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Outline how real-world default probabilities differ from risk-neutral ones and decide which belongs in a given credit risk analysis.

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

The idea

Real-world (physical) default probabilities are the actual chances that defaults occur, estimated from history or from models of the firm. Risk-neutral probabilities are the ones implied by market prices. They are higher because they embed the premium investors demand for bearing default risk. The rule of thumb is to use risk-neutral probabilities to value things, and real-world probabilities to estimate what will actually happen. Valuing a credit derivative, pricing a loan at market value, or computing a CVA charge for accounting purposes needs risk-neutral probabilities, because the price must be consistent with other traded instruments. Forecasting future losses, calculating expected loss for provisioning, credit VaR, stress testing and economic or regulatory capital need real-world probabilities. In the Merton model the two differ only in the asset drift. The risk-neutral case uses the risk-free rate, and the real-world case uses the actual expected asset return.

Formulas

Real-world distance to default

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

V
the market value of the firm's assets today
D
the default point, the face value of the debt
μ
the expected return on the assets, continuously compounded
σ
the volatility of the asset value
T
the horizon, in years

The real-world PD is N(−DD). Swapping μ for the risk-free rate is the only change needed to get the risk-neutral PD.

A practice question

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

A dealer is pricing a new 5-year credit default swap on a corporate name. Which default probabilities are appropriate?

  1. Real-world probabilities from historical default frequencies, because they reflect actual defaults.Why not: Real-world probabilities understate market prices, which include a risk premium.
  2. Either set, because the two are equal when recovery rates are the same.Why not: They differ because of the risk premium, whatever the recovery assumption.
  3. Risk-neutral probabilities, because the price must be consistent with market prices of other instruments. - correct
  4. Neither; derivative pricing does not use default probabilities.Why not: Credit derivative prices are built from default probabilities and recovery assumptions.

Valuation uses risk-neutral probabilities so that prices are consistent with traded instruments; forecasting uses real-world ones.

A harder one

A bank (i) quotes a price to a client for CDS protection and (ii) sets next year's loan-loss provision for the same name. Which combination of default probabilities is appropriate?

  1. Risk-neutral for the CDS price; real-world for the loan-loss provision. - correct
  2. Real-world for the CDS price; risk-neutral for the loan-loss provision.Why not: Reverses the two uses: real-world for pricing and risk-neutral for provisioning.
  3. Risk-neutral for both, since the market view is the best estimate of both.Why not: Using risk-neutral probabilities for provisioning overstates expected losses.
  4. Real-world for both, since actual defaults drive the payoff in each case.Why not: Using real-world probabilities for pricing would offer protection below the market price.

Pricing must be consistent with the market (risk-neutral); provisioning must forecast realised losses (real-world).

The hardest tier

In a Merton model a firm has asset value 100, debt face value 80 maturing in one year, asset volatility 25%, a risk-free rate of 3% and an expected asset return of 10%. What is its real-world one-year probability of default?

  1. 18.7%Why not: Uses the risk-free rate as the asset drift, which gives the risk-neutral probability.
  2. 9.8%Why not: Omits the volatility drag, sigma² over two, from the drift.
  3. 12.1% - correct
  4. 7.8%Why not: Uses d1 instead of d2 as the distance to default, adding sigma x sqrt T.

d2 = [ln(100/80) + (0.10 - 0.03125) x 1] / 0.25 = 1.1676; PD = N(-1.1676) = 12.1%. With r replacing mu it would be 18.7%.

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