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51    .7(.1000  .08)   .82(.005) =  .0981  [2] =    .7(.0981   .08)   1.61(.005) =  .0846  [3] =    .7(.0846   .08)   .43(.005) =  .0854  (4] =    .7(.0854   .08)   1.02(.005) =  .0889  (51 =    .7(.0889   .08)   .79(.005) =  .0823 
Thus, the accumulated value at the end of five years for this interest rate path is 1000e«e°»e°«5e°««e°«23 = * = $1552 to the nearest dollar. In an actual problem of this type a large number of interest rate paths would be generated in like fashion. Statistical measures, such as the mean and standard deviation, could be computed for the distribution of accumulated values produced by the simulation. Example 10.12 An insurance company issues a oneyear guaranteed investment contract (GIC) which credits an annual effective rate of 8.5%. The company decides to invest the proceeds from the sale of the GIC in threemonth instruments at the beginning of each quarter. The rate of interest earned for the first quarter is 8.4% convertible quarterly. It is assumed that future interest rates move according to the random walk model in which the probability of an upward movement in the rate of interest is .4, and the amount of upward and downward movements each quarter are .5% and .4% convertible quarterly, respectively. Find the probability that the company will lose money on the GIC. The binomial lattice in diis case is an open lattice as illustrated in Figure 10.7 and not die more typical closed lattice. Thus, there are 2 = 8 interest rate paths. We will work in quarterly rates of interest. The rate of interest for the first quarter is .021. The amount of each upward movement in die rate of interest is .00125, while the amount of downward movement is .001. The results for the right interest rates paths are summarized in Table 10.3. Example 10.11 $1000 is invested for five years at an unknown force of interest which changes annually. It is assumed that the force of interest follows an AR(1) process in which 8 = .08, k = .7, and a = .005. Five random numbers are generated which produce the following values from the standard normal distribution: +.82, 1.61, +.43, +1.02, .79. The force of interest for the year just ending was .10. Find the accumulated value of the investment at the end of five years for this interest rate path. Five successive values of 8y are generated using formula (10.21) as follows: the opposite will be true Hp < Ml. The volatility of results is controlled by the magnitude of the possible upward and downward movements. The third type of method is a stochastic method. Probably the most common stochastic method used is to assume that successive values of 1 + if follow a lognormal distribution. This is consistent with the definition given in formula (10.18) and with the assumptions underlying the BlackScholes formula. One advantage with the stochastic method is the ability to readily adjust for the volatility of the interest rate. It can be shown that if common assumptions are used throughout, then this method becomes the limiting case of the binomial lattice method as the intervals in the binomial lattice shrink to zero. This happens in much the same fashion as the manner in which the BlackScholes formula becomes the limiting case of the random walk model for the pricing of options. One problem with both the probabilistic and stochastic methods is that interest rates may appear in some of the interest rate paths after a number of periods which seem unreasonably high or unreasonably low. One way of dealing with this problem is to lower the assumed volatility assumption in the movement of interest rates. Another way is to put bounds on how high or low die rates can go. However, neither method of dealing with the problem is ideal. Time series analysis is useful in connection with the construction of scenarios. This forces the analyst to consider past data in the construction of scenarios. However, it is also important to remember the admonition that "the future will not duplicate the past." Thus, some of the scenarios may well involve conditions that have not previously occurred. Finally, it is important not to ignore the importance of intermediate results by focusing too heavily on results at the end of the projection period. For example, for a particular scenario if we show a positive result at the end of the projection period, but are in a significant negative position midway through the period, we may not have a favorable scenario. At a minimum, a closer analysis of this particular scenario is in order. The above is a brief descriptive discussion of scenario testing. It is difficult to be very specific in discussing the technique, since there are probably as many different ways of conducting scenario testing as there are analysts doing it (maybe even more!). Nevertheless, this section should be useful in providing the reader widi a general appreciation of the various methods used and some of the considerations involved. A discussion of scenario testing in greater depth appears in a paper by M. F. Jetton (1988) listed in the bibliography. This paper includes more analytical consideration of the three methods described above.
Find 5r. .100 .105 .095 .104 .096 .100 a) iii.u 0[4]. b) If var [5[,]] = .0001, find cov[5[3], Sfg,]. 12. An investment fiind eamed 6% effective during the past year. For each of the next two years die yield on die investment fiind .02 + *(/,.,  . .06 + k{i, i  . .10+ *(;, , . a) Show diat b) Show diat !, is equally likely to be: ,06) 06). £[a(2)] = (1.06)2 + i.(.0O32)*. var[a(2)] = 1(.02158336 + .021573 + .01079168it2). 10.4 An asset pricing model 13. Using die data in Tables 10.1 and 10.2, compute die market price of risk for: a) Common stocks. b) Corporate bonds. c) Treasury bills. 14. For die common stock in Example 10.8, fmd die amount by which die expected value of the price of the stock at the end of one year was increased by the assumption it was a stock with above average market risk. 10.5 An option pricing model 19. Verify algebraically that the BlackScholes formula for a call given by formula (10.38) and die BlackScholes formula for a put given by formula (10.40) satisfy putcall parity. 20. Find the value of the European put with the same expiry date and exercise price as the European call in Example 10.9. 21. Find the following limiting values for the value of a call option: a) S = 0. b) S is very large in relation to E. c) E = 0. d) E is very large in relation to S. e) /7 = 0. f) is very large. 22. In order to illustrate the sensitivity of the BlackScholes formula, rework Example 10.9 changing tlie following parameters onebyone while holding all the other parameters constant. (Note: You will need a table for values of the cumulative 10.3 Dependent rates of interest 7. Esdmate 5(,] for/ = 6, 7, 8 given die data in Example 10.6, assuming ftiture actual values are equal to estimated values. 8. a) Show diat formula (10.25) simplifies to formula (10.22) if = 0b) Show diat formula (10.26) simpUfies to formula (10.23) if *2 = 0. 9. Estimate for die AR(2) process in Example 10.7 based on die data for Z = 1 2, 3, 4, assuming errors are distributed normally with mean = 0 and assuming that the population variance is equal to the sample variance. 10. Assume diat die value of for die AR(2) process in Example 10.7 is actually equal to .0002. a) Find var[5j,j]. b) Find cov [5f,], 5[,2]] 11. It is known diat 5(,] is normally distiibuted and follows as AR(1) process. The following values are given: Z Actual 5[j] Estimated 5 15. a) Use the dividend discount model developed in Section 7.10 to find the implied annual rate of dividend increase for the common stock in Example 10.8, if present values are computed at the rate of interest produced by the Capital Asset Pricing Model. b) Is this implied annual rate of dividend increase consistent widi the answer to Example 10.8? c) Assuming the real riskfree rate of interest is 3%, find the excess of the annual rate of dividend increase over the annual rate of inflation. 16. A business firm decides to use the Capital Asset Pricing Model to evaluate two projects A and B. Project A has normal risk with /3=1, while Project has high risk with /3 = 2. Each project is expected to return the same dollar amount at the end of one year and nothing thereafter. The risk free rate of interest is 5% and the market risk premium is 1%. If die two projects are combined into one project, find /3 for the combined project. 17. Stock A has = .5 and investors expect it to return 7%. Stock has /3 = 1.5 and investors expect it to return 15%. a) Find the risk free rate of interest. b) Find (J/la, assuming equal correlation coefficients between the market portfolio and Stocks A and B. 18. An investment is projected to return $110 at the end of one year and $121 at the end of two years. The riskfree rate of interest is 5%, the market risk premium is 10%, and /3 for this investment is .5. a) Find the present value of the investinent at the risk adjusted rate. b) Find die certaintyequivalent dollar returns at the end of each year.
Values of v", (1+0", , , Vs and a table of constants are tabulated at the following rates of interest: 0.5%, 1%, 1.5%,2% 25% 3% 3.5%,4%,4.5%,5%,6%, 7%, 8%,9%, 10%, 12%.  INTEREST TABLES AT 0.5% Constants  Function  Value   .005000  ,<2)  .004994  ,<4)  .004991  ,<12)  .004989   .004988   .004975   .004981   .004984   .004987   .004988   .995025   .997509   .998754  1/12  .999584   1.005000   1.002497   1.001248  (l+0"2  1.000416   1.001248  7,<)  1.001873  /7/(12)  1.002290   1.002498  d(2)  1.003748  d<>  1.003123  /7d<2)  1.002706   1.002498 
 v"  (i + ,r      .99502  1.00500  .9950  1.0000  1.000000   .99007  1.01003  1.9851  2.0050  .498753   .98515  1.01508  2.9702  3.0150  .331672   .98025  1.02015  3.9505  4.0301  .248133   .97537  1.02525  4.9259  5.0503  .198010   .97052  1.03038  5.8964  6.0755  .164595   .96569  1.03553  6.8621  7.1059  .140729   .96089  1.04071  7.8230  8.1414  .122829   .95610  1.04591  8.7791  9.1821  .108907   .95135  1.05114  9.7304  10.2280  .097771   .94661  1.05640  10.6770  11.2792  .088659   .94191  1.06168  11.6189  12.3356  .081066   .93722  1.06699  12.5562  13.3972  .074642   .93256  1.07232  13.4887  14.4642  .069136   .92792  1.07768  14.4166  15.5365  .064364   .92330  1.08307  15.3399  16.6142  .060189   .91871  1.08849  16.2586  17.6973  .056506   .91414  1.09393  17.1728  18.7858  .053232   .90959  1.09940  18.0824  19.8797  .050303   .90506  1.10490  18.9874  20.9791  .047666   .90056  1.11042  19.8880  22.0840  .045282   .89608  1.11597  20.7841  23.1944  .043114   .89162  1.12155  21.6757  24.3104  .041135   .88719  1.12716  22.5629  25.4320  .039321   .88277  1.13280  23.4456  26.5591  .037652   .87838  1.13846  24.3240  27.6919  .036112   .87401  1.14415  25.1980  28.8304  .034686   .86966  1.14987  26.0677  29.9745  .033362   .86533  1.15562  26.9330  31.1244  .032129   .86103  1.16140  27.7941  32.2800  .030979   .85675  1.16721  28.6508  33.4414  .029903   .85248  1.17304  29.5033  34.6086  .028895   .84824  1.17891  30.3515  35.7817  .027947   .84402  1.18480  31.1955  36.9606  .027056   .83982  1.19073  32.0354  38.1454  .026215   .83564  1.19668  32.8710  39.3361  .025422   .83149  1.20266  33.7025  40.5328  .024671   .82735  1.20868  34.5299  41.7354  .023960   .82323  1.21472  35.3531  42.9441  .023286   .81914  1.22079  36.1722  44.1588  .022646   .81506  1.22690  36.9873  45.3796  .022036   .81101  1.23303  37.7983  46.6065  .021456   .80697  1.23920  38.6053  47.8396  .020903   .80296  1.24539  39.4082  49.0788  .020375   .79896  1.25162  40.2072  50.3242  .019871   .79499  1.25788  41.0022  51.5758  .019389   .79103  1.26417  41.7932  52.8337  .018927   .78710  1.27049  42.5803  54.0978  .018485   .78318  1.27684  43.3635  55.3683  .018061   .77929  1.28323  44.1428  56.6452  .017654 
Table of compound interest functions
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