The Mathematics of Financial Modelingand Investment Management

22-Credit Risk Model Derivs Page 689 Wednesday, February 4, 2004 1:12 PM

Credit Risk Modeling and Credit Default Swaps 689

T– t = 1 year

r =5%

Applying equation (22.5), the equity value based upon the above

example is,

ln140 – ln100 + (0.05 – 0.3^2 ) × 1

d 2 = --------------------------------------------------------------------------------------= 1.4382

0.3 1

d 1 = 1.4382 – 0.30 = 1.1382

Et()= 140 ×N(1.1382)– e –0.05 × 100 ×N(1.4382)

= $46.48 million

and market debt value, by equation (22.6) is

DtT ( , )= At()– Et()= 140 – 46.48 = $93.52 million

Hence, the yield of the debt is, by equation (22.7):

ln100 – ln93.52

y= ----------------------------------------- = 6.70%

1

which is higher than the 5% risk-free rate by 170 basis points. This “credit

spread” reflects the 1-year default probability from equation (22.4):

p= 1 – N(1.4382)= 12.75%

and the recovery value of

At()( 1 – Nd() 1 ) = $17.85

if default occurs.

From above, we can see that, as the asset value increases, the firm is

more likely to remain solvent, the default probability drops. When default

is extremely unlikely, the risky debt will be surely paid off at par, the risky

debt will become risk free, and yield the risk-free return (5% in our exam-

ple). In contrast, when default is extremely likely (default probability

approaching 1), the debt holder is almost surely to take over the company,

the debt value should be the same as the asset value which approaches 0.