1801_0607sem1

1801_0607sem1 - THE UNIVERSITY OF HONG KONG I DEPARTMENT OF...

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Unformatted text preview: THE UNIVERSITY OF HONG KONG I DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT1801 Probability and Statistics: Foundations of Actuarial Science December 15, 2006 Time: 9:30 am. - 11:30 a.m. Candidates taking examinations that permit the use of calculators may use any cal- culator which fulfils the following criteria: (a) it should be self—contained, silent, battery-operated and pocket-sized and (b) it should have numeral—display facilities only and should be used only for the purposes of calculation. It is the candidate’s responsibility to ensure that the calculator operates satisfactorily and the candidate must record the name and type of the calculator on the front page of the examination scripts. Lists of permitted/prohibited calculators will not be made available to candidates for reference, and the onus will be on the candidate to ensure that the calculator used will not be in violation of the criteria listed above. Answer ALL SIX questions. Marks are shown in square brackets. 1. There are r distinguishable visitors inside a lift which can reach n floors. Suppose that every visitor is equally likely to make an exit on each of the n floors, independent of the other visitors. (a) How many distinct exit patterns are there? [2 marks] (b) Let m be a positive integer between 1 and n. Derive the probability that nobody makes an exit on any of the lowest m floors. [3 marks] (c) Give an expression for the probability that there is at least one floor on which nobody makes an exit. [4 marks] [Hint You may quote the Inclusion—Exclusion Formula without proof: lP’(AlUA2U---UAn) =Z(—1)i*1 Z 1F(Ail nAizn-unAax "J j=l i1<i2<"'<ij for any events A1,A2, . . . ,An.] ((1) Calculate the expected number of floors on which nobody makes an exit. [4 marks] 2. A candidate is requested to solve a problem and is given a multiple choice of 19 answers, among which one is correct. Suppose that the candidate either finds an analytic way to solve the problem or makes a random guess at the correct answer. The probability that he uses an analytic method is p, in which case he has a positive probability of a to arrive at the correct answer. 5&AS: STAT1801 Probability and Statistics: Foundations of Actuarial Science 2 (a) It is found that the candidate has given the correct answer. What is the probability that he answered the question by analysis rather than a random guess? [3 marks] (b) A second candidate, who is equally capable as, and acts independently of, the first candidate, adopts a strategy identical to the first one’s for solving the same problem. Both candidates are found to give the correct answer. What is the probability that they both answered the questions by analysis? [4 marks] 3. A straight rod of unit length is broken at random into two segments, the longer of which is then broken at random into two further segments. Identifying the straight rod with the interval [0,1], we can represent the first breaking point by a random variable X uniformly distributed over [0, 1] and the second breaking point by a random variable Y which is uniformly distributed over [0,X] ifX > 1/2 and over [X, 1] ifX < 1/2. (a) Show that, for y E (0, 1), M if 1/2 < a: < 1, 1P<Ysy|X=w>= m 0} ifo < z <1/2. l—a: [4 marks] (b) Show that (X , Y) has the joint density function % if1/2<x<land0<y<x, 1 f($aZ/)= 1 ifO<x<1/2and:c<y<1, —a: 0 otherwise. [4 marks] (c) Show that the three segments of the rod can form the three sides of a triangle if either X < 1/2 and 1/2 < Y < X + 1/2, or (ii) X >1/2 andX—1/2<Y<1/2. [4 marks] [Hint The three segments can form a triangle if and only if the total length of any two segments must exceed that of the remaining segment. ] (d) Show that the three segments of the rod can be used to construct a triangle with probability 21n 2 — 1. [6 marks] 4. Let X1, X2, . . . be a sequence of independent and identically distributed ran- dom variables with a common mean ,u and common variance 02. Let N be a random positive integer, independent of X1,X2, . .. , with ]E [N] = z/ and Var(N) = r2. Define S = X,, e(n) = IE[S I N = n] and v(n) = Var(S| Nzn), forn: 1,2,.... S&AS: STAT1801 Probability and Statistics: Foundations of Actuarial Science 3 (a) Write down explicit expressions for the functions e(n) and [4 marks] (b) Show that e(N) has mean ,u,1/ and variance #272. [4 marks] (c) Show that v(N) has mean 021/. [2 marks] ((1) Deduce from (b) and (c), or otherwise, that S has mean [w and variance #272 + 021/. [6 marks] 5. A mysterious organism is capable of resurrections, so that it can start a new life immediately after death and repeat this cycle indefinitely. Let Xi be the duration of the 2th life of the organism so that Sn 2 EL Xi gives the time of its nth resurrection. Define So = 0 by convention. Assume that X1, X2, . . . are independent unit—rate exponential random variables with the density function f() e”, w>0, $ : 0, 3:30. Let t > 0 be a fixed time point. (a) Find the mean lifetime, that is IE [X 1], of the organism. [2 marks] (b) It is known that Sn has the gamma density function n—l —:c x e ', z > 0, grim) = (n — 1). 0, a: g 0. Show that 22:1 9,.(55) = 1 for :1: > 0. [2 marks] (c) Suppose that the organism is living its Nth life at time t, so that N is a positive random integer. (i) Show that [3 marks] (ii) Deduce from that 00 t IP(XNSx)=lP(tSX1§$)+Z/ P(t—s§-Xn§x)gn_1(s)ds. 71:2 0 [6 marks] (iii) Deduce from (b) and (c)(ii) that X N has the density function 0, :r S 0, 11(37) = ate—’3, 0 < a: S t, (1 +t) 6‘75, 1: > t. [8 marks] S&AS: STAT1801 Probability and Statistics: Foundations of Actuarial Science 4 (iv) Show that E [X N] = 2 — e". [3 marks] [Hint You may find the following integrals useful: u u f :c {Ida = 1— (1 + u)e_" and / $2e—mdz = 2 — (2+ 2u +u2)e_“, 0 0 for any u > 0.] (v) Do your answers to (a) and (c)(iv) contradict each other? Explain. [3 marks] 6. Let X1,X2, . .. and Y1, Y2, . .. be two sequences of positive random variables such that lim lP’(Xn S .73) = for all :1: > 0, n—>oo and lim lP’(]Yn—1|>e)=0 for alle>0, 11—)00 where F is a continuous cumulative distribution function. Let a: > 0 and e 6 (0,1) be fixed. Define, for u > 0, An(u) = ll"(Xn g u, |Yn ~1|g e) and Bn(u) = P(Xn g u, |Yn — 1] > e). (a) Show that limpMoo Bn (b) Show that limnkoo An (c) Show that 0. [2 marks] [3 marks] lim lP’(XnYn S 3:) = lim ll’(XnYn g x, [Yn — 1| S e). fl-)OO n—)oo [4 marks] (d) Show that An( 9: )gIP(XnYnSx,]Yn—1|Se)£An( x 1+6 1—6 [4 marks] (e) Deduce from (b), (c) and (d) that XnYn converges in distribution to a random variable with distribution function F. r [6 marks] ...
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This note was uploaded on 03/18/2012 for the course STAT 1801 taught by Professor Mrchung during the Fall '10 term at HKU.

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1801_0607sem1 - THE UNIVERSITY OF HONG KONG I DEPARTMENT OF...

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