j π j P j,i ad (iii) [Similar to coursework] The equatios become π i =. i π =, π 2 = π + π 3, π 3 = π 2, π 4 = π 4, 4 π i =. 2 i= The solutio to them is (π, π 2, π 3, π 4 ) = (, α, α, 2α), α /2. 2 There is more tha oe solutio to these equatios ad so there is ot a uique statioary disributio. (b) [Similar to coursework] Let X i be the time of the ith trip from Birmigham to Machester. Let Y i be the time of the ith trip from Machester to Leeds. Let Z i be the time of the ith trip from Leeds to Birmigham. E(X i ) = (6 + 2)/2 = 9 miutes, 2 E(Y i ) = (6 + )/2 = 8 miutes, E(Z i ) = (8 + 2)/2 = miutes. By the Reewal Rewards theorem with reward R i = Y i, 2 the proportio of time drivig from Leeds to Birmigham is E(Z i ) E(X i ) + E(Y i ) + E(Z i ) 3 = = 27 = TURN OVER

4 Page 4 MTH734U (26) Questio 3. (a) [Similar to examples preseted i lecture] (i) If X Expoetial(θ), the P(X > s + t X > s) = P(X > t). 2 (ii) P(X > s + t X > s) = (b) (i) The aswer is e !. 2 = P({X > s + t} {X > s}) P(X > s) P(X > s + t) P(X > s) = e θ(s+t) e θs = e θt = P(X > t). (ii) The aswer is e ! e !. 4 (iii) Let S i be the arrival time of the ith customer. Let U i, i =,..., 5 be i.i.d. Uiform[, ] distributed radom variables ad let U (i), i =,..., 5 be their order statistics. The, ( 5 ) ( 5 ) P I[S i /3] N() = 5 = P I[U (i) /3] i= i= ( 5 ) = P I[U i /3]. The idicator radom variables I[U i /3] are idepedet ad P(U i /3) = /3, so 5 i= I[U i /3] Biomial(5, /3). 5 The probability that exactly oe customer arrived i the first 2 miutes is ( ) 5 (/3) (2/3) 4 = 8/ i=

5 MTH734U (26) Page 5 Questio 4. (a) [Stated ad proved i lecture] (i) P(t) = e tg 2 (ii) P (t) = GP(t) P (t) = P(t)G (b) [Similar to coursework] (i) Let λ = 3, /µ = /4 µ = 4. P(X(h) = X() = ) = λh + o(h) q, = λ. P(X(h) = 2 X() = ) = o(h) q,2 = P(X(h) = X() = ) = (µh + o(h))( λh + o(h)) = µh + o(h) q, = µ. P(X(h) = 2 X() = ) = λh + o(h) q,2 = λ P(X(h) = X() = 2) = ( ) 2 (µh + o(h))( µh + o(h)) = 2µh + o(h) q 2, = 2µ. P(X(h) = X() = 2) = o(h) q 2, =. The rows of the G must sum to. We have λ λ G = µ λ µ λ = 2µ 2µ /3 mark for each etry of G (rouded to the earest iteger). (ii) The π i must satisfy the equatios 3π + 4π = () 3π 7π + 8π 2 = (2) 3π 8π 2 = (3) π + π + π 2 =. (4) TURN OVER

6 Page 6 MTH734U (26) These equatios ca be solved i the followig way, for example. From (), we have π = 4 3 π. From (3) we have π 2 = 3 8 π. Puttig the above expressios i (4) gives ( 4 π ) = 8 so π = π = 32 65, π 2 = (c) I the log ru the average umber of customers i the shop is π + π + 2 π 2 = π + 2π 2 =

7 MTH734U (26) Page 7 Questio 5. (a) [Stated i lecture] B(t) N(, t). 3 (b) Derived i lecture] B(t) N(, t) is easily see to be a Gaussia process. I detail, suppose that α i ad t i, i =, 2,..., are costats ad time poits. The i= α i B(t i ) = i= α i(b(t + t i ) B(t )) = i= α ib(t + t i ) + ( i= α i)b(t ) is ormally distributed because B(t) is a Gaussia process. 2 Therefore, B(t) is a Gaussia process. Moreover, for s t, sice we kow Cov(B(t ), B(t 2 )) = mi(t, t 2 ), Cov( B(s), B(t)) = Cov(B(t + s) B(t ), B(t + t) B(t )) = Cov(B(t + s), B(t + t)) Cov(B(t ), B(t + t)) Cov(B(t + s), B(t )) + Cov(B(t ), B(t )) = (t + s) t t + t = s = Cov(B(s), B(t)). 3 B(t) is a Gaussia process with the same expectatio ad covariace fuctios as B(t), therefore it is idetical to B(t) 2. (c) [Coursework problem] (i) Let A be a subset of realizatios of B(t). The, ( ) S t lim P A = P (B(t) A), where x is the largest iteger less tha or equal to x. 3 (ii) The expected time util S t reaches a or b approaches E(τ) as by the Ivariace Priciple. 2 The time util S t reaches a or b is T (α,β), where α = a, β = b, ( ) which has expectatio E T (α,β) E(T (α,β) = = (a )(b ) = ab. 3 Therefore, E(τ) = ab. Ed of Paper.

Apart from this page, you are not permitted to read the contents of this question paper until instructed to do so by an invigilator.

B. Sc. Examination by course unit 216 MTH6934: Topics in Probability & Stochastic Processes[SOLUTIONS] Duration: 2 hours Date and time: To Be Determined Apart from this page, you are not permitted to read