Hi Shannon - Strictly speaking, the Merton gives us no direct way to do that conversion, is sort of the the point. The Merton is limited to inferring a PD from a future distribution, either 1- year or 5-year or whatever SINGLE time horizon (that is less than or equal to the maturity of the long term debt). If we want the 1-year, we need to "re-run" the Merton under a one-year drift & one-year volatility (i.e., using T = 1.0 year). Nothing stops us from running the Merton at both T = 1.0 and (eg) T = 5.0 years. Of course, if Merton gives us a 5-year cumulative PD, we can receive the 5-year PD estimate given by Merton, assume it to be a 5-year cumulative PD (which is an approximation and technically confuses the Merton output with a cumulative PD) and further assume constant annual conditional PD, and then use: 1-year conditional PD = 1 - (1- 5yr PD)^(1/5). Or for that matter, we can infer a hazard rate = LN(1 - 5 yr cum)*-1/T. So, we have can resort to our usual mechanics, but we've left the Merton to do that. Thanks,

Thank you David. To me this part was the most helpful, as I was trying to reconcile the deviates that we get by 1. LN(13.34/10)/9.6% sigma = Z of ~ 3.0 where 3.0 is the (standardized) distance to default and 2. Merton's step 2 wants real-world DD = (LN[V(0)/F(t)] + [mu - sigma^2/2]*T)/[sigma*SQRT(T)] It's great to finally see that they are equivalent... It's probably best to stick with Merton's formula as exam qustion may be specific to a period, say 3 months. In that case, the formula (No. 2 above) does the time scaling, while this may be overlooked while using the No. 1 above... Just a thought.... Thanks.

Troubleshoot - Terrific, glad to help. My only concern is pragmatic, I worry sometimes about candidates who know too much for the exam. This post started with an idea, in part, to reconcile (as equivalents differing only by perspective) the normal returns-based distance to default and the lognormal value-based DD: just as normal log returns --> lognormal price levels, we can compute either a normal DD or a lognormal DD. But, then, given a correct understanding, Shannon was rightly confused by this 2012 practice question (P2 Number 6): Because GARP's answer is, DD = (70 - 50)/(70*20%) = 1.429 standard deviations It frustrates me that the sample paper makes this mistake (my comments for GARP are here). You, understanding Merton, will be looking for the means to find the future expected firm value, but it's not there. We have to go with the flow, I guess. By not including any return (drift), the question is sort of dumb and we should not overthink it, and realize we can just compute a PV DD (even as it's not KMV and the question is technically incorrect). I hope that helps, thanks,

Hello David, According the Merton model....equity value of the firm can be calculated assuming that is a call option with underlying the value of the firm (S=V) and strike price the face value of the debt(X=F). Call pricing d1= {ln(S/X)+(r+0.5*sigma^2)(T-t)}/{sigma*srt(T-t)} .....call value =SN(d1)+Kexp(-r(T-t))N(d1-sigma*sqrt(T-t)) In the "equivalent" d1 in Merton model we replace X with Fexp(-r(T-t)) and eliminate r...Could you please assist why these replacements/eliminations occur (why not F in X position and why we eliminate r)? Thank you PS (I know that I missing sthing and looking for it...)

Hi PL, For some baffling reason, Stulz re-formats the BSM to render it more difficult. It is the same, see reconciliations here: http://www.bionicturtle.com/forum/threads/calculating-d1-by-hull-and-stulz.5584/ http://www.bionicturtle.com/forum/threads/merton-formula.5517/#post-15553 He does not get rid of riskfree rate (r): it's in P(T) which is the present value of the zero-coupon bond maturing at time (T). So, P(T) = Face*exp(-rT) = X*exp(-rT). And LN[(V/X*exp(-rT)] = LN(V) - LN[X*exp(-rT)] = LN(V/X) - LN[exp(-rT)] = LN(V/X) -rT; i.e., the riskfree rate can be explicitly outside or "inside" the LN(). I hope that helps, thanks!

Hi Niko, De Servigny Chapter 3 (p 72) says "In short, equity-based models are prone to overrection due to market bubbles." De Servigny references this paper by Peter Crosbie, who I think implicates Merton as highly (over) sensitive before explaining that the KMV modifications overcome it (I consulted to him ~10 years ago, when he was CEO of KMV right as they were selling to Moody's). I recall quite a bit of self-doubt about Merton (or even KMV variations) models in service of default prediction, some folks were highly skeptical and i'd imagine the skepticism persists today. If you believe the equity market capitalization contains material short-term noise (I surely do; more importantly, i think it's "infected" by non-specific factors), I don't think it's too hard to make the over-reaction point. I hope that helps, thanks,

We use KMV model to get EDF's for our portfolio. Naturally, it is as volatile, if not more, as the stock market. This would qualify as over-reaction. Just because stock market reacts in a certain way to a peice of news in the short term does not automatically make our portfolio significantly more or less risky.

Hi David, Great stuff. Thanks. I have that paper. It is difficult to find now the KMV papers since they were acquire by Moody's The reason I asked is the I am writing my MBA dissertation on structural models. I wanted to include a section with controversies. What I found so far is that during times of extreme market swings, i.e. volatility such as 2008-2009, Merton is virtually useless as it predicts inevitable defaults of majority of the firms based on the market volatility, which has nothing to do with firm's fundamentals. In its most basic form it is useless during high volatility as it "over-reacts".

David, Thanks for the great explanation above. This greatly helps me understand the topic. However, I still don’t understand why you use the formula V(0)*exp(mu - asset variance^2/2)*T instead of V(0)*exp(mu)*T to calculate V(t). Can you expand the explanation a little further? Thanks.

Hi David, nevermind my question. I found out from my FRM P1 note. The difference is closely related to the difference between arithmetic mean and geometric mean. The mean return will always be slightly less than the expected return just as the geometric mean will always be slightly less than the arithmetic mean.

Hi David, There was a question from the 2011 GARP FRM practice exam where we were given the following: Current market value of firm = 4,500 Expected market value of the firm one year from now = 5,000 Short term debt = 1,000 Long term debt = 1,300 Annualized volatility of firm assets = 22% Calculate distance-to-default (DD). Shouldn't we calculate the DD as below? Default point = 1,000 + 0.5*1,300 = 1,650 DD=ln(5,000/1,650) / 0.22 = 5.04 But the answer was given as follows: (5,000 - 1,650)/(5,000*0.22)=3.045 That's a big difference. Can you advise which method should we follow on the exam? Thanks.

Distance to default is how many standard deviations (asset volatility) is the final asset value is from the default point, which is correctly shown in the answer given. This is as per Moody's KMV model. I am not sure why you are taking natural logs here. That's not KMV methodology.

Hi Troubleshooter, Thanks. I understand that "Distance to default is how many standard deviations (asset volatility) is the final asset value is from the default point". The Moody's KMV is built on the Merton model, which assumes lognormal distribution of stock price, isn't it? KMV only differs from the Merton model by the following: 1) KMV uses their proprietary data based to map the calculated DD, instead of using the Z-table (which underestimate fat tail distributions), 2) instead of assuming one zero coupon debt, KMV calculates default threshold as STD+0.5*LTD (or the other formula if LTD/STD >1.5). Other than these 2 points, KMV and Merton models are the same. Isn't it? I see many formulas for DD: DD = [Ln(V) - Ln(Default Threshold) + (ROA-0.5*variance)*T]/[volatility*SQRT(T)]; this is based on lognormal distribution assumption? DD = (Expected asset return % - Default threshold % ) / volatility % DD = (Asset value $ - Liability value $)/Volality$ David used the lognormal formula in the above question: Distance to the default threshold of 10 under lognormal =Ln(13.34/10)=28.8%. And this is 28.8%/9.6% = 3 standard deviation away from the default threshold.

Hi sleepybird, Yes, exactly: you have illustrated the issue. Actually, I wrote to GARP about it just last Wednesday, again for the 3rd or 4th time, because they are confused and their confusion continues to cause some confusion. Although this question (i.e., GARP 2011 P2.2.8) does not ask for a PD, if we want the PD for this scenario, and if we want to use the normal CDF (naturally!), then the PD is returned NOT with the 3.045 DD but rather the PD ~= N[ (LN(4500/1650) + ((5% - 0.5*20%^/2)*1)) / (22%*1) ] = N[5.039] ~= 0.000023%. I agree that the lognormal value-based DD is 3.045 which corresponds to a normal returns-based DD of 5.039375. (this can be verified with XLS 6.c.1). While (IMO) both DDs are valid as one refers to the returns-based normal and the other the value or price-based lognormal, it is confusing because we are assigned (and the literature) generally operates on the normal returns-based DD (sleepybird is right in the sense that you never see it presented lognormally), such that your LN(.) is sensible and smart. But, as I mentioned in the last Focus Review, GARP does not seem to be aware of this difference (or rather the technical nuance): all occurrences in my historical exam sample merely , as in the question above, look for the lognormal value-based DD. I imagine my focus will get this "fixed" for the 2013 exam, but possibly not by the upcoming November. (frankly, I think the "fix" is easy: just label the DD above as a value-based lognormal version ... I'd think that would be enough to save you from wanting to convert?) Please note that I am not aware of your 3rd variation, I do not think this should ever arise, unless i missed something: DD = (Expected asset return % - Default threshold % ) / volatility % ... this would appear to be the same as the normal DD, but only the default threshold is "parsed out" from within the LN(firm value/threshold), which is math possible but un-necessary i thnk But i do definitely agree with you: GARP's historical tendency is to ask the question, as above, based on a ratio of future values, and although your translation into a returns-based normal is totally valid, it is too smart (frankly) for the current exam. Thanks!

Hi David, In contrast to the Merton model (which naively assumes single bullet debt structure), KMV uses linear combination of short term and long term debts. My question is, does this make the KMV model assume 2 default dates? KMV still uses the BSM formula, which is a pricing model for European options. Applying the same logic, using the BSM model to model probability default means the firm either defaults at T or no default, you can't default in between. Based on this, even though the KMV models uses combinations short term debt and long term debt to calculate the default threshold, this does not mean the model allows jumps (i.e., early default)? Am I correct? Thanks.

Hi sleepybird, Great question. (caveat: my direct experience with KMV is aged, I can't speak to recent updates). As you say, KMV employs the Merton model, up to a point, so in my view: No, KMV model does not assume 2 default horizons. The combination of short & long-term debts is a "realism" that accounts for the fact that long-term debt, by definition, is not immediately due. So, while assets < liabilities is (technically) balance sheet insolvency, it is not necessarily funding illiquidity. Put another way, KMV includes all of the short-term debt (< 1 year) because if that can't be repaid, default occurs; but less than all of the long-term debt because there is still time to repay it. Therefore, the distance to default (DD) measure--which, to remind is similar to d2 in BSM but not exactly due to the use of a real-world drift expectation--assumes a single-horizon (e.g., a weakness of KMV, cited in de Servigny, is the lack of its ability to account for migrations, a general issue of structural/KMV models that do not simulate) However, importantly, KMV does not finish with PD = N(-DD). Rather, KMV maps DD to a historical (empirical) database of cumulative PDs for the horizon. In this way, although the DD is limited to a horizon, it is "translated" (in the case of multi-year) into a cumulative PD that does attempt to consider the entire interval. Your question is also great because it highlights how KMV/Merton is not BSM: while BSM is used to derive the equity value, after that step, there is no risk-neutral derivative valuation. Each of: DD, Merton's PD(-DD) and KMV's DD -- map to --> historical PDs .... are physical (intiutive), future tail distribution estimations. From Peter Crosbie's excellent primeron KMV

Hello, in your example, can you tell me how you arrive at $10 as your default threshold: (LN(13.34/10)/9.6% sigma = Z of ~ 3.0 where 3.0 is the (standardized) distance to default) Is it the amount of ST debt divided by something?

Hi blahbe - I'll add a note to the original post: the pictured example borrows the example in FRM-assigned de Servigny Chapter 3 where the firm's capital structure includes long-term debt of $6.0 billion and short-term liabilities of $7.0 billion. A Merton Model approach (although it strictly speaking assumes one debt class) would use 7+6 = $13 billion as the default threshold because that is the face (par) value of the debt; but de Servigy is illustrating a KMV-based variation where the default threshold is more realistically somewhere between short-term liabilities and total liabilities. In that KMV approach, unless LT/ST debt is > 1.5, then default threshold = ST + 0.5*LT debt, thanks,