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1 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 the contents of this question paper until instructed to do so by an invigilator. You should attempt ALL questions. Marks awarded are shown next to the questions. Calculators are NOT permitted in this examination. The unauthorised use of a calculator constitutes an examination offence. Complete all rough workings in the answer book and cross through any work that is not to be assessed. Possession of unauthorised material at any time when under examination conditions is an assessment offence and can lead to expulsion from QMUL. Check now to ensure you do not have any notes, mobile phones, smartwatches or unauthorised electronic devices on your person. If you do, raise your hand and give them to an invigilator immediately. It is also an offence to have any writing of any kind on your person, including on your body. If you are found to have hidden unauthorised material elsewhere, including toilets and cloakrooms it shall be treated as being found in your possession. Unauthorised material found on your mobile phone or other electronic device will be considered the same as being in possession of paper notes. A mobile phone that causes a disruption in the exam is also an assessment offence. Exam papers must not be removed from the examination room. Examiner(s): Dr Dudley Stark, Dr Jens Starke TURN OVER

2 Page 2 MTH6934 (216) Question 1. (a) [Stated in lecture] N(t) = max{n 1 : S n t}. 5 (b) [Proved in lecture] M(t) = E(N(t)) = P(N(t) n). 2 n=1 We know that Therefore, {N(t) n} = {S n t}. 3 M(t) = P(S n t) = F n (t). 2 n=1 n=1 (c) [Proved in Lecture/Similar to Coursework] The p.d.f. of an Exponential(θ) distributed random variable is f(x) = θe θx and the c.d.f. is F (x) = 1 e θx. 2 µ = 1 µ x xf(x) dx = [1 F (u)du = θ = θ xθe θx dx = θ 1 1. x x [1 (1 e θu ]du e θu du = θ(θ 1 (1 e θx )) = 1 e θx = F (x). 5

3 MTH6934 (216) Page 3 Question 2. (a) (i) [Stated in lecture] It is the semi-markov process with transition matrix P i,j 1 for which the process always stays in a state exactly one unit of time before making a transition. 1 (ii) [Stated in lecture] The equations for the stationary distribution are π i = j π j P j,i 2 and (iii) [Coursework problem] The equations become π i = 1. 1 i π 1 = π 2, π 2 = π 1, π 3 = π 4, π 4 = π 3, 4 π i = 1. 2 i=1 The solution to them is (b) [Similar to coursework] (π 1, π 2, π 3, π 4 ) = (α, α, 1/2 α, 1/2 α), α 1/2. 2 Let X i be the time of the ith trip from Birmingham to Manchester. Let Y i be the time of the ith trip from Manchester to Birmingham. E(X i ) = (6 + 12)/2 = 9 minutes, 2 E(Y i ) = (6 + 1)/2 = 8 minutes, 2 By the Renewal Rewards theorem with reward R i = Y i, 2 the proportion of time driving from Manchester to Birmingham is E(Y i ) E(X i ) + E(Y i ) 3 = = 8 17 = TURN OVER

4 Page 4 MTH6934 (216) Question 3. (a) [Similar to examples presented in lecture] (i) If X Exponential(θ), then P(X > s + t X > s) = P(X > t). 3 (ii) P(X > s + t X > s) = (b) (i) The answer is e /5!. 3 = P({X > s + t} {X > s}) P(X > s) P(X > s + t) 2 P(X > s) = e θ(s+t) e θs 2 = e θt 1 = P(X > t). 1 (ii) The probability that exactly one customer arrived in the first 2 minutes is ( ) 5 (1/3) 1 (2/3) 4 = 8/

5 MTH6934 (216) Page 5 Question 4. (a) [Stated and proved in lecture] (i) P(t) = e tg 2 (ii) P (t) = GP(t) 2 P (t) = P(t)G 2 (b) [Similar to coursework] (i) Let λ = 3, 1/µ = 1/4 µ = 4. 1 P(X(h) = 1 X() = ) = λh + o(h) q,1 = λ. P(X(h) = 2 X() = ) = o(h) q,2 = P(X(h) = X() = 1) = (µh + o(h))(1 λh + o(h)) = µh + o(h) q 1, = µ. P(X(h) = 2 X() = 1) = λh + o(h) q 1,2 = λ P(X(h) = 1 X() = 2) = ( ) 2 (µh + o(h))(1 µh + o(h)) 1 = 2µh + o(h) q 2,1 = 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 entry of G (rounded to the nearest integer). (ii) The π i must satisfy the equations 3π + 4π 1 = 1 (1) 3π 7π 1 + 8π 2 = 1 (2) 3π 1 8π 2 = 1 (3) π + π 1 + π 2 = 1. 1 (4) TURN OVER

6 Page 6 MTH6934 (216) These equations can be solved in the following way, for example. From (1), we have π = 4 3 π 1. From (3) we have π 2 = 3 8 π 1. Putting the above expressions in (4) gives ( 4 π ) = 1 8 so π 1 = π = 32 65, π 2 =

7 MTH6934 (216) Page 7 Question 5. (a) [Stated in lecture] B(t) N(, t). 4 (b) Derived in lecture] B(t) N(, t) is easily seen to be a Gaussian process. In detail, suppose that α i and t i, i = 1, 2,..., n are constants and time points. Then n i=1 α i B(t i ) = n i=1 α i(b(t + t i ) B(t )) = n i=1 α ib(t + t i ) + ( n i=1 α i)b(t ) is normally distributed because B(t) is a Gaussian process. 2 Therefore, B(t) is a Gaussian process. 2 Moreover, for s t, since we know Cov(B(t 1 ), B(t 2 )) = min(t 1, 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)). 5 B(t) is a Gaussian process with the same expectation and covariance functions as B(t), therefore it is identical to B(t) 2. (c) [Coursework problem] Let A be a subset of realizations of B(t). Then, ( ) S nt lim P A = P (B(t) A), n n where x is the largest integer less than or equal to x. 5 End of Paper.