Hello, Every once in a while, I see an old practice question that simply uses the formula: spread = PD*LGD. Is there ever a case where this is not a viable answer to a problem? I know this is extremely vague, but sometime it seems like the most simple solution is best and other times it seems like the problems have extra steps. Thanks! Shannon

Hi Shannon, I agree with you. This is Hull's approximation: spread = PD*LGD --> default intensity (aka conditional PD) ~ spread / (1 -recovery rate). If you are given these 2/3 variables (eg, given spread and LGD/RR, what is estimated PD?), i would always look for the simplest approximation first! But i would just watch the language of the question, right? I think the compound frequency often resolves. For example, if given Rf = 4% and risky =6%, and LGD = 100% (recovery = 0%), if question said "assume annual," I would reach for = 1 - (1.04)/(1.06) per the "no-arbitrage approach" and get 1.887%. But if continuous, this would get you 1 - exp(4%)/exp(6%) which is near to the approximation given by (6% - 4%)/(1 - 0) ~= 2%. The exam tends to make such difference moot to the method. Good luck!

Similar to what Shanlane asked, I noticed questions that do not specify that in Schweser end of book questions. When presented with a question like this: (Q28 Credit Risk) Suppose the rate on Company A's one-year zero-coupon bond is 10% and the one-year T-bill rate is 8.0%. Assume the T-bill is riskless and the probability of default of Company A's bond is 10%. What is the LGD of company A's bond? The natural way to start would be to utilize Hull since it is presented most commonly in the text PD = Credit Spread/(1-RR) arriving at 20%. However, we are expected to know to use the more precise (1+rF) * (1-PD) + [1+r(A)] * PD * (1 - LGD) = 1+rF to arrive at 18.18%. Both answers were presented. Any more thoughts on this?

Hi ebb, Same thoughts, I am sorry I can't be more helpful: Hull's is an approximation, while the more precise expression clearly has a testable precedent--although I assume you want to have "(1+r(A)) * (1-PD)" rather than "rF" in the first term?--as it applies the classic no-arbitrage idea; i.e., the expected returns should equate. The more precise formula, in my humble opinion, is the far more intuitive since the left hand side is just a weighted expectation. Ironically, the 2012 AIMS, far as i can tell do not explicitly identify Hull's approximation. Imprecisely, in my opinion, GARP includes "Estimate the probability of default for a company from its bond price." On a literal read of the AIMs, GARP arguably includes the precise method and omits the approximation [sic]. But, nevermind that AIM imprecision: we know from the test experience that both are "on the table."

Thanks for the correction. That was a careless typo. How would you extend this equation beyond one year?

If compound frequency is annual, with: [1+r(A)]^n * (1-PD)^n + [1+r(A)]^n *[1- (1-PD)^n] * (1 - LGD) = [1+rF]^n i.e., (1-PD)^n is cumulative prob of repayment and [1- (1-PD)^n] is n-year cumulative prob of default (cumulative PD) ... same no arbitrage idea. On the right is expected riskfree return (n) years. Ignoring liquidity preferences, expected (weighted) return on left have the same expected return (we should be ex ante indifferent); i.e., payoff if no default + payoff if default ... although (fwiw) some don't like to compound the default; some have argued that [1+r(A)]^n * PD * (1 - LGD) should be PD * (1 - LGD), to which I've replied that recovery (1-LGD) is not precise anyway more important than the formula is just the concept. A good test of the concept is: can you show the same multi-year no-arbitrage formula for PD, but assume continuous compounding instead of annual?

In Q27 2010 GARP Practice Exam, the equation given was this: (1+rF)^t = (1+rA)^t*[(1-df)^t + RR*{1-(1-df)^t}] It is a little different with the df also compounded with t as well. I understand LGD is imprecise but would you say the same about df?

Hi ebb, i dumbly forgot in my haste to make the payment and default cumulative; i.e., I revised mine above to match. I agree with the practice exam formula. For example, if the annual PD is 5% and annual probably of repayment (p) is 95%, then I can't just use 95% over multiple years, must be cumulative repayment prob = 95%^N, then the only other outcome is a default along the way, which is 1 - 95%^N; i.e., 1 - (1-PD)^N. Thanks,