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Use the Merton model to compute distance to default and the default probability.

FRM® Part II · Credit Risk Measurement and Management · Estimating Default Probabilities · FRM2.CR.9.m

The idea

In the Merton model default happens only at maturity, when the asset value falls below the face value of the debt. Distance to default is the number of standard deviations by which the expected log asset value exceeds the default point: DD = (ln(V0 ÷ D) + (μ − σ²÷2) × T) ÷ (σ × √T). Here μ is the asset drift, and under the risk-neutral measure μ equals r, so that DD becomes d2. The probability of default is the normal probability N(−DD), so a larger distance to default means a lower default probability. DD rises with lower leverage, lower asset volatility and higher expected return. Real-world probabilities use the actual expected return μ, while risk-neutral ones use r, so for μ above r the real-world PD is lower. Typical errors are using d1 instead of d2, forgetting the σ²÷2 term, using σ instead of σ scaled by the square root of T when T exceeds one year, and reading N(DD) as the default probability.

Formulas

Merton distance to default

DD = (ln(V0 ÷ D) + (μ − σ² ÷ 2) × T) ÷ (σ × √T)

V0
the asset value today
D
the face value of the debt, the default point
μ
the drift of the asset value: the expected return for a real-world figure, r for a risk-neutral one
σ
the asset volatility
T
the horizon in years

Default only at T, with lognormal assets. With μ = r this is d2; forgetting the σ² ÷ 2 term or the √T overstates the distance to default when T exceeds a year.

Merton default probability from distance to default

PD = N(−DD)

N
the standard normal cumulative distribution function
DD
the distance to default

Assumes normally distributed log asset value at the horizon. N(DD) is the survival probability, not the default probability.

KMV-style distance to default

DD = (V − DP) ÷ (V × σ)

V
the market value of the firm's assets
DP
the default point, short-term liabilities plus half of long-term liabilities
σ
the volatility of the asset value

A simple count of standard deviations of asset value above the default point. KMV then maps DD to a default frequency from historical data, not from the normal distribution; leaving σ out gives only the proportional cushion.

A practice question

From the question bank, with the reason each wrong option is wrong.

In the Merton model, distance to default is best described as:

  1. the number of standard deviations by which the expected log asset value exceeds the default point at the horizon. - correct
  2. the ratio of the market value of assets to the face value of debt.Why not: This is a leverage ratio, not a distance measured in standard deviations.
  3. the number of days until the firm's equity value reaches zero.Why not: Distance to default is measured in standard deviations, not time.
  4. the probability that asset value stays above the debt at maturity.Why not: That is the survival probability; distance to default is the argument of the normal function.

DD = [ln(V0/D) + (mu - sigma²/2) T] / (sigma sqrt T), and PD = N(-DD).

A harder one

A firm has asset value 250, debt of face value 200 due in one year, asset volatility 20% and an expected asset return of 8%. Using the Merton model with the real-world drift, what is the one-year probability of default?

  1. 6.48%Why not: Omits the sigma² over two term in the drift.
  2. 5.31%Why not: Uses d1 instead of d2 as the distance to default.
  3. 7.84% - correct
  4. 92.16%Why not: Uses N(DD), the survival probability, instead of N(-DD).

DD = [ln(250/200) + (0.08 - 0.02)] / 0.20 = 1.4157; PD = N(-1.4157) = 7.84%.

The hardest tier

A firm has asset value 100 and asset volatility 30%, and the risk-free rate is 2%. Using the risk-neutral Merton model with a one-year horizon, by how many percentage points does the default probability fall when the face value of debt is cut from 70 to 60 through an equity issue (asset value and volatility unchanged for simplicity)?

  1. 6.6Why not: Omits the sigma² over two term from the drift in both calculations.
  2. 8.2 - correct
  3. 60.8Why not: Reports the relative fall in percent instead of the fall in percentage points.
  4. 0.5Why not: Reports the change in distance to default instead of the change in probability.

DD = 1.1056 at D = 70 gives PD 13.4%; DD = 1.6194 at D = 60 gives PD 5.3%; the fall is 8.2 points.

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