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### edu-2009-05-mfe-exam

Course: MAT 2070, Spring 2011
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of Society Actuaries Casualty Actuarial Society Canadian Institute of Actuaries Exam MFE Actuarial Models Financial Economics Segment Friday, May 15, 2009 2:00 p.m. 4:00 p.m. MFE INSTRUCTIONS TO CANDIDATES 13. When the supervisor tells you to do so, break the seal on the book and remove the answer sheet. 1. Write your candidate number here ____________. Your name must not appear. 2. Do not break the...

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Exam MFE/3F Spring 2009Answer KeyQuestion #Answer1E2B3B4D5E6D7D8B9E10C11C12E13A14E15C16A17A18A19D20C11. Answer: EWe have S0 = 10, = 0.05, = 0.3, r = 0.05, and h = 1. By (10.10),u = exp[(r )h + h ] = exp[(0.05
May 2007 Exam MFE Solutions1.Answer = (B)Let D = the quarterly dividend.Using formula (9.2), put-call parity adjusted for deterministic dividends, we have4.50 = 2.45 + 52.00 D e 0.01 D e 0.025 50 e 0.03 = 54.45 D ( 0.99005 + 0.97531) 50 0.970446 .R
COURSE 5MORNING SESSIONAPPLICATION OF BASIC ACTUARIAL PRINCIPLESSECTION A-WRITTEN ANSWER*BEGINNING OF EXAMINATION*COURSE 5MORNING SESSION1.(4 points)(a)Describe the coverage in a business overhead expense disability income policy.(b)You are gi
Exam C - May 2005*BEGINNING OF EXAMINATION*1.You are given:(i)A random sample of five observations from a population is:0.2(ii)0.71.11.3You use the Kolmogorov-Smirnov test for testing the null hypothesis, H 0 , that theprobability density func
Question #4Key: AThe random variable ln( S0.5 / 0.25) has a normal distribution with parameters = (0.015 0.352 / 2)(0.5) = 0.044375 2 = 0.352 (0.5) = 0.6125.The upper limit for the normal random variable is0.044375 + 1.645 0.6125 = 0.45149 .The upp
*BEGINNING OF EXAMINATION*1.For a dental policy, you are given:(i)Ground-up losses follow an exponential distribution with mean .(ii)Losses under 50 are not reported to the insurer.(iii)For each loss over 50, there is a deductible of 50 and a poli
FALL 2006EXAM C SOLUTIONSQuestion #1Key: EWith n + 1 = 16, we need the 0.3(16) = 4.8 and 0.65(16) = 10.4 smallest observations. They are0.2(280) + 0.8(350) = 336 and 0.6(450) + 0.4(490) = 466.The equations to solve are:220.3 = 1 and 0.65 = 1
FALL 2005EXAM C SOLUTIONSQuestion #1Key: DS (300) = 3 /10 (there are three observations greater than 300)H (300) = ln[ S (300)] = ln(0.3) = 1.204 .Question #2Key: AE ( X | ) = Var ( X | ) = = v = E ( ) = ; a = Var ( ) = 2 ; k = v / a = 1/ nnZ
Exam C SolutionsSpring 2005Question #1Key: D1(1 + x) 4F(x)compare to:0.5180, 0.20.8800.2, 0.40.9230.4, 0.60.9490.6, 0.80.9640.8, 1.0The CDF is F ( x) = 1 Observation (x)0.20.70.91.11.3Maximum difference0.5180.6800.5230.3490.1
SPRING 2007EXAM MLC SOLUTIONSQuestion # 1Key: Ep70 =p70 0.95== 0.9896p71 0.962375 x dx71= e 0.107 = 0.89854 p71 = e5 p70 = 0.9896 0.8985 = 0.889Question # 2Key: BAx = / ( + ) = / ( + 0.08 ) = 0.3443 = 0.0422Ax = + 2()Var aT =2=
Exam MLC Spring 2007 FINAL ANSWER KEYQuestion # Answer Question # Answer1 2 3 4 5 6 7 8 9 10 11 12 13 14 15E B D E C A E E E C A D C * D16 17 18 19 20 21 22 23 24 25 26 27 28 29 30B D C D C B C B A B A A C A D* The exam problem was defective. Mechan
cole des sciences de la gestion UQMDpartement des sciences conomiquesECO1012 Microconomie 1PaulEns. : Paul RousselLABO #4Problme #1 Le partage du fardeau de la taxeComme nous lavons vu en classe, cest llasticit-prix de loffre et de la demande qui d
Universit des Sciences et Technologies de LilleeU.F.R. de Mathmatiques Pures et AppliqueseeBt. M2, F-59655 Villeneuve dAscq CedexaIntroduction auCalcul des ProbabilitseProbabilits ` Bac+2 et plus si anits. . .eaeCharles SUQUETDEUG MIAS 2 et
Probabilits &amp; Statistiques L1: ExercicesDecember 28, 2008Dnombrements IExercice 1 Permutations 11. On permute les lettres du mot BAN C .(i) Nb de mots(ii) Nb de mots commenant par B2. On doit asseoir 7 personnes discernables sur 7 chaises discernab
MAT 1720 - Examen intraProfesseur: Sabin Lessard27 octobre 2008 - 15:30-17:30DIRECTIVE PDAGOGIQUE: Aucune documentation ni calculatrice.Question 1 (30 points)Une urne contient 5 boules numrotes de 1 5. On en tire 4 successivementau hasard AVEC REMIS
MAT 2070-H09PROBABILITS 1EXAMEN 1Date : Vendredi 13 fvrier 2009Nom :Prnom :Code permanentGroupe10 (Serge Alalouf)30 (Hassan Youns)INSTRUCTIONS1. Prendre grand soin de ne pas dsassembler les feuilles du prsent cahier (6 pages),qui doit tre remi
Solution Manual for:Introduction to Probability Models: Eighth Editionby Sheldon M. Ross.John L. WeatherwaxOctober 26, 2008IntroductionChapter 1: Introduction to Probability TheoryChapter 1: ExercisesExercise 8 (Bonferronis inequality)From the in
CHAPITRE 1Rappels sur le calcul direntiel ` une variable.eaCest ` Wilhelm Gottfried Leibnitz (1646 1716) et ` Isaac Newton (1642 1727) que nous devonsaalinvention du calcul direntiel et intgral. Dj` depuis ses dbuts, il sest avr un outil indispensa
CHAPITRE 2Fonctions de plusieurs variables relles, drives partielles.eeeTr`s souvent les fonctions rencontres sont dpendantes non pas dune seule variable, mais plutt deeeeoplusieurs. Par exemple, si nous frappons sur la membrane dun tambour, elle
CHAPITRE 3Continuit.eCe chapitre sera bref. Nous y dcrirons la notion de continuit pour les fonctions de plusieurs variables.eeBeaucoup de rsultats sur les fonctions de plusieurs variables ne sont vris quavec lhypoth`se que celleseeeeci et leurs
CHAPITRE 4Approximation linaire, le gradient et les drives directionnelles.eeeSoit y = f (x), une fonction relle dune seule variable. La drive de f au point x = c esteeedydx=cdfdx= limch0f (c + h) f (c)het ceci peut tre rcriteey =dy
CHAPITRE 5R`gle de chaines et galit des drives partielles mixtes.eeeeeNous allons premi`rement noncer la r`gle de chaines pour une fonction f (u, v ) de deux variables u, veeeelles-mmes fonction de deux autres variables x, y .eeProposition 5.
CHAPITRE 6Maximums et minimums relatifs, optimisation.Nous allons initialement considrer dans ce chapitre que des fonctions de deux variables pour loptimisa etion sans contrainte. Cependant la thorie peut tre prsente dans un cadre plus gnral, mais alor
CHAPITRE 7Rappel sur lintgrale simple.eLes prochains chapitres traiteront de lintgration. Dans un premier temps, nous rappellerons ce questelintgrale simple (lintgration pour les fonctions dune seule variable relle), ainsi que le thor`me fondaeeee
CHAPITRE 8Intgrales doubles.eDans ce chapitre, nous dnirons lintgrale double dune fonction f (x, y ) sur une rgion borne du planeeeeet nous prsenterons quelques-unes de ces proprits. Ensuite nous verrons comment calculer ces intgraleseeeeau mo
CHAPITRE 9Intgrales triples.eDans ce chapitre, nous dnirons lintgrale triple dune fonction f (x, y, z ) sur une rgion borne de R3eeeeet nous prsenterons quelques-unes de ces proprits. Ensuite nous verrons comment calculer ces intgraleseeeeau m
CHAPITRE 10Jacobien, changement de coordonnes.eDans ce chapitre, nous allons premi`rement rappeler la dnition du dterminant dune matrice. Nouseeenous limiterons au cas des matrices dordre 2 2 et 3 3, bien que les rsultats noncs sont vrais dans une
CHAPITRE 11Applications de lintgrale multiple.eCe chapitre sera tr`s bref. Il existe un grand nombre dapplications de lintgrale multiple. Il sut deeepenser aux notions desprance et de variance en probabilits ou encore des quations intgrales. Beaucou
CHAPITRE 12Intgrales impropres, fonctions gamma et bta et transforme de Laplace.eeeDans ce chapitre, nous revenons aux intgrales simples, mais cette fois soit lintervalle dintgration, soiteela fonction ` intgrer, soit les deux ne sont pas borns. T
LSU - ISDS - 3115
CHAPTER 1: OPERATIONS AND PRODUCTIVITYTRUE/FALSE1. Some of the operations-related activities of Hard Rock Caf include designing meals and analyzing them for ingredient cost and labor requirements. True (Global company profile, easy) The production proce
RIT - PHY - 303
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Kettering - MECH - 115
00 introductionmy expectationsyour expectationsgradingon-line quizzeswhat I expect from you: active learningbefore classread chapter before we beginwarm-up quiz (on-line) on reading*during classthink/pair/share quizzes*be prepared with question
Kettering - MECH - 115
01 Motiondiagramsuse movie to studymotionfixed time betweenframesdon't pan the camera!usually 1/30th of a secstrobe pictures arefasterfaster objectshave greater positionchange between framesWhich car is going faster, A or B? Assume there are
Kettering - MECH - 115
02 Velocity &amp;accelerationrv(average) velocity: tvector displacement rtime interval tdraw the velocity vector vvthe same size as the displacement vectordon't forget the difference in meaning, unitsvelocity:how fast it's goingAND in what dir
Kettering - MECH - 115
03 Motion in1 dimensioncoordinate systemoriginpositive direction: conventionx: right y: upposition-time graphsposition on vertical axis(dependent variable)even if position ishorizontaltime on horizontal axis(independent variable)can pick off
Kettering - MECH - 115
04 position, velocity and accelerationif v is constant s=v s tdisplacementchange in positionarea under v(t)workbook:2.2b page 2-3displacement when v isn't constanttfs f = si v s t dttiintegral = areaunder curvebetween curve andaxisintegrat
Kettering - MECH - 115
05 problem solving: constant amodel: particle modelvisualize: shift between these as neededmotion diagrampictorial representation: labelv fs = v is a s tsolve12s f = si v is t a s t 2assess22v fs = v is 2 a s s units? sensible? believable?
Kettering - MECH - 115
06 Instantaneousaccelerationa s=d vsdt v s = a s dtfollow up:workbook 2.28page 2-16Rank in order, from largest to smallest,the accelerations aA aC at points A C.A) aA &gt; aB &gt; aCB) aC &gt; aA &gt; aBC) aC &gt; aB &gt; aAD) aB &gt; aA &gt; aCWhich velocity-vers
Kettering - MECH - 115
07 Vectors and componentsvector: magnitude &amp; directionscalar: magnitude onlyreview:follow up:Workbook 3.1-3Page 3-1Which figure showsA1 A 2 ?A1 A 2 A 3 ?multiplying a vector by a scalar. by a positive scalar. by a negative scalarfollow up: 3.