This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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,
batteryoperated and pocketsized 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’(AlUA2UUAn) =Z(—1)i*1 Z 1F(Ail nAiznunAax "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<YsyX=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 ~1g 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—1Se)£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] ...
View
Full
Document
This note was uploaded on 03/18/2012 for the course STAT 1801 taught by Professor Mrchung during the Fall '10 term at HKU.
 Fall '10
 MrChung
 Statistics, Probability

Click to edit the document details