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49

if,--

Thus, the standard deviation is .,/.029581 = .17199. 4, Mean

In 3 above we found that i = .056541. Thus, from formula (10.11) we have f?]] = «51.056541 = 4-2523.

Standard deviation

We have (1 +/t)" = = =.904837, so that k = .105171.

Applying formula (10.14) we have

var[a5] = .383244.

Thus, the standard deviation is /383244 = .6191.

Example 10.4 Assume that Efif] = .08 for t - 1, 2, 3. Also assume that 1 + if follows a lognormal distribution with = .0001. Find a 95% confidence interval for the accumulated value of an investment of 1 at the end of three years.

The random variable log(l + /,) has a normal distribution widi parameters M = log(1.08) = .076961

= .0001 .

Applying formula (10.17), we find that \oga(i) has a normal distribution with parameters

M = 3(.076961) = .230883

<t = 3(.0001) = .0003. The standard 95% confidence interval for Iogja(3) is given by

At + 1.96 a = .230883 + 1.96v.0003

or (.196935, .264831). The corresponding 95% confidence interval for a(3) would 196935 264831 (1.21766, 1.30321). This is a 95% confidence interval around (1.08) = 1.25971.

10.3 DEPENDENT RATES OF INTEREST

In Section 10.2 we assumed that the rates of interest /, in each successive period are independent. In Section 10.3 we consider some results if this assumption is removed.

Dependent rates certainly have intuitive appeal. For example, if the rate of interest in one period is considerably higher than a long-term average rate, then it is reasonable to assume that the following periods rate is more likely to be higher than average than it is to be lower than average. The same assumption would also seem reasonable for rates lower than average.

E[a{5)] = (1.067159)5 = 1.38403.

Standard deviation

From formula (10.16) we have

var[l + = ei2+.oi(g.oi , .011445.

Now 0(5) has a lognormal distribution with parameters 5/i and 5ff, so that from formula (10.16)

var[a(5)] = 5(.12+.01)(.05 .098212.

Thus, the standard deviation is v-098212 = .31339. As an exercise, the reader will be asked to confirm this answer by the alternate approach of using formula (10.4a).

2. Mean

From formula (10.5) we have

tsll = 51.067159 = 6-1023.

Standard devintion

From formula (10.8) we have

var[ijj] = .881737.

Thus, the standard deviation is /.881737 = .9390.

3. Mean

We first must analyze the distribution of (1 + /,)". We have oZeO + i,y = - logg(l + (,), which has a normal distribution with mean -fi and variance . Thus, (1 +;,)- has a lognormal distribution with parameters - and a.

From formula (10.15) we have

E[(i+i,r] = e-o+oos = .946485

Applying formula (10.9) gives us die mean

E[a\5)] = (.946485)* = .759572.

Note diat (1 + ()" = .946485, so dial / = .056541, which is a different rate of interest dian in 1 and 2 above.

Standard dfvifftjftrj

From formula (10.16) we have

var[(l + ,,)-] = e-.i2 + .oi(g.oi 1) .009003.

Now a\5) has a lognormal distribution widi parameters -5 and 5a, so diat from formula (10.16)

var[a-(5)] = e5(-.i2+,oi)(.05 1) .029581.



= =

var .003

We will denote actual values by and estimated values by 5 we have

From formula (10.21)

"[41

= S + /fe(5(3,-S)

! which gives

.075 = .09 + (8( ]-.09)

= .0675.

Example 10.7 It is known that follows an AR(2) process with a mean equal to .08. The foUowing values are given:

Z Actual Estimated 6]

1 2 3 4

.100 .110 .090 .095

.086 .094 .102 .092

Find the estimated value for Sj5j.

If we apply formula (10.24) for z = 3 and 4, we have

.102 = .08 + it,(.l 1 - .08) + -t2(.10 - .08)

.092 = .08 + *,(.09 - .08) + /2(11 - -08) "which simpUfies to

.03;t, + .022 = .022 .01*, + . = .012. These are two equations in two unknowns, which can be solved to give = .6 and k2 = .2. Note diat die values of satisfy die diree necessary (Conditions. Thus, we have

5,, = .08 + .6(.095 - .08) + .2(.09 - .08) = .091.

10.4 AN ASSET PRICING MODEL

In Section 9.5 we discussed the effect which risk and uncertainty have on die rate of interest. In this section we extend that discussion and introduce the

ky+k2<\ kj- ki<l -\ <k2<l.

Again, note that if ky = kj - 0, we have independence and the results ofl Section 10.2 apply.

The derivations of the formulas for the variance and covariance of the AR(1) and AR(2) processes are given in the books by G.E.P. Box and G. M. 1 Jenkins (1976) and R. B. Miller and D. W. Weichern (1977) listed in die bibliography. Formulas for the values of various financial functions based on the AR(1) and AR(2) processes are developed in two extensive papers by H. H. Panjer and D. R. Bellhouse (1980 and 1981) listed in the bibliography. These results are quite involved and are beyond the scope of this book.

The above cited approach to implementing the AR(1) and AR(2) processes are analytical in nature. However, anodier approach which can readily be used is simulation. Other models involving dependency can also be constructed and then implemented with simulation. It is important that any model be tested with empirical data before using it to see how well the model handles actual known results.

Example 10.5 Assume that the long-term average effective rate of interest in a particular situation is 6% and that the prior years rate is 9%. Compare the pattern produced by formula (10.19) over a three-year period, assuming each years rate turns out to exactly equal its estimated rate based on the long-term average rate and the prior years rate: (1) ifk = .2, and (2) if ~ .8.

1. We have

i, = .06 + .2(.09 - .06) = .066 (2 = .06 + .2(.066 - .06) = .0612 » = .06 + .2(.0612 - .06) = .06024. The regression of values to the mean rate of 6% is quite rapid.

2. We have

jl = .06 + .8(.09 - .06) = .084 /2 = .06 + .8(.084 - .06) = .0792 13 = .06 + .8(.0792 - .06) = .07536. Here the regression is much slower.

The purpose of this example is to illustrate the imphcit pattern produced by different values of Of course, in practice each years actual rate will quite likely differ from the estimated rate based on the long-term average rate and die prior years rate, so diat the smooth pattern above will be much more bumpy. However, the effect of the choice of k would be just as pronounced in terms of the magnitude of the regression toward the long-term average rate as in the above illustfadon.

Example 10.6 It is known that follows an AR(1) process f vith mean - .09, variance = .003, and covariance between adjacent values = .002. The estimated value of is .075. Find the actual value

We divide formula (10.23) by formula (10.22) to obtain

.002 2 3 "



Figure 10.1 Graphical representation of the Capital Asset Pricing Model

When /3 = 1 the covariance in the numerator equals the variance in the denominator. This condition can be inte eted as the security having the same level of rislc as the market portfolio. The inte retations ioT 0 < fij < I and > I reflect below-average or above-average risk, respectively.

As previously indicated, the Capital Asset Pricing Model was designed to estimate the expected yield rates on common stocks. Thus, the risk premium for the market portfolio, [ ] - is usually assumed to be a value close to 8.4% as indicated in Table 10.1.

The model has been subjected to extensive empirical testing and has performed reasonably well, particularly in view of its simplicity. However, it has been shown that other factors in addition to systematic risk have an influence on actual realized yield rates, e.g. seasonal factors and size factors. Also, the systematic risk for particular securities has been found to vary over time.

Nevertheless, despite these imperfections, the model has proven to be quite useful in practical applications. In fact, the model is so widely used that values of are calculated for a large number of common stocks which are periodically updated and published. They are available from any major brokerage firm.

Although the Capital Asset Pricing Model is generally used in connection with common stock analysis, it has broader applicability. For example, we could use a risk premium for 1 bonds and apply it to individual 1 bonds. From Table 10.1 the appropriate risk premium for the market portfolio in this case would be approximately 1.7%. Some analysts might prefer to use a rate based on more recent experience than going back as far as 1926. Since bond prices are considerably less volatile than common stock

Now W is an uncertain cash flow, but V is fixed and does not covary

with r

Thus, the expression becomes

I + rjr+ cov

and making a scalar adjustment we have

cov[W, ]

lp--f

fP - 7

We define

prices (see Table 10.2), a shorter period to u.se to determine the average experience would not seem to be unreasonable.

In Section 9.5 we discussed two alternative ways of reflecting risk: (1) through a risk premium in the interest rate, and (2) through an adjustment in the expected payments. It is possible to use the results above to develop an analytical approach to this second alternative.

To illustrate such an approach, consider an investment for one period. Let £[WI be the expected value of the uncertain cash flow at the end of the period, which we discount at the yield rate r which includes an appropriate risk premium. Then the present value, denoted by V is

1 + r

We now seek to quantify the risk in the payment instead of in the rate of interest and then discount at the risk-free rate of interest. Let the equivalent adjusted payment, which is assumed certain to be paid, be denoted by W. We then have

V = . (10.32)

We seek to find an expression for W and establish that W < E[W\.

In order to simplify the notation, we will set E[r] = r, [ \ = / , and var[r,] = We substitute formula (10.29) into formula (10.31) to obtain

= 1 + r = 1 + + ff,(f.p - r).

Substituting formula (10.30) the expression becomes

cov[r, r] 1 +rjr+ -i~r/).



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