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Course: ACTSC 431, Fall 2011
School: Waterloo
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to Answers Practice Questions 1 ACTSC 431/831, FALL 2011 1. (a) FY (y ) = Pr{Y y |N = 0} Pr{N = 0} + Pr{Y y |1 N 5} Pr{1 N 5} + Pr{Y y |N &gt; 5} Pr{N &gt; 5} 0, y &lt; 0, y/100 y/200 1 0.5 e 0.2 e , y 0. = (b) The probability is Pr{Y &gt; 500} = 1 FY (500) = 0.019786. (c) Pr{0 &lt; Y &lt; 300} = FY (300) FY (0) = 0.63048. (d) The mean of the total loss is E (Y ) = 0.3 0 +...

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to Answers Practice Questions 1 ACTSC 431/831, FALL 2011 1. (a) FY (y ) = Pr{Y y |N = 0} Pr{N = 0} + Pr{Y y |1 N 5} Pr{1 N 5} + Pr{Y y |N > 5} Pr{N > 5} 0, y < 0, y/100 y/200 1 0.5 e 0.2 e , y 0. = (b) The probability is Pr{Y > 500} = 1 FY (500) = 0.019786. (c) Pr{0 < Y < 300} = FY (300) FY (0) = 0.63048. (d) The mean of the total loss is E (Y ) = 0.3 0 + 0.5 100 + 0.2 200 = 90. (e) The 20th percentile of Y is 0.2 = 0. (f) The 80th percentile of Y is the root of equation: FY (y ) = 0.8, y > 0. Solving the equation, we obtain that the 80th percentile of Y is 0.8 = 153.87. 2. (a) Note that 0 Y L 70. FY L (y ) = 0, Pr{X 80 y } = 80+y , 200 Pr{40 + 0.75(X 120) y } = 1, y < 0, 0 y < 40, y +50 , 40 y < 70, 150 y 70. (b) Note that Y P = Y L |X > 80 and 0 < Y P 70. FY P (y ) = 0, y Pr{X 80 y | X > 80} = 120 , Pr{40 + 0.75(X 120) y | X > 80} = 1, y 0, 0 < y < 40, y 10 , 40 y < 70, 90 y 70. (c) The probability is FY L (50) FY L (40) = 1/15. (d) The probability is FY P (70) FY P (30) = 5/12. (e) The mean of the per loss of the policy is E (Y L ) = or E (Y L ) = 0 120 160 1 1 (40 + 0.75(x 120)) dx + dx 200 200 80 120 200 1 + 70 dx = 29 200 160 (x 80) SY L (y )dy = 29. (f) The median of Y P is the solution to FY P (y ) = 0.5, 40 < y < 70. Hence 0.5 = 55. (g) The 90th percentile of Y P is 0.9 = 70. 3. Note that the cdf and sf of X are 0, x < 0, x , 0 x < 200, F ( x) = 200 1, x 200, 1, x < 0, 200 x and S (x) = , 0 x < 200, 200 0, x 200. 1 (a) The cdf of Y P is given by 0, y 0, y S (y + 50) , 0 < y < 150, = F Y P (y ) = 150 1 , y > 0, S (50) 1, y 150, y 0, 0, which is the cdf of the uniform distribution U (0, 150). (b) The cdf of Y L is given by FY (y L ) = 0, y < 0, 0, y < 0, y + 50 = , 0 y < 150, 200 F (y + 50), y 0, 1, y 150. (c) The cdf of Y is given by FY (y ) = 0, y < 0, y F (y ), y < 50, , 0 y < 50, = 200 1, y 50, 1, y 50. (d) The expectation of the per-loss variable is given by E (Y L ) = 56.25 200 1 50 (x 50) 200 dx = (e) We have E (Y ) = E (X 50) = 050 S (y )dy = 43.75 or E (Y ) = E (X 50) = 50 50 2 2 0 xf (x)dx +50 (1 F (50)) = 43.75, and E (Y ) = E [(X 50) ] = 2 0 yS (y )dy = 2083.33 or E (Y 2 ) = E [(X 50)2 ] = 050 x2 f (x)dx + 502 (1 F (50)) = 2083.33. Hence, the variance of the limit loss variable is V ar(Y ) = 169.27. (f) By (a), we know that Y P has the uniform distribution U (0, 150), hence e(50) = L) X 50) E (Y P ) = 75 or e(50) = E [(F (50)+ ] = 1E (Y(50) = 75. 1 F (g) We have V ar[(X 50)2 |X > 50] = V ar[(Y P )2 ] = E [(Y P )4 ] (E [(Y P )2 ])2 . By (a), we know that Y P has the uniform distribution U (0, 150), hence E [(X 50)2 |X > 50] = E [(Y P )2 ] = V ar(Y P ) + E [(Y P )]2 = 7500 or E [(X 50)2 |X > (x50)2 f (x)dx 50] = E [(Y P )2 ] = 0 y 2 fY P (y )dy = 7500 or E [(X 50)2 |X > 50] = 50 1F (50) = 3 P4 7500. Furthermore, E [(Y ) ] = 4 0 y SY P (y )dy = 101250000. Thus, V ar[(X 50)2 |X > 50] = 4.5 107 . 4. (a) We know that X1 + X2 still has a normal distribution. Thus, V aR (X1 + X2 ) = 2 2 1 + 2 + 1 + 2 + 21 2 1 (). (b) Note that 1 < < 1. Thus, it is easy to verify that V aR (X1 + X2 ) V aR (X1 ) + V aR (X2 ) for 0.5 < 1, and V aR (X1 + X2 ) > V aR (X1 ) + V aR (X2 ) for 0 < < 0.5. (c) Since X1 + X2 still has a normal distribution, we have T V aR (X1 + X2 ) = 1 ( 2 2 1 + 2 + 1 + 2 + 21 2 ( )) . 1 (d) Since 1 < < 1, (x) > 0 for all x (, ), it is easy to see that T V aR (X1 + X2 ) T V aR (X1 ) + T V aR (X2 ) for all 0 < < 1. 2
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Waterloo - ACTSC - 431
Practice Questions 1 ACTSC 431/831, FALL 20111. Let Y be the total loss in an insurance portfolio and N be the number of claims in theportfolio. If N = 0, then Y = 0. If 1 N 5, then Y has an exponential distributionwith mean 100. If N &gt; 5, then Y has a
Waterloo - ACTSC - 431
Answers to Practice Questions 2 ACTSC 431/831, FALL 20111. Let X1 = 1/ X and X2 = eX .(a) We have E (X1 ) =1f (x)dx0xX= and E (X2 ) = E (eX ) = MX (1) = 16/9.Thus, E (Y ) = 0.2E (X1 ) + 0.8E (X2 ) = 1.7767.(b) The probability is Prcfw_Y &gt; E (Y )
Waterloo - ACTSC - 431
Practice Questions 2 ACTSC 431/831, FALL 20111. Suppose that loss Y has a two-point mixture distribution with the following cdf:FY (y ) = 0.2F1 (y ) + 0.8F2 (y ) for all y,where F1 (y ) and F2 (y ) are the distribution functions of 1X and eX , respecti
Waterloo - ACTSC - 431
Answers to Practice Questions 3 ACTSC 431/831, FALL 20111. (a) Prcfw_X &lt; 1 = 0.75.(b) The distribution of X is neither DFR nor IFR.(c) X 4 has a heavier tail than X .2. (a) We have a = 600. Then, V ar(X ) = 35.42.(b) We havexFX (x) =fX (t)dt =0,
Waterloo - ACTSC - 431
Practice Questions 3 ACTSC 431/831, FALL 20111. The conditional hazard rate function of loss X , given = , is h(x|) = x3 . The distributionof is a gamma distribution G(2, 4).(a) Calculate the probability that the loss is less than one.(b) Justify if t
Waterloo - ACTSC - 431
ANSWERS TO PRACTICE QUESTIONS 4 ACTSC 431/831, FALL 20111. Use the pgf of X to prove that X is innitely divisible.2. (a) Let v = Prcfw_Xj 50. Use PN P (z ) = PN L (1 + v (z 1) to prove that N P isinnitely divisible.i. V ar(N P ) = 2108.05.(b)ii. Cov
Waterloo - ACTSC - 431
Practice Questions 4 ACTSC 431/831, FALL 20111. Prove that if X has a mixed Poisson distribution and the mixing distribution is innitelydivisible, then X is innitely divisible.2. Let N L be the number of losses. The size of the j th loss is Xj . Assume
Waterloo - ACTSC - 431
Practice Questions 5 ACTSC 431/831, FALL 20111. The aggregate losses for an insurer is S , which has the following pdffS (x) =2500/x5 , x 5;0,otherwise.The premium charged by the insurer is equal to (1+)E [S ], which is called the expectedvalue pri
Waterloo - ACTSC - 431
ANSWERS TO PRACTICES QUESTIONS 5 ACTSC 431/831, FALL 20111. We have625, x 5,x4250020x 5 dx = ,x35502 2500xdx (E (S )2 = .5x95FS (x) = Prcfw_S x = 1 E (S ) =V ar(S ) =Thus, Pr cfw_S (1 + )E [S ] = Pr S E [S ] + V ar(S ) = 0.95 gives =
Waterloo - ACTSC - 431
ANSWERS TO PRACTICE QUESTIONS 6 ACTSC 431/831, FALL 20111. (a) The probability isPrcfw_N 3 = 1, N 5 = 2 = Prcfw_N 3 = 1 Prcfw_N 2 = 1 = 0.02747.12121212(b) The probability is Prcfw_N 1 = 0|N 1 = 3 =1252(c) The probability is2Prcfw_N1 =0 Prcf
Waterloo - ACTSC - 431
Practice Questions 6 ACTSC 331, FALL 20111. The surplus process of an insurer is Ut = u + 1.2 t Nt Xi , t 0, where u 0 is thei=1initial surplus; the claim number process cfw_Nt , t 0 is a Poisson process with rate1; and the claim sizes cfw_X1 , X2 , .
Waterloo - ACTSC - 431
Project ACTSC 831, FALL 2011Project deadline: 4:00 pm, Friday, December 9. Please submit a hard copy of your projectto M3 4012 or email a PDF le of your project to jcai@uwaterloo.ca by the deadline.Project requirement: Projects must be typed using LaTe
Waterloo - ACTSC - 431
Review Notes for Loss Models 1 ACTSC 431/831, Fall 2011Part 1 Random Variables and Distributional Quantities The distribution function (df ) or cumulative distribution function (cdf ) andsurvival function (sf ) of a random variable (rv) X are dened by
Waterloo - ACTSC - 431
Review Notes for Loss Models 1 - ACTSC 431/831, FALL 2011Part 2 Severity Models1. Severity models are distributions that are used to model the distribution of theamount of a claim/loss.2. The distribution function F (y ) of a random variable Y is call
Waterloo - ACTSC - 431
Review Notes for Loss Models 1 - ACTSC 431/831, FALL 2011Part 3 Frequency Models1. Frequency models are used to model the number of events or claims/losses.2. A counting random variable N is a nonnegative integer-valued random variablewith pf pk = Pcf
Waterloo - ACTSC - 431
Review Notes for Loss Models 1 - ACTSC 431/831, FALL 2011Part 4 Frequency and Severity with Coverage ModicationsLet X be the ground-up loss for an insurance policy or an insurer and assume that X is acontinuous r.v. with cdf F (x), sf S (x), and pdf f
Waterloo - ACTSC - 431
Review Notes for Loss Models 1 - ACTSC 431/831, FALL 2011Part 5 Aggregate Loss ModelsRoughly speaking, an aggregate loss model is used to describe the total loss of an insurance portfolio in a xed time period.1. Individual Risk Model: There are n polic
Waterloo - ACTSC - 431
Review Notes for Loss Models 1 - ACTSC 431/831, FALL 2011Part 6 The Classical Continuous Time Ruin ModelIn this part, times are measured in years, unless stated otherwise.1. Poisson Process: Let Nt be the number of claims up to time t or the number of
Waterloo - ACTSC - 431
ACTSC 431/831 LOSS MODELS 1, FALL 2011Instructor: Jun Cai, M3 4012, 519-888-4567 ext. 36990, jcai@uwaterloo.caInstructors Oce Hours: Monday, 14:0016:00.Lectures: 11:3012:20 on Monday, Wednesday, and Friday in MC 2066.Course Webpage: UWACE. Course info
Waterloo - ACTSC - 432
Parameter estimationActSCi 432/8321 / 31Distributions SummaryX is Binomial with parameters n (positive integer) andp (0, 1)p .f . f (x ) =nxp (1 p )nx for x = 0, 1, 2, 3, .nxE (X ) = np ,Var (X ) = np (1 p )X is Bernoulli with parameters p (0,
Waterloo - ACTSC - 432
Credibility IActSCi 432/8321 / 16Distributions SummaryX is Binomial with parameters n (positive integer) andp (0, 1)p .f . f (x ) =nxp (1 p )nx for x = 0, 1, 2, 3, .nxE (X ) = np ,Var (X ) = np (1 p )X is Bernoulli with parameters p (0, 1) X i
Waterloo - ACTSC - 432
Credibility IIActSCi 432/8321 / 22Greatest Accuracy Credibility Theory (GACT)Bayesian approach developed by H. BhlmannuSuppose we have a risk r.v. X that depends on a xed butunknown parameter . (Here can be a vector in general)For example, represe
Waterloo - ACTSC - 432
Credibility IIIActSCi 432/8321 / 20Linear Exponential Family - A quick reviewDenition (LEF)X LEF if the density is given byf (x ; ) =p (x )e r ()xq ()r () is called the canonical parameter.It turns out that the LEF class contains well-known dist
Waterloo - ACTSC - 432
Parameter estimation suuplement noteActSCi 432/8321/7Review: Unbiased estimationConsider an iid random sample X = (X1 , ., Xn ) from a modelspecied by its pdf (or pmf) f (x |)Here parameter is a vector or scalar to be estimated from XAn estimator n
Waterloo - ACTSC - 432
1!!&quot;#&quot;\$%!#&amp;'()*&amp;+%!)*&quot;%!,*&amp;(*\$!!-&amp;%.!/0)1!'&quot;*&quot;%-!!This Rate Manual has been approved by theSuperintendent of Insurance of the State of New Yorkand is effective September 1, 1993.FOURTH REPRINT: May 1, 2007Fourth Revision: March 3, 2010This r
Waterloo - ACTSC - 432
ActSc 432/832: Loss Models II (Fall 2011)1Course outlineInstructor: Joseph H.T. Kim (M3 3126, ext. 35539, jhtkim@uwaterloo.ca) Best to email me since I do notcheck my voice mail often.Lecture: 2:30-3:20 MWF (MC 4059)Teaching Assistants and oce hours
Waterloo - ACTSC - 432
Formula sheetThe Normal EquationsE (Xn+1 ) = 0 +Cov (Xi , Xn+1 ) =nj =1nj =1j E (Xj )j Cov (Xi , Xj )Empirical Bayesian Estimation Formulasnim (X Xi )2 ; i = j =1 ij ij=X;ni 1a = (m m1Distributions Summaryri=1r(ni 1)ivi = =1r(ni
Waterloo - ACTSC - 432
Name:ID:1. The random variable has pdf . It is known that =1000. You are given the following fiveobservations: 43 145 233 396 775 (sample mean = 318.4).Determine the method-of-moments estimate of .A.B.C.D.E.&lt;44&lt;4.24.2&lt;4.44.4&lt;4.6&gt;4.62. Suppo
Waterloo - ACTSC - 432
Name:ID:1. A random variable X is assumed to satisfy X|( = ) ~ Pareto (, = 1200), and thus= , x &gt; 0 and the prior distribution for is Exponential with mean 5, i.e. , &gt;0. Determine theposterior distribution of given X = 155 (i.e. say which type of dist
Waterloo - ACTSC - 432
Waterloo - ACTSC - 463
1. Use the following data to answer the questions below. Assume this company does not reserve for salvage or subrogation recoveries.Claim # Acc. Year Rep. Year1200520052200520063200620064200720075200720076200720087200820088200620
Waterloo - ACTSC - 463
1. Use the transactional data to create the cumulative data:Paid Triangle20052006200720081250070072050024700900900362300105048230012120020015755002417005001000365002004850012170090022951000242400140019003628001
Waterloo - ACTSC - 463
1. Use the transactional data (as of December 31, 2008) to develop cumulative accident year paidand incurred loss triangles.Claim #1112233455566777891010Acc Date12-Jan-0512-Jan-0512-Jan-0510-Mar-0510-Mar-0501-Nov-0501-Nov-05
Waterloo - ACTSC - 463
1. SolutionNote that there is a lot of extaneous information.Since we are interested in the value of 2006 at 24 months we need to look at data from 2008 which iscurrently at 24 months of development.Calculate the average case outstanding for 2008 at 2
Waterloo - ACTSC - 463
1. You are given the following information:AY20052006200720082009Incurred Paid Claims Paid ClaimsClaims @@@ Open Claim Open Claim12/31/2009 12/31/2009 12/31/2007 Counts @ Counts @(000's)(000's)(000's) 12/31/2009 12/31/2007117,160100,75864
Waterloo - ACTSC - 463
1. XYZ Insurance Company has been writing business since 1/1/2005 and has provided you withfollowing incurred loss data evaluated at 12/31/2009:Report YearAccident Year20052006200720082009DevelopmentInterval1-2 Years2-3 Years3-4 Years4-5 Yea
Waterloo - ACTSC - 463
1. SolutionCalculate ultimate AY and RY loss development factors:Cumulative LDFsDevelopmentInterval1-2 Years2-3 Years3-4 Years4-5 Years5-UltimateAccident CumulativeYearAY1.5002.3821.2501.5881.1001.2711.1001.1551.0501.050ReportYear
Waterloo - ACTSC - 463
1. XYZ Insurance Company has been writing business since 1/1/2005 and has provided you withfollowing incurred loss data evaluated at 12/31/2009:Report YearAccident Year2005200620072008200910,0004,0002,0001,000200512,0005,0002,5001,500200
Waterloo - ACTSC - 463
ACTSC 463/863Property and CasualtyLoss Reserving1ACTSC 463/863Chris Van Kooten&gt; FCAS, FCIA&gt; 8 Years of P&amp;C Insurance IndustryExperience&gt; VP, Corporate Actuarial at Economical&gt; 2011-2012 OCCA Executive2ACTSC 463/863Course Description&gt; This co
Waterloo - ACTSC - 463
ACTSC 463/863Property and CasualtyLoss Reserving1Development Triangles2Exercises&gt; Calculate the CY 2007 incurred losses&gt; Calculate the runoff in 2006&gt; Derive the RY incurred loss triangleClaim Snap ShotClaim #12345678910Acc Date12-J
Waterloo - ACTSC - 463
ACTSC 463/863Property and CasualtyLoss Reserving1Frequency-SeverityTechniques2Frequency-Severity&gt; Three basic approaches Simple development of counts andseverities Simple development with exposure andinflation Disposal Rate Technique3Freque
Waterloo - ACTSC - 463
ACTSC 463/863Property and CasualtyLoss Reserving1Case OutstandingDevelopment Technique2Case Outstanding Development&gt; Attempts to develop ultimate incrementalclaim payments based on historic levels ofcase reserve adequacy&gt; Assumptions Similar t
Waterloo - ACTSC - 463
ACTSC 463/863Property and CasualtyLoss Reserving1Recoveries2Recoveries&gt; Salvage/Subrogation&gt; Reinsurance Quota Share Treaty with pro-rata sharing of premiums and losses Per-Risk Excess of Loss Per-risk treaty where losses above a certain reten
Waterloo - ACTSC - 463
ACTSC 463/863Property and CasualtyLoss Reserving1ALAE &amp; ULAE2ALAE&gt; Allocated Loss Adjustment Expense Expense that can be directly associated with aparticular claim Legal Expert witness Police reports External adjuster fees Availability of da
Waterloo - ACTSC - 463
ACTSC 463/863Property and CasualtyLoss Reserving1Discounting Loss Reserves2Discounting&gt; Process of recognizing the time value ofmoney Required in Canadian reporting Outstanding reserves can be invested \$1 of future payments requires &lt;\$1 of rese
Waterloo - ACTSC - 463
ACTSC 463/863Property and CasualtyLoss Reserving1Examples From PreviousCalculate the ULAE reserve using both the traditional approach and the Kittel Refinement and thefollowing information:CY 2010 Paid LossesTotal Reserves at 12/31/2009Total Rese
Waterloo - ACTSC - 463
Property and Casualty Insurance Fall 2011ACTSC 463/8631Property and Casualty InsuranceACTSC 463/863Fall 2011InstructorE-MailOffice HoursChris Van Kooten, FCAS,FCIAvankootenc@gmail.comBy appointment onlyCourse DescriptionThis course is focuse
Waterloo - ACTSC - 232
ACTSC 970/ACC 770:Finance I Foundations of FinanceTony S. WirjantoM3 -3013, x 35210Email: twirjant at uwaterloo dot caOffice Hours: Th, Fr: 4:00-5:00 pm (or by appointment)Fall 2011Course SyllabusObjectivesThis is a first graduate course in finan
Waterloo - ACTSC - 232
Assignment 1ACTSC232 (Introduction to Actuarial Mathematics), Fall 2011This assignment consists of two parts. In Part I, the students are expected to construct oneExcel spreadsheet to answer the given questions, and submit the le electronically to the
Waterloo - ACTSC - 232
Assignment 2ACTSC232 (Introduction to Actuarial Mathematics), Fall 2011This assignment consists of two parts. In Part I, the students are expected to construct oneExcel spreadsheet to answer the given questions, and submit the le electronically to the
Waterloo - ACTSC - 232
Tutorial 1ACTSC 232, Fall, 20111. You are given the following probability distribution function (cdf) for T0 , the futurelifetime of a newborn (0):0,0.005t,F0 (t) = 0.3 + 0.0175(t 60),1,t&lt;00 t &lt; 6060 t &lt; 100t 100.(a) Verify that S0 (t) 1 F0 (
Waterloo - ACTSC - 232
SYLLABUSFall 2011ACTSC 232-001 Introduction to Actuarial MathematicsInstructor:Lectures:Tutorials:Oce hours:Pre-requisites:Dr. Chengguo Weng, M3 3136, ext.31132, c2weng@uwaterloo.ca12:30-01:20pm MWF, PHY 14505:30-06:20pm Monday, MC 20661:30-2:30
Waterloo - ACTSC - 232
Chapter 1. Introduction to life insuranceACTSC 232 Introduction to Actuarial MathematicsDepartment of Statistics and Actuarial ScienceUniversity of WaterlooFall 2011Instructor: Chengguo WengC. Weng (c2weng@uwaterloo.ca) p. 1/9Life insuranceA cont
Waterloo - ACTSC - 232
Chapter 2. Survival modelsACTSC 232 Introduction to Actuarial MathematicsDepartment of Statistics and Actuarial ScienceUniversity of WaterlooFall 2011Instructor: Chengguo WengC. Weng (c2weng@uwaterloo.ca) p. 1/42Review on Prob. &amp; StatRandom varia
Waterloo - ACTSC - 232
Tutorial 3ACTSC 232, Fall, 2011The rst two questions use this select table of mortalityl[x][ x][40] 100000[41] 99802[42] 99597[43] 99365[44] 99120l[x]+1 l[x]+2 l[x]+399899 99724 9952099689 99502 9928399471 99268 9903099225 99007 9874798964 9
Waterloo - ACTSC - 371
ActSc 371 F2011 - Assignment TwoInstructor: Dr. Lysa PorthPossible MarksPart 1: MultipleChoice (MC)33Student ScoreID Number: _Last Name:_First Name:_This assignment is due on November 2, 2011, at the beginning of class (i.e. 2:30 pm in MC2066).
Waterloo - ACTSC - 371
ActSc 371 Lecture 4 HandoutInstructor: Dr. Lysa PorthHANDOUT: Useful Financial Ratios (Ross et al. 2011)SHORTTERM SOLVENCY RATIOSCurrent ratio = Current assets Current liabilitiesQuick ratio = (Current assets Inventory) Current liabilitiesACTIVITY R
Waterloo - ACTSC - 371
9/11/2011ActSc 371 Corporate Finance 1Introduction to Corporate FinanceInstructor: Dr. Lysa PorthAir Canada Case StudyIntroductionLecture 1Sections 1.1 and 1.2 from Chapter 1: Introduction toCorporate Finance(Corporate Finance by Ross et al.)AC
Waterloo - ACTSC - 371
9/13/2011ActSc 371 Corporate Finance 1Introduction to Corporate FinanceInstructor: Dr. Lysa Porth1.3 The Corporate Firm1.4 Goals of the Corporate Firm1.5 Financial Institutions, Financial Markets, and the Corporation1.6 Trends in Financial Markets
Waterloo - ACTSC - 371
ActSc 371 Corporate Finance 1Instructor: Dr. Lysa PorthLecture 2Practice Questions11. Firms issue securities or financial instruments (orclaims) to raise capital. These claims are classifiedas:A) stocks or bondsIntroductionB) debt or equityC) c
Waterloo - ACTSC - 371
9/15/2011ActSc 371 Corporate Finance 1Introduction to Corporate FinanceInstructor: Dr. Lysa Porth2.1 The Balance Sheet2.2 Statement of Comprehensive Income2.3 Net Working CapitalLecture 3IntroductionSections 2.1-2.3 from Chapter 2: Introduction t