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# Credit Risk Management Problems

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Simple problems in Credit risk management

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### Credit Risk Management Problems

1. 1. Problems in Credit Risk Management September 16, 2009 By A. V. Vedpuriswar
2. 2. Problem <ul><li>There are 10 bonds in a portfolio. The probability of default for each of the bonds over the coming year is 5%.These probabilities are independent of each other. What is the probability that exactly one bond defaults? </li></ul>
3. 3. <ul><li>Required probability </li></ul><ul><li>= (10)(.05)(.95) 9 </li></ul><ul><li>= .3151 </li></ul><ul><li>= 31.51% </li></ul>
4. 4. Problem <ul><li>A CDS portfolio consists of 5 bonds with zero default correlation. One year default probabilities are : 1%, 2%, 5%,10% and 15% respectively. What is the probability that that the protection seller will not have to pay compensation? </li></ul>
5. 5. <ul><li>Probability of no default </li></ul><ul><li>= (.99)(.98)(.95)(.90)(.85) </li></ul><ul><li>= .7051 </li></ul><ul><li>= 70.51% </li></ul>
6. 6. Problem <ul><li>If the probability of default is 6% in year 1 and 8% in year 2, what is the cumulative probability of default during the two years? </li></ul>
7. 7. Solution <ul><li>Probability of default not happening in both years </li></ul><ul><li>=(.94) (.92) = = .8648 </li></ul><ul><li>Required probability = 1 － .8648 = .1352 </li></ul><ul><li>= 13.52% </li></ul>
8. 8. Problem <ul><li>The 5 year cumulative probability of default for a bond is 15%. The marginal probability of default for the sixth year is 10%. What is the six year cumulative probability of default? </li></ul>
9. 9. <ul><li>Required probability </li></ul><ul><li>= 1- (.85)(.90) </li></ul><ul><li>= .235 </li></ul><ul><li>= 23.5% </li></ul>
10. 10. Problem <ul><li>Calculate the implied probability of default if the one year T Bill yield is 9% and a 1 year zero coupon corporate bond is fetching 15.5%. </li></ul>
11. 11. Solution <ul><li>Let the probability of default be (1-p) </li></ul><ul><li>Returns from corporate bond = 1.155p + (0) (1-p) </li></ul><ul><li>Returns from treasury = 1.09 </li></ul><ul><ul><li>1.155p = 1.09 </li></ul></ul><ul><ul><li>p = 1.09/1.155 = .9437 </li></ul></ul><ul><li>Probability of default = .0563 = 5.63% </li></ul>
12. 12. Problem <ul><li>In the earlier problem, if the recovery is 80% in the case of a default, what is the default probability? </li></ul>
13. 13. Solution <ul><li>1.155p + (.80) (1.155) (1 － p) = 1.09 </li></ul><ul><li>.231p = 0.166 </li></ul><ul><li>p = .7186 </li></ul><ul><li>1 － p = .2814 </li></ul>
14. 14. Problem <ul><li>If 1 year and 2 year T Bills are fetching 11% and 12% and 1 year and 2 year corporate bonds are yielding 16.5% and 17%, what is the marginal probability of default for the corporate bond in the second year? </li></ul>
15. 15. Solution <ul><li>Yield during the 2 nd year can be worked out as follows: </li></ul><ul><li>Corporate bonds: (1.165) (1+i) = 1.17 2 </li></ul><ul><li> i = 17.5% </li></ul><ul><li>Treasury : (1.11) (1+i) = (1.12) 2 </li></ul><ul><li> i = 13.00% </li></ul><ul><li> p (1.175) + (1-p) (0) = 1.13 </li></ul><ul><li> p = .9617 </li></ul><ul><li> Default probability = 1 － .9617 </li></ul><ul><li>= 3.83% </li></ul>
16. 16. Problem <ul><li>The spread between the yield on a 3 year corporate bond and the yield on a similar risk free bond is 50 basis points. The recovery rate is 30%. What is the cumulative probability of default over the three year period? </li></ul>
17. 17. Solution <ul><li>Spread = (Probability of default) (loss given default) </li></ul><ul><li>or .005 = (p) (1-.3) </li></ul><ul><li>or p = = .00714 = .71% per year </li></ul><ul><li>No default over 3 years = (.9929) (.9929) (.9929) = .9789 </li></ul><ul><li>So cumulative probability of default = 1 – 9789 = .0211 = 2.11% </li></ul>
18. 18. Problem <ul><li>The spread between the yield on a 5 year bond and that on a similar risk free bond is 80 basis points. If the loss given default is 60%, estimate the average probability of default over the 5 year period. If the spread is 70 basis points for a 3 year bond, what is the probability of default over years 4, 5? </li></ul>
19. 19. Solution <ul><li>Probability of default over the 5 year period = = .0133 </li></ul><ul><li>Probability of default over the 3 year period = = .01167 </li></ul><ul><li>(1-.0133) 5 = (1-.01167) 3 (1-p) 2 </li></ul><ul><li>or (1-p) 2 = = .9688 </li></ul><ul><li>or 1 – p = .9842 </li></ul><ul><li>or p = .0158 = 1.58% </li></ul>
20. 20. Problem <ul><li>A four year corporate bond provides a 4% semi annual coupon and yields 5% while the risk free bond, also with 4% coupon yields 3% with continuous compounding. </li></ul><ul><li>Defaults may take place at the end of each year. </li></ul><ul><li>The recovery rate is 30% of the face value. </li></ul><ul><li>What is the risk neutral default probability? </li></ul>
21. 21. Solution <ul><li>Risk free bond </li></ul><ul><li>So expected value of losses = 103.65 – 96.21 = 7.44 </li></ul>Risk free bond Corporate bond Year Cash flow PV factor e -(.03)t PV PV factor e -(.05)t PV .5 2 .9851 1.9702 .9754 1.9508 1.0 2 .9704 1.9408 .9512 1.9024 1.5 2 .9560 1.9120 .9277 1.8554 2.0 2 .9418 1.8836 .9048 1.8096 2.5 2 .9277 1.8554 .8825 1.7650 3.0 2 .9139 1.8278 .8607 1.7214 3.5 2 .9003 1.8006 .8395 1.679 4.0 102 .8869 90.4638 .8187 83.51 103.6542 96.21
22. 22. <ul><li>Let the default probability per year = Q. </li></ul><ul><li>The recovery rate is 30 %. </li></ul><ul><li>So if the notional principal is 100, we can recover 30. </li></ul><ul><li>We can work out the present value of losses assuming the default may happen at the end of years 1, 2, 3, 4. </li></ul><ul><li>Accordingly, we calculate the present value of the risk free bond at the end of years 1, 2, 3, 4. </li></ul><ul><li>Then we subtract 30 being the recovery value each year. </li></ul><ul><li>We then calculate the present value of the losses using continuously compounded risk free rate. </li></ul>
23. 23. Solution (Cont…) <ul><li>PV factors </li></ul><ul><li>e -.015 = .9851 </li></ul><ul><li>e -.030 = .9704 </li></ul><ul><li>e -.045 = .9560 </li></ul><ul><li>e -.060 = .9417 </li></ul><ul><li>e -.075 = .9277 </li></ul><ul><li>e -.09 = .9139 </li></ul>
24. 24. Solution (Cont…) <ul><li>Time of default = 1 </li></ul><ul><li>PV of risk free bond = 2+2e -.015 +2e -.030 +2e -.045 + 2e -.060 +2e -075 + (102)e -.090 </li></ul><ul><li>= 2[1+.9851+.9704+.9560+.9417+.9277]+(102)(.9139) </li></ul><ul><li>= 11.56 + 93.22 = 104.78 </li></ul><ul><li>Time of default = 2 </li></ul><ul><li>PV of risk free bond </li></ul><ul><li>= 2[1+.9851+.9704 +.9560]+(102) (.9417) </li></ul><ul><li>= 7.823 + 96.05 = 103.88 </li></ul>
25. 25. Solution (Cont…) <ul><li>Time of default = 3 </li></ul><ul><li>PV of risk free bond = [1+ .9851] + (102) (.9704) </li></ul><ul><li>= 3.97 + 98.98 </li></ul><ul><li>= 102.95 </li></ul><ul><li>Time of default = 4 </li></ul><ul><li>PV of risk free bond </li></ul><ul><li>= 102 </li></ul>
26. 26. Solution (Cont…) <ul><li>So we can equate the expected losses: </li></ul><ul><li>i.e, 7.44 = 272.68Q </li></ul><ul><li>or Q = .0273 = 2.73% </li></ul>Default point (Years) Expected losses PV 1 (104.78 – 30)Q = 74.78Q (74.78)Qe -.03 = 72.57Q 2 (103.88 – 30)Q = 73.78Q (73.88)Qe -.06 = 69.58Q 3 (102.95 – 30)Q = 72.95Q (72.95)Qe -.09 = 66.67Q 4 (104 – 30)Q = 72 Q (72)Qe -.12 = 63.86Q 272.68 Q
27. 27. Problem <ul><li>A bank has made a loan commitment of \$ 2,000,000 to a customer. Of this, \$ 1,200,000 is outstanding. There is a 1% default probability and 40% loss given default. In case of default, drawdown is expected to be 75%. What is the expected loss? </li></ul>
28. 28. Solution <ul><li>Drawdown in case of default = (2,000,000 – 1,200,000) (.75) = 600,000 </li></ul><ul><li>Adjusted exposure = 1,200,000 + 600,000 = 1,800,000 </li></ul><ul><li>Loss given default = (.01) (.4) (1,800,000) = \$ 7,200 </li></ul>
29. 29. Problem <ul><li>A bank makes a \$100,000,000 loan at a fixed interest rate of 8.5% per annum. </li></ul><ul><li>The cost of funds for the bank is 6.0%, while the operating cost is \$800,000. </li></ul><ul><li>The economic capital needed to support the loan is \$8 million which is invested in risk free instruments at 2.8%. </li></ul><ul><li>The expected loss for the loan is 15 basis points per year. </li></ul><ul><li>What is the risk adjusted return on capital? </li></ul>
30. 30. Solution <ul><li>Net profit = 100,000,000 (.085 - .060 - .0015) – 800,000 + (8,000,000) (.028) </li></ul><ul><li> = 23,50,000 – 800,000 + 224,000 </li></ul><ul><li> = \$1,774,000 </li></ul><ul><li>Risk adjusted return on capital = 1.774/8 = .22175 = 22.175% </li></ul>
31. 31. Problem <ul><li>Using the Merton Model, calculate the value of the firm’s equity. </li></ul><ul><li>The current value of the firm’s equity is \$60 million and the value of the zero coupon bond to be redeemed in 3 years is \$50 million. </li></ul><ul><li>The annual interest rate is 5% while the volatility of the firm value is 10%. </li></ul>
32. 32. Solution <ul><li>Formula is: St = V x N(d) – Fe -r(T-t) x N (d-  T-t) </li></ul><ul><li>d = </li></ul><ul><li>V = value of firm </li></ul><ul><li>F = face value of zero coupon debt </li></ul><ul><li> = firm value volatility </li></ul><ul><li>r = interest rate </li></ul>
33. 33. Solution <ul><li>S = 60 x N (d) – (50)e -(.05)(3) x N (d-(.1)  3) </li></ul><ul><li>d = </li></ul><ul><li>= = 2.005 </li></ul><ul><li>S = 60 N (2.005) – (50) (.8607) N (2.005 - .17321) </li></ul><ul><li>= 60 N (2.005) – (43.035) N (1.8318) </li></ul><ul><li>= (60) (.9775) – (43.035) (.9665) </li></ul><ul><li>= \$17.057 million </li></ul>
34. 34. Problem <ul><li>In the earlier problem, calculate the value of the firm’s debt. </li></ul>
35. 35. Solution <ul><li>Dt = Fe -r(T-t) – p t </li></ul><ul><li>= 50e -.05(3) – p t </li></ul><ul><li>= 43.035 – p t </li></ul><ul><li>Based on put call parity </li></ul><ul><li>p t = C t + Fe -r(T-t) – V </li></ul><ul><li>Or p t = 17.057 + 43.035 – 60 = .092 </li></ul><ul><li>D t = 43.035 - .092 = \$42.943 million </li></ul><ul><li>Alternatively, value of debt </li></ul><ul><li>= Firm value – Equity value = 60 – 17.057 </li></ul><ul><li>= \$42.943 million </li></ul>
36. 36. <ul><li>Consider the following figures for a company. What is the probability of default? </li></ul><ul><ul><li>Book value of all liabilities : \$2.4 billion </li></ul></ul><ul><ul><li>Estimated default point, D : \$1.9 billion </li></ul></ul><ul><ul><li>Market value of equity : \$11.3 billion </li></ul></ul><ul><ul><li>Market value of firm : \$13.8 billion </li></ul></ul><ul><ul><li>Volatility of firm value : 20% </li></ul></ul>Problem
37. 37. <ul><li>Distance to default (in terms of value) = 13.8 – 1.9 = \$11.9 billion </li></ul><ul><li>Standard deviation = (.20) (13.8) = \$2.76 billion </li></ul><ul><li>Distance to default (in terms of standard deviation) = 4.31 </li></ul><ul><li>We now refer to the default database. </li></ul><ul><li>If 5 out of 100 firms with distance to default = 4.31 actually defaulted, probability of default = .05 </li></ul>Solution
38. 38. <ul><li>Given the following figures, compute the distance to default: </li></ul><ul><ul><li>Book value of liabilities : \$5.95 billion </li></ul></ul><ul><ul><li>Estimated default point : \$4.15 billion </li></ul></ul><ul><ul><li>Market value of equity : \$ 12.4 billion </li></ul></ul><ul><ul><li>Market value of firm : \$18.4 billion </li></ul></ul><ul><ul><li>Volatility of firm value : 24% </li></ul></ul>Problem
39. 39. <ul><li>Distance to default (in terms of value) = 18.4 – 4.15 = \$14.25 billion </li></ul><ul><li>Standard deviation = (.24) (18.4) = \$4.416 billion </li></ul><ul><li>Distance to default (in terms of standard deviation) = 3.23 </li></ul>Solution