Chapter-22Credit-Risk資管所廷課件

1、Chapter 22Credit Risk資管所 陳竑廷Chapter 22Credit Risk資管所 陳Agenda22.1 Credit Ratings22.2 Historical Data22.3 Recovery Rate22.4 Estimating Default Probabilities from bond price22.5 Comparison of Default Probability estimates22.6 Using equity price to estimate Default ProbabilitiesAgenda22.1 Credit Ratings

2、Credit RiskArise from the probability that borrowers and counterparties in derivatives transactions may default.Credit Risk22.1 Credit RatingsS&P AAA , AA, A, BBB, BB, B, CCC, CC, CMoodyAaa, Aa, A, Baa, Ba, B, Caa, Ca, CInvestment grade Bonds with ratings of BBB (or Baa) and abovebestworst22.1 Credi

3、t RatingsS&P bestw22.2 Historical DataFor a company that starts with a good credit rating default probabilities tend to increase with timeFor a company that starts with a poor credit rating default probabilities tend to decrease with time 22.2 Historical DataFor a comDefault IntensityThe uncondition

4、al default probability the probability of default for a certain time period as seen at time zero39.717 - 30.494 = 9.223%The default intensity (hazard rate)the probability of default for a certain time period conditional on no earlier default100 30.494 = 69.506%0.09223 / 0.69506 = 13.27%Default Inten

5、sityThe unconditiChapter-22Credit-Risk資管所廷課件Q(t) : the probability of default by time t(22.1) Q(t) : the probability of defa22.3 Recovery RateDefined as the price of the bond immediately after default as a percent of its face valueMoody found the following relationship fitting the data:Recovery rate

6、 = 59.1% 8.356 x Default rateSignificantly negatively correlated with default rates22.3 Recovery RateDefined as Source :Corporate Default and Recovery Rates, 1920-2006Source :22.4 Estimating Default ProbabilitiesAssumptionThe only reason that a corporate bond sells for less than a similar risk-free

7、bond is the possibility of defaultIn practice the price of a corporate bond is affected by its liquidity.22.4 Estimating Default Proba(22.1)(22.1)11R11-1-Taylor expansion11R11-1-Taylor expansionA more exact calculationSuppose that Face value = $100 , Coupon = 6% per annum , Last for 5 yearsCorporate

8、 bondYield : 7% per annum $95.34Risk-free bondYield : 5% per annum $104.094The expected loss = 104.094 95.34 = $ 8.75A more exact calculationSupposQ : the probability of default per year288.48Q = 8.75Q = 3.03%0 1 2 3 4 5e -0.05 *3.5Q : the probability of default22.5 Comparison of default probability

9、 estimatesThe default probabilities estimated from historical data are much less than those derived from bond prices22.5 Comparison of default prHistorical default intensityThe probability of the bond surviving for T years is(22.1)Historical default intensityThChapter-22Credit-Risk資管所廷課件Default inte

10、nsity from bonds A-rated bonds , Merrill Lynch 1996/12 2007/10The average yield was 5.993% The average risk-free rate was 5.289% The recovery rate is 40%(22.2)Default intensity from bonds A0.11*(1-0.4)=0.0660.11*(1-0.4)=0.066Real World vs. Risk Neutral Default ProbabilitiesRisk-neutral default proba

11、bilitiesimplied from bond yields Value credit derivatives or estimate the impact of default risk on the pricing of instrumentsReal-world default probabilitiesimplied from historical dataCalculate credit VaR and scenario analysisReal World vs. Risk Neutral De22.6 Using equity prices to estimate defau

12、lt probabilityUnfortunately , credit ratings are revised relatively infrequently.The equity prices can provide more up-to-date information22.6 Using equity prices to eMertons ModelIf VT D , ET = VT - DMertons ModelIf VT D , ET =V0 And 0 cant be directly observable.But if the company is publicly trad

13、ed , we can observe E0.V0 And 0 cant be directly oMertons model gives the value firms equity at time T asSo we regard ET as a function of VTWe writeOther term without dW(t) , so ignore itMertons model gives the valueReplace dE , dV by (*) (*) respectivelyWe compare the left hand side of the equation

14、 above with that of the right hand side(22.4)Replace dE , dV by (*) (*) reExampleSuppose thatE0 = 3 (million) r = 0.05 D = 10 E = 0.80 T = 1Solvingthen getV0 = 12.400 = 0.2123N(-d2) = 12.7%ExampleSuppose thatN(-d2) = 12SolvingSolvingF(x,y)2+G(x,y)2=(D2)2+(E2)2F(x,y)=A2*NORMSDIST(LN(A2/10)+(0.05+B2*B2/2)/B2) -10*EXP(-0.05)*NORMSDIST(LN(A2/10)+(0.05+B2*B2/2)/B2-B2)G(x,y)=NORMSDIST(LN(A