7.4Multiperiod Binomial Valuation

Risk-neutral valuation methods can be applied whenever perfect tracking is possible.

When continuous and simultaneous trading in the derivative, the underlying security,

and a risk-free bond can take place and when the value of the tracking portfolio changes

smoothly (i.e., it makes no big jumps) over time, perfect tracking is usually feasible.

To the extent that one can trade almost continuously and almost simultaneously in the

real world, and to the extent that prices do not move too abruptly, a derivative value

obtained from a model based on perfect tracking may be regarded as a very good

approximation of the derivative’s fair market value.

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246Part IIValuing Financial Assets

When large jumps in the value of the tracking portfolio or the derivative can occur,

perfect tracking is generally not possible. The binomial process that we have been dis-

cussing is an exception where jumps occur and perfect tracking is still possible.

How Restrictive Is the Binomial Process in a Multiperiod Setting?

Initially at least, it appears as though the assumption of a binomial process is quite

restrictive—more restrictive in fact than the continuous trading and price movement

assumptions that permit tracking. Why, then, are we so focused on binomial processes

for prices, rather than focusing on the processes where prices move continuously? We

prefer to discuss tracking with binomial processes instead of continuous processes

because the latter, requiring advanced mathematics, is more complex to present. More-

over, binomial processes are not as restrictive as they might at first seem.

Although binomial processes permit only two outcomes for the next period, one

can define a period to be as short as one likes and can value assets over multiple peri-

ods. Paths for the tracking portfolio when periods are short appear much like the paths

seen with the continuous processes. Hence, it is useful to understand how multiperiod

binomial valuation works. Another advantage of multiperiod binomial models is that

the empirical accuracy of the binomial pricing model increases as time is divided into

finer binomial periods, implying that there are more binomial steps.

Numerical Example of Multiperiod Binomial Valuation

Example 7.12 determines the current fair market price of a derivative whose value is

known two binomial periods later. However, it will be necessary to program a computer

or use a spreadsheet to value derivatives over large numbers of short binomial periods.

In Example 7.12, the derivative is an option to buy airplanes. The “underlying asset,”

basically a financial instrument that helps determine risk-neutral probabilities, is a set

of forward contracts on airplanes. The binomial tree for the underlying forward price

(above the nodes) and derivative values (below the nodes) is given in Exhibit 7.11.

Example 7.12:Valuing a Derivative in a Multiperiod Setting

As the financial analyst for American Airlines, you have been asked to analyze Boeing’s offer

to sell options to buy 200 new 777 airplanes 6 months from now for $83 million per plane.

You divide up the 6 months into two 3-month periods and conclude—after analyzing his-

torical forward prices for new 777s—that a reasonable assumption is that new forward

prices, currently at $100 million per airplane, can jump up by 10 percent or down by 10 per-

cent in each 3-month period.By this, we mean that new 3-month forward prices tend to be

either 10% above or 10% below the new 6-month forward prices observed 3 months earlier.

Similarly, spot prices for 777s tend to be 10% above or 10% below the new 3-month for-

ward prices observed 3 months earlier.Thus, the Boeing 777 jet option is worth $38 million

per plane if the forward price jumps up twice in a row and $16 million if it jumps up only

once in the two periods.If the forward price declines 10 percent in the first 3-month period

and declines 10 percent again in the second 3-month period, the option Boeing has offered

American will not be exercised because the prespecified 777 purchase price of the option

will then be higher than the price American would pay to purchase 777s directly.What is the

fair price that American Airlines should be willing to pay for an option to buy one airplane,

assuming that the risk-free rate is 0%?

Answer:Exhibit 7.11 graphs above each node the path of the new forward prices that (as

described above) are relevant for computing risk-neutral probabilities.Below each node are

the option values that one either knows or has to solve for.The leftmost point represents

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Strategy, Second Edition

Chapter 7

Pricing Derivatives

247

today.Movements to the right along lines connecting the nodes represent possible paths that

new zero-cost forward prices can take.If there are two up moves, the forward (and spot) price

is $121 million per plane.Two down moves mean it is $81 million per plane.One up and one

down move, which occurs along two possible paths, results in a $99 million price per plane.

Solve this problem working backward through the tree diagram, using the risk-neutral val-

uation technique.Exhibit 7.11 labels the nodes in the tree diagram at the next period as U

and D(for up and down), and as UU, UD,and DD,for the period after that.Begin with a

look at the U, UU, UDtrio from Exhibit 7.11.

At nodeU,the forward price is $110 million per plane.Using the risk-neutral valuation

method, solve for the that makes 12199(1 ) 110.Thus, .5.The value of

the option at Uis

.5($38 million) .5($16 million) $27 million

At nodeD, the forward price is $90 million per plane.Solving 9981(1 ) 90 iden-

tifies the risk-neutral probability .5.This makes the option worth

.5($16 million) .5($0) $8 million

The derivative is worth either $27 million (node U) or $8 million (node D) at the next trad-

ing date.To value it at today’s date, multiply these two outcomes by the risk-neutral proba-

bilities.These are again .5 and 1 .5.Thus, the value of the derivative is

$17.5 million .5($27 million) .5($8 million)

Algebraic Representation of Two-Period Binomial Valuation

Let us try to represent Example 7.12 algebraically. Let V, V, and Vdenote the

uuuddd

three values of the derivative at the final date. To obtain its node Uvalue V, Exam-

u

ple 7.12 took (which equals .5) and computed

EXHIBIT7.11

Forward Price Paths (in $ millions) and Derivative Values (in $ millions) in a Two-Period Binomial Tree

6-month	3-month	Spot Price
Forward	Forward
		$121

	UU
$110	$38 = $121 – $83

	U
$100		$99
	?
		UD
	$90
?		$16 = $99 – $83
	D

?
	$81

DD

$0

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Strategy, Second Edition

248Part IIValuing Financial Assets

V(1 )V

V^uuud

^u1r

f

(Since rwas zero in the example, there was no reason to consider it, but it needs to

f

be accounted for in a general formula.) To obtain its node Dvalue V,the example

d

computed

V(1 )V

V^uddd

^d1r

f

To solve for today’s value V, Example 7.12 computed

		V(1 )V
V		ud
		1r
		f
		V(1 )VV(1 )V
		uuud(1 ) uddd
		(1r)2(1r)2
		ff
		2V2(1 )V(1 )2V
		uuuddd	(7.7)
		(1r)2
		f

This is still the discounted expected future value of the derivative with an expected

value that is computed with risk-neutral probabilities rather than true probabilities. The

risk-neutral probability for the value Vis the probability of two consecutive up

uu

moves, or ². The probability of achieving the value Vis 2(1 ), since two

ud

paths (down-up or up-down) get you there, each with probability of (1 ). The

probability of twodown moves is (1 )². Thus, the risk-neutral expected value of

the derivative two periods from now is the numerator in equation (7.7). Discounting

this value by the risk-free rate over two periods yields the present value of the deriv-

ative, V.Of course, the same generic result applies when extending the analysis to

three periods or more. To value the security today, take the discounted expected future

value at the risk-free rate.

Note that in principle, can vary along the nodes of tree diagrams. However, in

Example 7.12, was the same at every node because u, d, and r, which completely

f

determine , are the same at each node, making the same at each node. If this was

not the case, it is necessary to compute each node’s appropriate from the u, d, and

rthat apply at that node.

f