[ 2009 FRM Sample Exam ] Quantitative Analysis Q15

lalamei

 

15. A junior bond with a face value of 200 matures in 5 years. A senior bond on the same firms also matures in 5 years, and has a face value of 100. Assume sA=.5 and the riskfree rate=.04. Firm value is equal to 400. Using the Merton model, what is the value of the junior bond? (The following table includes figures that will reduce the time required to answer this but it also includes figures that are irrelevant to the problem and some that are strictly wrong).

 

d1	d2	N (d1)	N (d2)
1.978	0.860	.9761	.8051
1.978	0.860	.8051	.9761
0.9952	-.1228	.8413	.5478
0.9952	-1.505	.8413	.0661
0.9952	-.1228	.8413	.4522

 

 

A. 99.07

B. 174.55

C. 75.48

D. 16.63

lalamei

 

Correct answer is Afficeffice" />

A is correct.  To get this answer use the first and last row of the table to make the calculation of the Black-Scholes formula go faster.  Plug in the values given into the B-S formula and use the first row to get the value of the equity if the leverage were just 100 (the face value of the senior bond).  Subtract this off 400 to get the value of the senior bond.  Plug in the values of the B-S formula assuming a strike of 300 (the total debt) to get the value of the equity.  Use the last row of the table to make this go faster.   Subtracting the value of the equity and the senior debt from 400 gives us the answer, a.

B is incorrect because it is the value of the total debt of the firm, not the value of the junior debt alone.

C is incorrect because it is the value of the senior debt, not the junior debt.

D is incorrect because it uses the wrong line in the table to calculate the junior debt.  It uses the second line instead of the first.  One needs to use the B-S to calculate d1 and d2 and find their normal distribution values in the table.  All of the rows except the second line from the bottom have the correct values of d1 and d2 but they do not always have the correct normal distribution value that goes with d1 and d2.   In particular, the middle line is incorrect because a negative value for d2  has a normal distribution value that is less than .5  The second line is easily detected as wrong because the normal area for a large number should be larger than that of a small number ? here they are reversed.

Reference: Stulz, Risk Management and Derivatives, Chapter 18

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谢谢

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学习来了

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