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51
 .7(.1000- .08) .82(.005) = 0.0981 [2] = .7(.0981 - .08) 1.61(.005) = 0.0846 [3] = .7(.0846 - .08) .43(.005) = 0.0854 (4] = .7(.0854 - .08) 1.02(.005) = 0.0889 (51 = .7(.0889 - .08) .79(.005) = 0.0823

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 one-year 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 three-month 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 Black-Scholes 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 Black-Scholes 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 Black-Scholes formula for a call given by formula (10.38) and die Black-Scholes formula for a put given by formula (10.40) satisfy put-call 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 Black-Scholes formula, rework Example 10.9 changing tlie following parameters one-by-one 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 = 0-b) 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 risk-free 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 risk-free 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 certainty-equivalent 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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