The Investment Advisor Body of Knowledge + Test Bank - IMCA

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easier to work with a discount factor. In order to obtain the present value, we simply multiply the future value by the discount factor:

      (3.8)

      Because δ is less than one, Vt will necessarily be less than Vt+n. Different authors refer to δ or δn as the discount factor. The concept is the same, and which convention to use should be clear from the context.

      Geometric Series

      In the following two subsections we introduce geometric series. We start with series of infinite length. It may seem counterintuitive, but it is often easier to work with series of infinite length. With results in hand, we then move on to series of finite length in the second subsection.

      INFINITE SERIES

      The ancient Greek philosopher Zeno, in one of his famous paradoxes, tried to prove that motion was an illusion. He reasoned that, in order to get anywhere, you first had to travel half the distance to your ultimate destination. Once you made it to the halfway point, though, you would still have to travel half the remaining distance. No matter how many of these half journeys you completed, there would always be another half journey left. You could never possibly reach your destination.

      While Zeno's reasoning turned out to be wrong, he was wrong in a very profound way. The infinitely decreasing distances that Zeno struggled with foreshadowed calculus, with its concept of change on an infinitesimal scale. Also, an infinite series of a variety of types turn up in any number of fields. In finance, we are often faced with series that can be treated as infinite. Even when the series is long, but clearly finite, the same basic tools that we develop to handle infinite series can be deployed.

      In the case of the original paradox, we are basically trying to calculate the following summation:

      (3.9)

      What is S equal to? If we tried the brute force approach, adding up all the terms, we would literally be working on the problem forever. Luckily, there is an easier way. The trick is to notice that multiplying both sides of the equation by

has the exact same effect as subtracting
from both sides:

      The right-hand sides of the final line of both equations are the same, so the left-hand sides of both equations must be equal. Taking the left-hand sides of both equations, and solving:

      (3.10)

      The fact that the infinite series adds up to one tells us that Zeno was wrong. If we keep covering half the distance, but do it an infinite number of times, eventually we will cover the entire distance. The sum of all the half trips equals one full trip.

      To generalize Zeno's paradox, assume we have the following series:

(3.11)

      In Zeno's case, δ was

. Because the members of the series are all powers of the same constant, we refer to these types of series as geometric series. As long as |δ| is less than one, the sum will be finite and we can employ the same basic strategy as before, this time multiplying both sides by δ.

(3.12)

      Substituting

for δ, we see that the general equation agrees with our previously obtained result for Zeno's paradox.

Before deriving Equation 3.12, we stipulated that |δ| had to be less than one. The reason that |δ| has to be less than one may not be obvious. If δ is equal to one, we are simply adding together an infinite number of ones, and the sum is infinite. In this case, even though it requires us to divide by zero, Equation 3.12 will produce the correct answer.

      If δ is greater than one, the sum is also infinite, but Equation 3.12 will give you the wrong answer. The reason is subtle. If δ is less than one, then δ∞ converges to zero. When we multiplied both sides of the original equation by δ, in effect we added a δ∞+1 term to the end of the original equation. If |δ| is less than one, this term is also zero, and the sum is unaltered. If |δ| is greater than one, however, this final term is itself infinitely large, and we can no longer assume that the sum is unaltered. If this is at all unclear, wait until the end of the following section on finite series, where we will revisit the issue. If δ is less than –1, the series will oscillate between increasingly large negative and positive values and will not converge. Finally, if δ equals –1, the series will flip back and forth between –1 and +1, and the sum will oscillate between –1 and 0.

One note of caution: In certain financial problems, you will come across geometric series that are very similar to Equation 3.11 except the first term is one, not δ. This is equivalent to setting the starting index of the summation to zero (δ0 = 1). Adding one to our previous result, we obtain the following equation:

      (3.13)

      As you can see, the change from i = 0 to i = 1 is very subtle, but has a very real impact.

      Sample Problem

      Question:

      A perpetuity is a security that pays a fixed coupon for eternity. Determine the present value of a perpetuity, which pays a $5 coupon annually. Assume a constant 4 percent discount rate.

      Answer:

      FINITE SERIES

      In many financial scenarios – including perpetuities and discount models for stocks and real estate – it is often convenient to treat an extremely long series of payments as if it were infinite. In other circumstances we are faced with very long but clearly finite series. In these circumstances the infinite series solution might give us a good approximation, but ultimately we will want a more precise answer.

      The basic technique for summing a long but finite geometric series is the same as for an infinite geometric series. The only difference is that the terminal terms no longer converge to zero.

(3.14)

We can see that for |δ| less than 1, as n approaches infinity δn goes to zero and Equation 3.14 converges to Equation 3.12.

      In finance, we will mostly be interested in situations where |δ| is less than one, but Equation 3.14, unlike Equation 3.12, is still valid for values of |δ| greater than one (check this for yourself). We did not need to rely on the final term converging to zero this time. If δ is greater than one, and we substitute infinity for n, we get:

      (3.15)

      For the last step, we rely on the fact that (1 – δ) is negative for δ greater than one. As promised in the preceding subsection, for δ greater than one, the sum of the infinite geometric series is indeed infinite.

      Sample Problem

      Question:

      What

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