11
Multi-period binomial tree model

The one-period and two-period binomial trees presented in Chapters 9 and 10 had the advantage of introducing important concepts and procedures, such as replication, portfolio dynamics and risk-neutral formulas, in fairly simple setups. Now, we seek to generalize these ideas to a model with more than two time steps.

The main objective of this chapter is to introduce the general multi-period binomial tree model and its own challenges. The algorithmic approach we will use throughout the chapter will allow a straightforward computer implementation of the model and it will lay the foundations for the limiting case known as the Black-Scholes-Merton model. The specific objectives are to:

  • build a general binomial tree and relate the asset price observed at a given time step to the binomial distribution;
  • understand the difference between simple options and path-dependent options;
  • establish the dynamic replicating strategy to price an option;
  • apply risk-neutral pricing in a multi-period model.

11.1 Model

Fix n ≥ 1, the number of time steps or periods. In an n-period model there are n + 1 time points: 0, 1, 2, …, n. One should think of time 0 as today while time 1, time 2, etc. are in the future. Often, time 0 will represent the issuance date of a derivative whose maturity occurs at time n.

When time is expressed without units, the timeline is as follows:

A diagram that shows a timeline pointing from left to right with time points from zero to n marked on it.

As is the case for most discrete-time financial models, it is often more convenient to express time points 0, 1, 2, …, n in units of true time such as a year. Suppose we have a time horizon [0, T] in mind, with 0 being the initial time and T the end of the investment period (in years). We divide the time interval [0, T] in n periods given by (0, T/n), (T/n, 2T/n), …, ((n − 1)T/n, T). The whole timeline is represented graphically by

A diagram that shows a timeline pointing from left to right with time points from zero to T marked on it. The first five time points are shown to be 0, T over n, 2T over n, 3T over n, and 4T over n. The last two are left parenthesis n minus 1 into T over n and T.

Alternatively, defining h = T/n as the length of a time step, then each time interval (0, h), (h, 2h), …, ((n − 1)h, nh) can be represented graphically by

A diagram that shows a timeline pointing from left to right with time points from zero to nh marked on it. The first five time points are shown to be 0, 1h, 2h, 3h, and 4h. The last two are left parenthesis n minus 1 into h and nh.

In this chapter, we will express time both ways.

Finally, an n-period binomial model is a frictionless financial market with only two basic assets:

  • a risk-free asset (a bank account or a bond) which evolves according to the risk-free interest rate r;
  • a risky asset (a stock or an index) with a known initial value and unknown future values.

Investors in this market act and make transactions only at those n + 1 time points and not in between.

11.1.1 Risk-free asset

The risk-free asset is modeled by a (deterministic) process B = {Bk, k = 0, 1, 2, n − 1, n} with B0 = 1 and where BkBk − 1, for each k = 1, 2, ..., n. If r is the constant continuously compounded (annual) interest rate, then

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and, if r is the periodically compounded (annual) interest rate, i.e. compounded n times per year, then

numbered Display Equation

In other words,

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is independent of k: it is the constant one-period discount factor. Of course, units of h and r must coincide for everything to make sense.

11.1.2 Risky asset

The risky asset is modeled by a stochastic process S = {Sk, k = 0, 1, 2, …, n − 1, n}, where S0 is a known (deterministic) quantity and

numbered Display Equation

For simplicity, we assume that {Uk, k = 1, 2, …, n − 1, n} are independent and identically distributed random variables taking values in {u, d} with (real-world or actuarial) probabilities {p, 1 − p}. Once again, we assume that d < u and 0 < p < 1.

As before, in order to verify the no-arbitrage assumption, we must make sure that

(11.1.1)numbered Display Equation

when interest is compounded continuously, or

numbered Display Equation

when interest is compounded at each period. In plain words, one dollar accumulated at the risk-free rate over each period does not earn systematically more (or less) than an equivalent investment in the risky asset.

It should be clear that in this multi-period setup, we restrict ourselves to the simplified version of the binomial tree model as discussed in Chapter 10. Consequently, the n-period binomial tree model is based upon five parameters (r, S0, u, d, p) and it is recombining. Because the tree is recombining, the 2n possible trajectories lead to only n + 1 possible prices at time n.

The following illustration shows all possible trajectories of the risky asset price (time series) in a three-step binomial tree:

A tree with S subscript zero at the center from where two arrows lead to u S subscript zero and d S subscript zero, respectively. From u S subscript zero, two arrows lead to u squared S subscript zero and ud S subscript zero, respectively. From d S subscript zero, two other arrows lead to ud S subscript zero and d squared S subscript zero, respectively. From u squared S subscript zero two arrows lead to u cubed S subscript zero and u squared d S subscript zero. From ud S subscript zero, two arrows lead to u squared d S subscript zero and ud squared S subscript zero. From d squared S subscript zero, two arrows lead to u d squared S subscript zero and d cubed S subscript zero, respectively.

The sample space and the price of the risky asset

It is sometimes useful to describe the full probability model on which the binomial tree is constructed. For an n-period binomial model, one can choose its sample space Ωn as follows:

numbered Display Equation

It is then clear that each state of nature ω ∈ Ωn, that is each scenario, corresponds to a full trajectory of the risky asset S (a path in the tree). For example, if n = 3 as in the previous illustration, then the possible paths and prices at maturity are

numbered Display Equation

Example 11.1.1 Construction of a three-step binomial tree

Assume that, each week, a stock price can increase by 3% or decrease by 2%. If the initial stock price is 50, build the corresponding three-step binomial tree.

For each time point k = 1, 2, 3, the corresponding random variable Sk, modeling the stock price in k weeks from now, can take one of the following values:

numbered Display Equation

where i = 0, 1, …, k. The corresponding tree is

A tree with 50 at the center from where two arrows lead to 51.50 and 49, respectively. From 51.50, two arrows lead to 53.045 and 50.47, respectively. From 49, two other arrows lead 50.47 and 48.02, respectively. From 53.045, two arrows lead to 54.6364 and 51.9841. From 50.47, two arrows lead to 51.9841 and 49.4606. From 48.02, two arrows lead to 49.4606 and 47.0596. The path from 50 to 51.50 to 50.47 and to 49.4606 appears shaded.

where the highlighted path {50, 51.50, 50.47, 49.46} is a possible realization (time series) for the weekly stock price process S.

As illustrated by the previous trees (with n = 3), after k periods/steps there are k + 1 (different) nodes or, in other words, k + 1 different time-k prices. But overall, after k periods/steps, there are 2k different price trajectories for the process S = {Sk, k = 0, 1, 2, …, n − 1, n}. One of these particular paths was illustrated in the preceding and the next trees: one upward movement followed by two downward movements.

At maturity (after n periods), the rationale is similar: there are n + 1 different prices for the risky asset but a total of 2n paths leading to those values. Such a distinction is fundamental when pricing path-dependent derivatives as opposed to simple vanilla derivatives (see Section 11.1.3).

A tree with S subscript zero at the center from where two arrows lead to u S subscript zero and d S subscript zero, respectively. From u S subscript zero, two arrows lead to u squared S subscript zero and ud S subscript zero, respectively. From d S subscript zero, two other arrows lead to ud S subscript zero and d squared S subscript zero, respectively. From u squared S subscript zero two arrows lead to u cubed S subscript zero and u squared d S subscript zero. From ud S subscript zero, two arrows lead to u squared d S subscript zero and ud squared S subscript zero. From d squared S subscript zero, two arrows lead to u d squared S subscript zero and d cubed S subscript zero, respectively. The path from S subscript zero to u S subscript zero to ud S subscript zero and then to ud squared S subscript zero appears shaded.

Two popular choices for u and d

There are two popular specifications for u and d in the finance literature:

  • Cox-Ross-Rubinstein (CRR) model:

    numbered Display Equation
  • Jarrow-Rudd (JR) model:

    numbered Display Equation

where σ stands for the (annualized) volatility of the risky asset return. In practice, σ is in the range of [0.1, 0.4]. These formulas for u and d are practical because:

  1. they automatically adjust with the length of the time step h;
  2. they will become useful later to relate the binomial tree to the lognormal distribution (and hence the Black-Scholes formula).

Therefore, once n (or h) is fixed, the CRR or JR trees have four parameters: S0, r, σ and p.

11.1.2.1 Probability distribution

We now look at the probability distribution of the risky asset time-k price, where k = 1, 2, …, n. We can write

numbered Display Equation

where the random variable Ik counts the number of upward movements in the trajectory after k periods. Mathematically, it is defined as

numbered Display Equation

If there are i upward movements after k time steps, then

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where i can be as little as 0 but at most equal to k (k up-moves in k trials).

It is clear that Ik follows a binomial distribution (with respect to ) with parameters (k, p):

numbered Display Equation

which means that

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and consequently

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Example 11.1.2 Probability distribution in a three-step binomial tree

We follow example 11.1.1 and we add that the probability of an up-move is 58%. We want to determine the probability distribution of the random variables S1, S2 and S3.

The probability distribution of S1 is simply

numbered Display Equation

Also, we find that

numbered Display Equation

because there are two possible paths leading to a price of 50.47 (up & down, down & up). Moreover, . Finally,

numbered Display Equation

where again the constant 3 comes from the three possible paths leading to the price 51.9841 (same reasoning applies to 49.4606).

11.1.3 Derivatives

We now introduce a derivative (a third asset) in our model which is an asset whose value V is derived from the risky asset S. Therefore, the derivative’s price is modeled by a stochastic process V = {Vk, k = 0, 1, …, n}, where once again V0 is today’s value and V1, V2, …, Vn are (random) future values. It is assumed that the derivative matures at time n which means that Vn is a random variable representing the payoff.

We will consider two types of derivatives:

  • simple (vanilla) options, whose final payoff Vn is some function g of the final asset price, i.e. Vn = g(Sn). Examples include standard call and put options, binary and gap options, etc.;
  • exotic or path-dependent options, whose final payoff Vn is some function g of the entire trajectory, i.e. Vn = g(S1, S2, …Sn). Examples include Asian, lookback and barrier options.

Again, when dealing with call (resp. put) options, we will use C (resp. P) instead of V to represent the option’s price.

11.1.4 Labelling the nodes

The notation of the form vuu, xd or sud (with superscripts) was appropriate for a one-period or a two-period binomial tree to emphasize on the path taken by the price of the risky asset. Unfortunately, for n large, this notation is not convenient.

Therefore, we will use a more efficient notation relying on the fact that the underlying tree is recombining. The idea is that a node in the tree is uniquely identified by:

  1. time;
  2. number of upward movements needed to reach this node.

Therefore, for k = 1, 2, …, n and j = 0, 1, …, k, the pair (k, j) uniquely identifies the node at time k reached by a total of j upward movements (and therefore kj downward movements) after k steps. This notation will be useful to simplify and formalize some, but not all, computations and eventually for coding the binomial tree model with a programming language.

The following tree illustrates how each node is identified in a three-step binomial tree. We see that, for each time point k, nodes are numbered j = 0, 1, …, k, from bottom to top.

A tree with left parenthesis 0, 0 right parenthesis at the center from where two arrows lead to left parenthesis 1, 1 right parenthesis and left parenthesis 1, 0 right parenthesis, respectively. From left parenthesis 1, 1 right parenthesis, two arrows lead to left parenthesis 2, 2 right parenthesis and left parenthesis 2, 1 right parenthesis, respectively. From left parenthesis 1, 0 right parenthesis, two other arrows lead to left parenthesis 2, 1 right parenthesis and left parenthesis 2, 0 right parenthesis, respectively. From left parenthesis 2, 2 right parenthesis, two arrows lead to left parenthesis 3, 3 right parenthesis and left parenthesis 3, 2 right parenthesis. From left parenthesis 2, 1 right parenthesis, two arrows lead to left parenthesis 3, 2 right parenthesis and left parenthesis 3, 1 right parenthesis. From left parenthesis 2, 0 right parenthesis, two arrows lead to left parenthesis 3, 1 right parenthesis and left parenthesis 3, 0 right parenthesis.

To code the tree in a spreadsheet, or in an array-oriented programming language, we can also organize the tree as a table/matrix as illustrated next, where j corresponds to rows and k to columns. Note that nodes are numbered j = 0, 1, …, k from top to bottom which is much more convenient for referencing array cells.

j/k 0 1 2 3
0 (0, 0) (1, 0) (2, 0) (3, 0)
1 (1, 1) (2, 1) (3, 1)
2 (2, 2) (3, 2)
3 (3, 3)

As a first application of this notation, we can define Skj) as the realization/value of the random variable Sk corresponding to the j-th node. For a given path, in which we observe j upward movements from time 0 to time k, we define: for each j = 0, 1, …, k,

numbered Display Equation

Simple vanilla options are easier to handle than path-dependent options as there is no need to analyze all possible paths: only the final asset price matters. Therefore, we define Vkj) as the realization/value of the random variable Vk corresponding to the j-th node.

We have the following relationship at maturity: for each j = 0, 1, …, n,

numbered Display Equation

Our goal, in the next sections, is to determine the derivative’s price for all times and nodes prior to maturity. In other words, we will compute Vkj), for all k = 0, 1, …, n − 1 and j = 0, 1, …, k.

Moreover, when dealing with call (resp. put) options, we will use C (resp. P) instead of V to represent the option price. In particular, for a call option with exercise price K and maturing after n time steps, we have

numbered Display Equation

whereas for a similar put option, we have

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The evolution of the risky asset price and the corresponding derivative price is depicted in the following tree over three periods:

A box labeled V subscript zero equals TBD, with another label S subscript 0 outside it, appears at the center of a tree. From there, two arrows lead to boxes v subscript 1 of 1 equals TBD and v subscript 1 of 0 equals TBD, respectively. Outside of these two boxes are the labels S subscript 1 of 1 and S subscript 1 of 0. From S subscript 1 of 1, two arrows lead to boxes labeled V subscript 2 of 2 equals TBD and V subscript 2 of 1 equals TBD, respectively. Outside these two boxes are the labels S subscript 2 of 2 and S subscript 2 of 1. From S subscript 1 of 0, two other arrows lead to boxes labeled V subscript 2 of 1 equals TBD and V subscript 2 of 0 equals TBD, respectively. Outside these two boxes are the labels S subscript 2 of 1 and S subscript 2 of 0. From S subscript 2 of 2 two arrow lead to V subscript 3 of 3 and V subscript 3 of 2. Outside these two boxes are the labels S subscript 3 of 3 and S subscript 3 of 2. From S subscript 2 of 1, two arrows point to V subscript 3 of 2 and V subscript 3 of 1. Outside these are the labels S subscript 3 of 2 and S subscript 3 of 1. From V subscript 2 of 0, two other arrows lead to V subscript 3 of 1 and V subscript 3 of 0. Outside these are labels S subscript 3 of 1 and S subscript 3 of 0.

Example 11.1.3 Call option payoff in a general tree

The initial stock price is 32 and it may go up by 6% or down by 4% every period. Let us represent the evolution of the stock price by a three-step binomial tree along with the payoff of an at-the-money call option.

We have S0 = 32 and Sk = 32 × 1.06 j × 0.96kj, for j = 0, 1, …, k and k = 1, 2, 3. Moreover, the payoff of the call option is

numbered Display Equation

Those values are illustrated in the following tree:

A box labeled C subscript zero equals TBD, with another label 32 outside it, appears at the center of a tree. From there, two arrows lead to boxes C subscript 1 of 1 equals TBD and C subscript 1 of 0 equals TBD, respectively. Outside of these two boxes are the labels 33.92 and 30.72. From 33.92, arrows point to boxes C subscript 2 of 2 equals TBD and C subscript 2 of 1 equals TBD, respectively. Outside these two boxes are the labels 35.9552 and 32.5632. From 30.72, two other arrows lead to boxes labeled C subscript 2 of 1 equals TBD and C subscript 2 of 0 equals TBD, respectively. Outside these two boxes are the labels 32.5632 and 29.4912. From 35.9552, two arrows lead to boxes labeled 6.112512 and 2.516992. Outside these two boxes are the labels 38.112512 and 34.516992. From 32.5632, two arrows point to boxes labeled 2.516992 and 0. Outside these are the labels 34.516992 and 31.260672. From 29.4912, two other arrows lead to two boxes, each labeled 0. Outside these two boxes are labels 31.260672 and 28.311552.

11.1.5 Path-dependent payoffs

Exotic or path-dependent options such as Asian or lookback options cannot be treated as easily in the n-step binomial tree. One has to be very careful with the current notation (labeling of the nodes) because it focuses on the risky asset price at any given node, not taking into account the different paths reaching that node. Unless we rely on advanced techniques,1 the only choice we have when computing the final payoff of an exotic option is to consider each possible path separately. This is illustrated in the following example.

Example 11.1.4 Payoff of a fixed-strike Asian call option

Using the context of example 11.1.3, we now introduce a fixed-strike Asian call option with strike price K = 31.50. Determine the behavior of the random variable V3.

Asian options are path-dependent derivatives so we have to be careful and consider all eight possible payoffs (eight possible trajectories) in the three-step binomial tree of example 11.1.3. For this option, we cannot use the algorithmic notation Vkj). This Asian option has a payoff given by whose realizations/values are computed in the next table. Note that the random variable is based on an arithmetic average.

Path S0 S1 S2 S3 Average Payoff
uuu 32 33.92 35.9552 38.112512 35.995904 4.495904
uud 32 33.92 35.9552 34.516992 34.7973973 3.29739733
udu 32 33.92 32.5632 34.516992 33.6667307 2.16673067
udd 32 33.92 32.5632 31.260672 32.5812907 1.08129067
duu 32 30.72 32.5632 34.516992 32.600064 1.100064
dud 32 30.72 32.5632 31.260672 31.514624 0.014624
ddu 32 30.72 29.4912 31.260672 30.490624 0
ddd 32 30.72 29.4912 28.311552 29.507584 0

For example, the path corresponding to udu is highlighted in the following tree:

A tree with 32 at the center from where two arrows lead to 33.92 and 30.72, respectively. From 33.92, two arrows lead to 35.9552 and 32.5632, respectively. From 30.72, two other arrows lead 32.5632 and 29.4912, respectively. From 35.9552, two arrows lead to 38.112512 and 34.516992. From 32.5632, two arrows lead to 34.516992 and 31.260672. From 29.4912, two arrows lead to 31.260672 and 28.311552. The path from 32 to 33.92 to 32.5632 and to 34.516992 appears shaded.

For this path, the computation of the average goes as follows:

numbered Display Equation

So, the payoff in this scenario is given by

numbered Display Equation

Computations are done the same way for other trajectories.

As we saw in the previous example, pricing exotic options in an n-period binomial tree can be tedious, especially when n is large. Therefore, unless stated otherwise, the rest of this chapter will focus on simple vanilla options, even if most of the upcoming material could apply to path-dependent options with some adaptation to the notation.

11.2 Pricing by replication

Replicating a derivative in a multi-step binomial tree is similar to what we did in a two-step binomial tree. We will decompose the tree into 1 + 2 + 3 + … + n = n(n + 1)/2 one-period sub-trees and, for each, we will solve a system of two equations with two unknowns.

However, before looking at the replication procedure, we need to define and analyze trading/investment strategies, the portfolio value process, etc.

11.2.1 Trading strategies/portfolios

We now look at the concept of trading or investment strategies, also called portfolios, of which replicating strategies are a sub-group. Let us first define by Δk (resp. Θk), the number of units of the risky asset (resp. risk-free asset) that we hold during the k-th period, that is from time k − 1 to time k or during the time interval [k − 1, k). It is important to note that both Δk and Θk are determined at time k − 1, therefore using the information about the tree available up to that time point. This is illustrated on the following timeline:

A diagram that shows a timeline from left to right with the points k minus 1 and k marked on it. There is an arrow pointing from k minus 1 to k, labeled left parenthesis theta subscript k, delta subscript k.

The evolution of the number of units held in the risk-free asset and in the risky asset is represented by the corresponding (stochastic) processes Θ = {Θk, k = 1, 2, …, n} and Δ = {Δk, k = 1, 2, …, n}. Note that the time-index starts at k = 1, since Θ1 and Δ1 are the quantities held in the portfolio during the first period (from time 0 to time 1). A trading strategy is given by (Θ, Δ).

An investment strategy (Θ, Δ) is said to be static if Θk and Δk are constant over time, that is

numbered Display Equation

This is also known as a buy-and-hold strategy. Otherwise, when the strategy (Θ, Δ) is updated periodically as asset prices evolve, the strategy is said to be dynamic.

Example 11.2.1 Static and dynamic investment strategies

We know from Chapter 3 that to replicate a forward contract we need to hold one unit of stock and borrow the present value of K. Therefore, using the previously defined notation, this trading strategy is given by

numbered Display Equation

and

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This is a static strategy.

An example of a dynamic investment strategy is when we update our portfolio to maintain a fixed proportion of, say, 60%–40% (of the portfolio value) in stocks and bonds. Suppose a stock and a bond trade at $100 each and we want/need to invest $10,000. Therefore, at time 0, we choose Δ1 = 60 and Θ1 = 40, for a total investment of

numbered Display Equation

After one period, if for example the stock price is worth 150 and the bond 110, then the portfolio will be worth 60 × 150 + 40 × 110 = 13400. However, the position in the stock is worth 9000, which corresponds to 9000/13400 = 67.16% of the portfolio value at time 1. To maintain, our 60–40 strategy, we need to sell some shares of stock and buy more bonds, so that we ultimately set

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and

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In other words, we sell 60 − 53.6 shares of stock and buy 48.73 − 40 units of the bond. Those are the quantities set up at time 1 and prevailing during the second period. Because the portfolio has been rebalanced following the information observed at time 1, we say the investment strategy is dynamic.

Similar trades would need to be performed in the future, depending on the values of the basic assets at those times.

11.2.2 Portfolio value process

Let us now define by Πk the value of a trading strategy at time k (before rebalancing). Thus, for each k = 1, 2, …, n, set

(11.2.1)numbered Display Equation

This portfolio value is computed before rebalancing takes place. In particular, Πn = ΘnBn + ΔnSn is the value of the portfolio at maturity, a time where no rebalancing is needed.

At time k, based upon asset prices Bk and Sk, we will choose (Θk + 1, Δk + 1) for the upcoming period, i.e. the (k + 1)-th period (from time k to time k + 1). This is the rebalancing procedure.

To complete the definition of the portfolio value process, started in equation (11.2.1), we define the portfolio initial value as follows:

numbered Display Equation

It is the value of the portfolio at time 0 (inception).

The notation for trading strategies and its corresponding portfolio value is complicated by the fact that the portfolio quantities Θk and Δk are set up at time k − 1 and remain unchanged during the whole k-th period. They affect the portfolio value at both time k − 1 (after rebalancing) and time k (before rebalancing).

Moreover, we can see that the portfolio value at time k, that is Πk, depends on the risky asset price, so it is also a random variable. Therefore,

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is a stochastic process and it is often called the portfolio value process.

Example 11.2.2 Portfolio value process of the 60–40 portfolio

We continue example 11.2.1 where we had S0 = B0 = 100. Recall that to hold 60% in stocks and 40% in bonds, we need to set Δ1 = 60 and Θ1 = 40 initially, so that

numbered Display Equation

One period later, i.e. at time 1, in the scenario that S1 = 150 and B1 = 110, the portfolio would be worth

numbered Display Equation

before any rebalancing takes place.

To maintain the 60–40 proportion (of the new portfolio value of 13, 400), we need to update our portfolio quantities. Based upon the information available at time 1 (in the given scenario), we set Δ2 = 53.6 and Θ2 = 48.73 for the second period (from time 1 to time 2).

Finally, at time 2, the portfolio value will be given by

numbered Display Equation

which depends on the realization of the random variable S2.

11.2.3 Self-financing strategies

A trading strategy (Θ, Δ) is said to be self-financing if, when it is rebalanced, no money is injected or withdrawn: the portfolio value, before and after rebalancing, is equal. Mathematically, this means that, at each time step k = 1, 2, …, n − 1,

(11.2.2)numbered Display Equation

where the left-hand side is in fact Πk as given in equation (11.2.1). Thus, a self-financing strategy does not generate any cash flows (in or out) between inception and maturity. Only at maturity or inception does it require/allow money to be injected or withdrawn from it, making it comparable to the cash flows of most derivatives (premium at inception, payoff at maturity, no other cash flow). Note that equation (11.2.2) is an equality of random variables.

The profit or loss during the k-th period, for any investment portfolio, is given by

numbered Display Equation

For a self-financing strategy, using the self-financing condition of equation (11.2.2), we get

(11.2.3)numbered Display Equation

This provides another interpretation of the self-financing condition: the variation in the portfolio value comes from changes in values from both assets within the portfolio. This interpretation is the one to have in mind when we consider continuous-time models.

Example 11.2.3 Self-financing condition in the 60–40 portfolio

We continue example 11.2.2 and want to verify that the investment strategy satisfies the self-financing condition at time 1, at least in the given scenario.

Indeed, at time 1, in the given scenario, we have

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and

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So, the self-financing condition is satisfied because the gain/loss of the portfolio corresponds to the sum of the gains/losses made on each asset. Alternatively, still in the given scenario, we have

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In other words, before and after rebalancing, the portfolio is worth the same.

11.2.4 Replicating strategy

The main objective now is to determine the initial value V0 of a derivative having final payoff Vn. The key idea is that we need to set up a (self-financing) trading strategy such that

numbered Display Equation

in each possible scenario. Moreover, if the strategy is self-financing, which means it has no cash flows except at time 0 and time n (just like the derivative we want to replicate), the strategy is called a replicating strategy for this derivative.

By the no-arbitrage assumption, we can conclude that the portfolio process Π mimics the price of the derivative at each time point and in each possible scenario. Consequently, we have

numbered Display Equation

for all k = 0, 1, …, n. In particular, the initial values are equal: V0 = Π0. The price of the derivative is equal to the cost of the replicating portfolio.

11.2.5 Backward recursive procedure

Recall that in what follows, we consider only simple options. Using the labeling of nodes presented in section 11.1.4, let

numbered Display Equation

be the realized number of units of each asset an investor holds during the k-th period and in the scenario corresponding to the one-period sub-tree with root (k − 1, j). In other words, if at time k − 1 we have observed so far j upward movements, then during the next period (the k-th), we will hold Θkj) units of the risk-free asset and Δkj) units of the risky asset. Then, the time-(k − 1) value of the corresponding portfolio in this scenario is

(11.2.4)numbered Display Equation

Graphically, we have

A tree with left parenthesis k minus 1, j right parenthesis at the center from where two arrows, each labeled theta subscript k of j, delta subscript k of j, point to left parenthesis k, j plus 1 right parenthesis and left parenthesis k, j right parenthesis.

In this case, we emphasize that, for each fixed k = 1, 2, …, n, the possible values for j are 0, 1, …, k − 1.

For k = 0, the only possible value for j is of course 0, which is in agreement with the fact that Θ1 and Δ1 are deterministic values, i.e. are fixed at time 0.

Example 11.2.4 Relating the notation from Chapters 9 and 10

In the one-step binomial tree, the investment strategy set up at inception to replicate the realizations vu and vd of a payoff V1 was the pair (x, y). Using this chapter’s notation, we thus have

numbered Display Equation

In the two-step binomial tree, the quantities (xd, yd) were chosen at time 1, in the scenario S1 = sd, to replicate the values vdu and vdd, while the quantities (xu, yu) were chosen, in the scenario S1 = su, to replicate vuu and vud. With the new notation, in the scenario S1 = sd (i.e. no up-move so far), we have

numbered Display Equation

and, in the scenario S1 = su (i.e. one up-move so far), we have

numbered Display Equation

The one-period sub-tree with root at node (k − 1, j) is given by:

A tree with a box labeled V subscript k minus 1 of j at the center from where two arrows, each labeled theta subscript k of j, delta subscript k of j, point to boxes labeled V subscript k of j plus 1 and V subscript k of j. Outside the three boxes are labels S subscript k minus 1 of j, S subscript k of j plus 1 and S subscript k of j, respectively.

In this tree, to replicate the value of the derivative, it should be clear that we need to solve the following system of equations:

numbered Display Equation

where the only two unknowns at this point/stage are Θkj) and Δkj). As in Chapter 10, the solution is

numbered Display Equation

which can be further simplified to obtain

(11.2.5)numbered Display Equation

Again, to avoid arbitrage opportunities, we must have

(11.2.6)numbered Display Equation

where Πk − 1j) is given in equation (11.2.4).

The full n-period binomial tree and replicating strategy for payoff Vn is obtained by pasting together all the one-period sub-trees, i.e. for each k = 1, …, n and j = 0, 1, …, k − 1.

11.2.6 Algorithm

Replication and valuation in the general n-step binomial tree are summarized by the following algorithm:

  1. At maturity (time n), in each possible scenario, we must:

    1. compute the payoff for each j = 0, 1, …, n

      numbered Display Equation
    2. compute the corresponding (replicating) portfolio quantities

      numbered Display Equation

      as given in (11.2.5) for j = 0, 1, …, n − 1.

    3. compute the value of the replicating portfolio (at time n − 1):

      numbered Display Equation
    4. compute the value of the derivative:

      numbered Display Equation
  2. Then, we work backward and recursively: for each time k = n − 1, n − 2, …, 1, and in each possible scenario, i.e. for each j = 0, 1, ..., k, we must:

    1. from the previous step, retrieve the derivative’s value Vkj);
    2. compute the corresponding (replicating) portfolio quantities: for each j = 0, 1, …, k − 1, compute

      numbered Display Equation

      as given in (11.2.5).

    3. compute the value of the replicating portfolio (at time k − 1):

      numbered Display Equation
    4. compute the value of the derivative:

      numbered Display Equation

Note that in the very last step, i.e. for k = 1, we compute (Θ1(0), Δ1(0)) and then set V0(0) = Π0(0) or, written as before, V0 = Π0.

Example 11.2.5 Pricing a put option

Consider a European put option on a stock with spot price 50, maturity 2 years and exercise price 52. Assume the stock price evolves according to a two-period binomial tree, where each time step corresponds to 1 year and during which the price increases or decreases by 20%. In this model, the continuously compounded annual interest rate is 5%. Let us find the replicating portfolio of this option.

We have T = 2, K = 52, S0 = 50, u = 1.2, d = 0.8, n = 2, h = 1 and r = 0.05. We have graphically:

A tree with 50 at the center from where two arrows lead to 60 and 40, respectively. From 60, two arrows lead to boxes labeled P subscript 2 of 2 equals 0 and P subscript 2 of 1 equals 4, respectively. Outside these boxes are the labels 72 and 48. From 40, two other arrows lead to boxes labeled P subscript 2 of 1 equals 4 and P subscript 2 of 0 equals 20, respectively. Outside of these boxes are the labels 48 and 32.

Using the expressions in (11.2.5), we find the replicating strategies for the second period: in the sub-tree with root (1, 0), i.e. for S1 = 40,

numbered Display Equation

whereas, in the sub-tree with root (1, 1), i.e. for S1 = 60,

numbered Display Equation

So, the option price after 1 period (at time k = 1) is given by

numbered Display Equation

During the first period, the portfolio quantities are given by

numbered Display Equation

In conclusion, the initial price of the option is

numbered Display Equation

and more generally we have obtained

A box labeled 4.19, with another label 50 outside it, appears at the center of a tree. From there, two arrows, each labeled left parenthesis 24.3156, negative 0.402 right parenthesis lead to boxes labeled 1.41 and 9.46, respectively. Outside of these two boxes are the labels 60 and 40. From 60, two arrows, each labeled left parenthesis 12 e to the power of negative 0.1, negative 1 over 6 right parenthesis, lead to boxes labeled 0 and 4, respectively. Outside these two boxes are the labels 72 and 48. From 40, two other arrows, each labeled left parenthesis 52 e to the power of negative 0.1, negative 1 right parenthesis, lead to boxes labeled 4 and 20, respectively. Outside these two boxes are the labels 48 and 32.

Example 11.2.6 Replicating a GMMB in a binomial tree

A GMMB was issued by your insurance company several years ago. It has an annual premium of 2% with a guaranteed minimum payout of 102. As of today, the sub-account balance is $98.74 and the guarantee will apply 3 years from now. Given that the reference stock actually trades for 50 and increases by 7% or decreases by 5% every year, let us describe the replicating strategy covering the loss on the guarantee. Assume the risk-free rate is 3% (annually compounded) and use a binomial tree with annual time steps.

As we did in Chapter 10, replicating the loss on a variable annuity is similar to replicating options. Recall that the payoff is computed on the sub-account (after premiums and withdrawals), whereas replication is performed with the reference asset. Therefore, we will first build the binomial tree for the reference stock index and for the sub-account value. The tree for the reference stock is constructed using

numbered Display Equation

for each j = 0, 1, 2, 3, with S0 = 50, u = 1.07 and d = 0.95. The evolution of the reference asset is shown in the next table:

j/k 0 1 2 3
0 50 47.5 45.125 42.86875
1 53.5 50.825 48.28375
2 57.245 54.38275
3 61.25215

Then, the sub-account balance of the policyholder gets credited with the returns of the reference asset whereas premiums are withdrawn. Recursively, as we saw in Chapter 8, we have

numbered Display Equation

which is equivalent to

numbered Display Equation

knowing that A0 = 98.74. Therefore, the values Akj) of the sub-account balance are:

j/k 0 1 2 3 Loss
0 98.74 91.92694 85.5839811 79.6786864 22.3213136
1 103.538764 96.3945893 89.7433626 12.2566374
2 108.570748 101.079366 0.92063368
3 113.847286 0

The minimum guaranteed amount is G = 102 and hence the loss for the insurer is computed in the last column as max (102 − A3, 0). Therefore, the insurance company must implement an investment strategy to replicate the payoff values {22.3213136, 12.2566374, 0.92063368, 0} in the final nodes of the tree. Indeed, when the sub-account balance is $79.68 upon maturity, the company needs to provide for an extra $22.32 in order for the policyholder to receive at least 102.

We will illustrate how to replicate in a single sub-tree. The whole replicating strategy and its cost are provided in the tables below.

For example, the one-period sub-tree with root (2, 0), i.e. at time 2 with no up-move so far, is given by

A tree with a box labeled pi subscript 2 of 0 equals TBD at the center from where two arrows, each labeled theta subscript 3 of 0, delta subscript 3 of 0, point to boxes labeled V subscript 3 of 1 equals 12.2566374 and V subscript 3 of 0 equals 22.3213136. Outside of these two boxes are labels S subscript 3 of 1 equals 48.28375 and S subscript 3 of 0 equals 42.86875.

Using the formulas in (11.2.5), we get

numbered Display Equation

and

numbered Display Equation

Hence,

numbered Display Equation

In order to cover a random loss of either $22.32 or $12.26, the insurance company must have $15.16 to short-sell 1.86 shares of stock and invest 93.34 at the risk-free rate.

Repeating the same procedure for each sub-tree, we can determine the replicating strategy. This is given in the next table.

Δkj) Θkj)
j/k 1 2 3 1 2 3
0 −1.0314657 −1.85866596 −1.85866596 55.1968642 93.3444493 93.3444493
1 −0.66425063 −1.85866596 36.1230717 93.3444493
2 −0.13401952 7.51238289

Finally, the realizations Πkj) of the replicating portfolio value process are given in the last table. Note that this also corresponds to the no-arbitrage values of the insurer's loss.

j/k 0 1 2 3
0 3.62357939 7.85814956 15.1568247 22.3213136
1 1.66935538 4.56242872 12.2566374
2 0.2979397 0.92063368
3 0

11.3 Pricing with risk-neutral probabilities

As in the one- and two-period models, we can reorganize equation (11.2.6) (combined with the positions of equation (11.2.5)) in order to write the option price in node (k − 1, j) as a linear combination of its values one step ahead: for all k = 0, 1, …, n − 1 and j = 0, 1, …, k,

(11.3.1)numbered Display Equation

where

numbered Display Equation

is the one-period discount factor (applying from time k to time k − 1) and

numbered Display Equation

Since the model is arbitrage-free (cf. (11.1.1)), then 0 < q < 1 and we can interpret q as a (conditional) risk-neutral probability. Again, the expression in (11.3.1) can be interpreted as a risk-neutral expectation, i.e. using risk-neutral weights. Specifically, for each k = 0, 1, …, n − 1, we can write

numbered Display Equation

and, using iterated expectations, we further get

(11.3.2)numbered Display Equation

As in Chapters 9 and 10, it is important to recall that the (conditional) risk-neutral probability q has nothing to do with the (conditional) real-world probability p: it is only a convenient way to rewrite the value of the replicating portfolio as an expectation.

In conclusion, valuation in the general n-step binomial tree using risk-neutral formulas is summarized by the following algorithm:

  1. At time n, calculate the payoff of the derivative Vnj) = g(Snj)) for each j = 0, 1, ..., n.
  2. Then, proceed recursively, from time k = n − 1 back to time k = 0: compute Vkj) using equation (11.3.1) for each j = 0, 1, 2, …, k.

Example 11.3.1 Pricing a put option with risk-neutral formulas

Let us have a second look at example 11.2.5, now using risk-neutral formulas. We start by computing the risk-neutral probability of an up-move:

numbered Display Equation

The values of the payoff are

numbered Display Equation

Then, for (k, j) = (1, 0) and (k, j) = (1, 1), and then for (k, j) = (0, 0), using the relationship

numbered Display Equation

we obtain

numbered Display Equation

and finally

numbered Display Equation

We already know that the random variable Ik follows a binomial distribution with parameters (k, p) for each k = 1, 2, …, n. This is denoted as where we use to emphasize that each up-move has a (conditional) probability p. The probability measure is thus known as the real-world or actuarial probability measure.

But if instead each up-move had a (conditional) probability q, then we would have where emphasizes the use of q instead of p. The probability measure is known as the risk-neutral probability measure. It results that, for j = 0, 1, 2, …, k,

(11.3.3)numbered Display Equation

Combining equation (11.3.2) with equation (11.3.3), we deduce that the initial price of a simple option with payoff Vn = g(Sn) can be written as a risk-neutral binomial sum:

(11.3.4)numbered Display Equation

Example 11.3.2 Pricing a put option with risk-neutral formulas (continued)

Let us use equation (11.3.4) to simplify the computations of example 11.3.1. We know that q = 0.628177741 and hence

numbered Display Equation

Example 11.3.3 Replicating a GMMB in a binomial tree (continued)

We illustrate how to use risk-neutral formulas in example 11.2.6. Given the equivalence between replicating portfolios and risk-neutral formulas, we can easily combine the two approaches to determine the replicating strategy the actuary should implement during the first time period.

First, we compute the risk-neutral probability of an up-move for the reference stock. We find

numbered Display Equation

Second, the replicating strategy that applies in the first period requires to find the no-arbitrage price of the GMMB at time 1, i.e. corresponding to nodes (1, 0) and (1, 1). Using risk-neutral formulas, we get

numbered Display Equation

and

numbered Display Equation

Hence, initially the actuary must hold

numbered Display Equation

and

numbered Display Equation

as already obtained in example 11.2.6.

Applying equation (11.3.4) to a call option with payoff (SnK)+, we find that the initial price can be written as

numbered Display Equation

However, in many scenarios, the payoff is equal to zero. Indeed, when the realization of Sn is less than K, the corresponding term in the summation is equal to zero. Therefore, if we define

numbered Display Equation

as the smallest i such that the payoff is strictly positive, then we can get rid of the positive part and write2

(11.3.5)numbered Display Equation

We can do similar computations for a put option. The next example shows a numerical application to a 12-step binomial tree.

Example 11.3.4 Pricing a 1-year call option

Assume the 1-year rate is 5% (continually compounded) and the evolution of a stock price is represented by a 12-period binomial tree: every month, the stock price can increase by 1% or decrease by 0.7%. If the current stock price is 52, let us compute the no-arbitrage price of a 1-year call option with strike 54.

Although the problem seems daunting, the call option is a vanilla option whose price will be easy to compute with equation (11.3.5). We start by computing the risk-neutral (conditional) probability of an up-move:

numbered Display Equation

We compute the payoff in the extremal upper nodes when the price of the underlying is big enough, in which case the option will mature in the money. The possible outcomes for the payoff and the corresponding risk-neutral probabilities are given in the following table:

j C12j)
12 4.59490157 0.00651255
11 3.60865075 0.04073238
10 2.63900019 0.11676431
9 1.68567049 0.20286004
8 0.74838692 0.23789574
≤ 7 0 irrelevant

Consequently,

numbered Display Equation

A risk-neutral probability measure

The probability measure is an example of what is called a risk-neutral probability measure. It is such that:

  1. is equivalent to , which means here that 0 < q < 1;
  2. the stochastic process {Sk/Bk, k = 0, 1, 2, …, n} is such that:

    numbered Display Equation

    for all k = 0, 1, …, n − 1.

The second condition says that the process given by Sk/Bk is a -martingale.

11.4 Summary

Multi-period binomial tree model
  • Risk-free interest rate: r.
  • Risk-free asset price: B = {Bk, k = 0, 1, 2, …, n}, where B0 = 1 and, for k = 1, 2, …, n,

    numbered Display Equation
  • Risky asset price: S = {Sk, k = 0, 1, 2, …, n − 1, n}, where S0 is a constant and, for k = 1, 2, …, n,

    numbered Display Equation

    where the Ujs are iid random variables taking values in {u, d} with (real-world) probabilities {p, 1 − p}. Consequently, for each k = 1, 2, …, n,

    numbered Display Equation
  • No-arbitrage condition:

    numbered Display Equation
  • Derivative (written on S): V = {Vk, k = 0, 1, 2, …, n − 1, n}, where V0 is a constant (to be determined) and where Vn = g(Sn) (simple payoff) or Vn = g(S1, S2, …Sn) (exotic path-dependent payoff).
  • Important characteristics of the multi-period model:
    • – at time k: there are k + 1 possible stock prices but there are 2k different paths leading to those prices;
    • – it contains one-period binomial sub-trees.
  • Labelling the nodes:
    • – the pair (k, j) corresponds to the node at time k reached by a total of j upward movements (after k steps), where k = 0, 1, 2, …, n and j = 0, 1, …, k;
    • Skj) = ujdkjS0: realization of Sk in node (k, j);
    • – for a simple payoff Vn = g(Sn) with realization Vnj) = g(Snj)) = g(ujdnjS0) in node (n, j), the realizations of the time-k derivative prices Vk are given by Vkj) in node (k, j);
    • – for exotic path-dependent payoffs: we cannot use this notation.
Trading strategy/portfolio
  • Trading strategy/portfolio  

    Portfolio: a pair (Θ, Δ), where Θ = {Θk, k = 1, 2, …, n} and Δ = {Δk, k = 1, 2, …, n} are such that:

    • – number of units of B held from time k − 1 to time k (k-th period): Θk;
    • – number of units of S held from time k − 1 to time k (k-th period): Δk.
  • Number of units of the risk-free asset set up in node (k − 1, j): Θkj).
  • Number of units of the risky asset set up in node (k − 1, j): Δkj).
  • Portfolio value process: Π = {Πk, k = 0, 1, …n}, where

    numbered Display Equation

    and where, for k = 1, 2, …, n,

    numbered Display Equation
  • Self-financing condition: (Θ, Δ) is self-financing if, at each time step k = 1, 2, …, n − 1,

    numbered Display Equation

    or equivalently

    numbered Display Equation
Pricing by replication
  • A self-financing trading strategy (Θ, Δ) is a replicating strategy for V if

    numbered Display Equation

    Consequently, Vk = Πk, for all k = 0, 1, …, n.

  • Replicating portfolio for Vn: for each sub-tree with root (k, j), we solve

    numbered Display Equation

    and find

    numbered Display Equation

Pricing with risk-neutral probabilities

  • Risk-neutral probability: , where

    numbered Display Equation
  • Risk-neutral pricing formula for a derivative with payoff Vn: for each k = 0, 1, …, n − 1,

    numbered Display Equation

    or, written differently,

    numbered Display Equation

    for all j = 0, 1, …, k.

  • Initial price: .
  • Risk-neutral probabilities are only a mathematical convenience used for no-arbitrage pricing.

11.5 Exercises

  1. In a four-period binomial tree, the stock price can increase by 3% or decrease by 1%. If the initial stock price is $100 and analysts have determined that the probability that the stock price increases is 72%, compute:

    1. probability that the stock price is 108.18 after four periods (three ups, one down);
    2. probability of observing the path 100, 103, 101.97, 105.0291 and 108.18.
  2. You model the evolution of stock prices using a three-period binomial tree. You are given:

    • u = 1.07 and d = 0.98;
    • S0 = 37

    Detail all possible scenarios/paths taken by the stock price over the next three periods. Then, compute the payoff in each possible scenario for the following derivatives:

    1. at-the-money call option;
    2. floating-strike lookback put option;
    3. fixed-strike Asian call option with strike $35;
    4. up-and-out put option with H = 41 and strike $38;
    5. asset-or-nothing call option with strike $43.
  3. In a 3-month CRR binomial tree, you are given:

    • volatility of the stock price is 25%;
    • S0 = 40;
    • risk-free rate is 3% (compounded continuously).
    1. Build the tree for the stock price where one time step corresponds to one month.
    2. What is the no-arbitrage price of an at-the-money 3-month put option?
    3. How many shares of stock do you need to buy/sell at time 0 to replicate the option price at time 1 (for the option in (b))?
    4. What is the no-arbitrage price of the option in (b) at time 2 given that the stock price went down twice?
  4. A share of stock trades for $45 and a 3-year zero-coupon bond can be bought for $87 (face value of 100). The stock price evolves according to a three-period binomial tree with u = 1.07 and d = 0.96 (h = 1) and the interest rate is constant. You have $1000 to invest.

    1. You buy 60% of your initial investment in stocks and 40% in bonds. You make no more adjustments to your portfolio over the course of time. Describe your strategy (Δk, Θk) mathematically and compute the portfolio value after 3 years.
    2. Suppose now you want to make sure that as the stock price evolves, you always hold 60% of the investment in stocks. Describe your strategy (Δk, Θk) mathematically and compute the portfolio value after 3 years.
  5. In a 20-period binomial tree, you are given that:

    • the risk-free asset is such that B0 = B1 = B2 = … = B20 = 1;
    • the risky asset price can increase by 0.4% or decrease by 0.2% on each period;
    • the initial stock price is 1000;
    • one period corresponds to 1 day;
    • according to analysts, the probability that the asset price goes up on any given day is 62%.

    A 20-day at-the-money put option is issued. What is the no-arbitrage price of this option?

  6. In a three-period binomial tree, you are given:

    • S0 = 100;
    • u = 1.04 and d = 0.98;
    • the risk-free asset is such that Bk = 1.01k, k = 0, 1, 2, 3;
    • one period is 4 months.

    A 1-year floating-strike lookback call option is issued.

    1. There are eight possible paths for the stock price. Describe them all.
    2. Compute the payoff of this option on each of the possible paths. Note that the maximum/minimum is computed over S1, S2, S3.
    3. You believe that over any given period, the probability that the stock price goes up is 65%. What is the expected payoff of the option?
    4. What is the risk-neutral probability of observing each path?
    5. What should be the no-arbitrage price of this option at time 0?
    6. What is the replicating strategy applying during the first period to ultimately replicate the option’s payoff at maturity?
  7. In the binomial tree of exercise 11.6, a down-and-out call option with strike $102 and barrier $95 is issued.

    1. Compute the payoff of the option on each of the possible paths.
    2. Using risk-neutral probabilities, compute the option’s initial price.
    3. At each node, determine the replicating portfolio. What do you observe whenever the stock price is close to the barrier (over or below)?
  8. The price of a stock can increase by 5% or decrease by 2%. A dividend of $2.50 will be paid two periods from now and the initial stock price is $30. You build a four-period binomial tree.

    1. Build the binomial tree for the ex-dividend stock price.
    2. Compute the payoff of a four-period 28-strike put option.
    3. Given that the stock price went down twice after two periods, compute the put option price at time 2 if the periodic interest rate is 1%.
  9. In a three-period binomial tree, you are given:

    • each period is 3 months;
    • the initial stock price is $57;
    • the stock price can increase by a factor of u = 1.03 or decrease by a factor of d = 0.99;
    • the 3-month interest rate observed at time 0 is 2% (annualized, compounded quarterly) ;
    • future 3-month interest rate, i.e. , is such that ;
    • when the interest rate goes up, the stock price goes down.
    1. Build the binomial tree for the stock, for the interest rate and for a simple bank account with B0 = 1.
    2. Verify that the no-arbitrage condition holds in each sub-tree.
    3. Using replicating portfolios, compute the initial no-arbitrage price of an at-the-money call option.
    4. Using risk-neutral probabilities, verify the option price with the replicating portfolio value in each node of your work in (c).
  10. Equity-indexed annuities (EIAs)

    You are using the binomial tree of exercise 11.6.

    1. Compute the payoff a three-period compound annual ratchet EIA whose participation rate is 80%.
    2. Find the participation rate β such that the no-arbitrage value of the benefit is equal to the initial investment.
  11. Guaranteed minimum maturity benefit (GMMB)

    A 5-year GMMB is issued to a policyholder on a reference portfolio whose value can increase by 7% annually or decrease by 3% over the same period. The annual premium is 1.75% whereas the initial investment is 100,000. Calculate the loss in each possible scenario if the guaranteed amount is 85% of the capital. Assume one period corresponds to a year.

  12. Guaranteed minimum withdrawal benefit (GMWB)

    There are 3 years remaining on a GMWB whose sub-account balance is currently $26,000. The policyholder is allowed to withdraw 10,000 at the end of each year. If the reference index has a volatility of 30% and you are using a CRR binomial tree, calculate the loss in each possible scenario when the annual premium is 2.25%. Assume one period corresponds to a year.

Notes

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