Going back to our numerical example, suppose we bought the American put for P(0)= 4.51. The problem we are facing at all times is the decision whether to exercise this option or not. (Refer to Figure 5.6 for the option prices computed at each stage.)
Consider first the strategy of exercising at the earliest possible time when the option price is equal to the available payoff.
Of course, we do not exercise at time 0 since the payoff is 0. Let BU
(resp. BD) be the set of all paths beginning with a u (resp. d) movement, and similarly define BUU, BU D, BDU, BDDand so on.
• Suppose the stock goes down in the first step, that is, consider ω ∈ BD. We exercise the option at time 1, cashing 10, which, as we saw, is higher than the expected profit from waiting.
• Suppose the stock goes up in the first step, so let ω ∈ BU. Here we distinguish three cases.
– Ifω ∈ BU DDDwe exercise at time 4, obtaining 16.17.
– Ifω = uuddd, ω = ududd or ω = uddud, we exercise at time 5 receiving 3.59.
– For other paths we do not exercise the option at all and receive zero.
Or, in other words, we exercise at time 5, where the payoff is zero.
We have defined a random variable assigning to eachω the time when we exercise the option:
τ1(ω) =⎧⎪⎪⎪⎨
⎪⎪⎪⎩
1 ω ∈ BD, 4 ω ∈ BU DDD, 5 otherwise.
This is an example of an important general notion, which we define next.
Definition 5.3
A random variableτ: Ω → {1, 2, . . . , N} is a stopping time if {ω : τ(ω) ≤ n} ∈ Fnfor all n.
The condition means that the decision to stop at time n will be based on the information available at that moment. This is even clearer in the following equivalent formulation, which you should verify.
Exercise 5.3 Prove thatτ: Ω → {1, 2, . . . , N} is a stopping time if and only if{ω : τ(ω) = n} ∈ Fnfor all n.
Going back to the example of the decision strategy for option exercise considered above, we can define a natural modification, related to the early exercise, of the process of the option values. These values fluctuate with time when we observe them along various scenarios. For example, ifω = udddu we have the sequence
P(n, ω) = (4.51, 1.23, 2.94, 6.94, 16.17, 3.59).
In such a scenario our strategy tells us to exercise at time 4. Imagine that we keep the money we have cashed, so the sequence is modified to become
Pτ(n, ω) = (4.51, 1.23, 2.94, 6.94, 16.17, 16.17).
For this particularω, τ(ω) = 4, and we left the sequence unchanged for n≤ 4, replacing the subsequent values by the value at time 4, so for n ≥ 4 we have P(n, ω) = P(τ(ω), ω).
This motivates the next definition.
Definition 5.4
For any sequence of random variables X(n) and any stopping time τ we define the stopped process Xτ(n) by
Xτ(n)= X(min{τ, n}),
or, more explicitly, by Xτ(n, ω) =
X(n, ω) if n≤ τ(ω), X(τ(ω), ω) if n > τ(ω).
In particular, Xτ(N) = X(τ). The stopped process Xτ inherits all the important properties of X.
Proposition 5.5
If X is adapted, then Xτis also adapted. If X is a martingale (supermartin-gale) the Xτis also a martingale (supermartingale).
Proof First we claim that we can write Xτin the form Xτ(n)= X(0) +
n j=1
1{ j≤τ}(X( j)− X( j − 1)). (5.1) The terms in the sum cancel out (the difference term on the right with j = 1 takes care of X(0)) so only the last term survives. The index of this survivor depends on the indicator function. Ifτ(ω) ≥ n, then it is one for all j so we have X(n) left. Otherwise the last j for which the indicator is one is
j= τ(ω) and the right-hand side reduces to X( j).
The sequence 1{ j≤τ}is predictable since
{ j ≤ τ} = Ω \ { j > τ} = Ω \ {τ ≤ j − 1} ∈ Fj−1.
Hence Xτ is adapted since all the terms on the right in (5.1) are Fn -measurable. So if X is a martingale then Xτ is a martingale transform, so by Proposition 4.31, Xτis also a martingale.
Next, again using the representation of Xτ(n) given in (5.1), we compute the conditional expectation of the increment:
E(Xτ(n+ 1) − Xτ(n)|Fn)= E(1{n+1≤τ}(X(n+ 1) − X(n))|Fn)
= 1{n+1≤τ}E(X(n + 1) − X(n)|Fn),
where we used the ‘take out what is known’ property of the conditional expectation (1{n+1≤τ} is Fn-measurable). Therefore the random variables E(Xτ(n+ 1) − Xτ(n)|Fn) andE(X(n + 1) − X(n)|Fn) have the same sign – they are either both greater than or equal to zero, or both less than or equal to zero. Hence if X is a supermartingale, that is,
E(X(n + 1) − X(n)|Fn)= E(X(n + 1)|Fn)− X(n) ≤ 0
then Xτis a supermartingale.
Remark 5.6
A submartingale is anFn-adapted sequence satisfyingE(X(n + 1)|Fn) ≥ X(n). The above proof also shows that if X is a submartingale, then the stopped process has the same property.
Going back to the numerical example, we seek a criterion for optimal-ity of the exercising strategy. Since the sum of money generated by such strategies can be random (in our special strategy the decision and the out-come depend onω), their comparison is difficult. Random variables are functions, and functions are rarely comparable. For this reason we need to associate a single number with each exercising strategy.
A natural candidate as optimality criterion is to maximise the mathemat-ical expectation of the payoff obtained at the exercise time. If the moment at which we exercise is denoted byτ, the money received in a particular scenarioω is the payoff I(τ(ω)). These sums of money emerge at different time instants, so for economic reasons we should discount them to make them comparable.
We choose the risk-neutral probability Q = (q, 1 − q) for the purpose of computing the expectation. To find the expected value of all discounted payments, note that we receive 10 forω ∈ BD, 16.17 for ω ∈ BU DDD, and 3.59 for ω = uuddd, ω = ududd, or ω = uddud so that
(1− q) 10
1+ R + q(1 − q)3 16.17
(1+ R)4 + 3q2(1− q)3 3.59
(1+ R)5 = 4.51 (5.2) which, remarkably, is the money we paid for the option. (We show the results up to two decimal points, so the equality is approximate, but we will prove that the equality is exact in general.)
As we will see, this is the best we can get. Such a claim could be verified numerically in our example since we have to deal with a finite number of possible strategies, but we will prove it in general.
First we give a formal definition of the optimal exercise strategy.
Definition 5.7
A stopping timeτ is optimal if
E(Z(τ)) = max{E(Z(ν)) : ν is a stopping time}. (5.3) Analysing exercising strategies in general, consider again the discounted payoff Z(n) = I(n) of the American put option. The Snell envelope Y of Z gives the value of our option at each trading date. If it satisfies Y(n)> Z(n), then it is not reasonable to exercise at time n, since we would be replacing a security worth Y(n) by a smaller amount Z(n). So, a sensible exercise time
τ should satisfy the condition Y(τ) = Z(τ). If such an opportunity emerges, then from the economic point of view we should grab it as soon as possible due to the time value of money. Below we prove that this intuition gives a correct solution.
Proposition 5.8
The random variableτmindefined by
τmin= inf{n : Y(n) = Z(n)}
with inf(∅) = N, is an optimal stopping time. In addition
Y(0)= E(Z(τmin)). (5.4)
Proof First note thatτminis a stopping time since {τmin= n} =
n−1
k=0
{Y(k) > Z(k)} ∩ {Y(n) = Z(n)}
and each of the sets on the right belongs toFn since the random variables involved are suitably measurable. For instance, for k< n,
{Y(k) > Z(k)} = (Y(k) − Z(k))−1((0, ∞)) ∈ Fk.
We will show that the stopped Snell envelope Yτmin is a martingale. Again using the representation (5.1) of the stopped process we have
Yτmin(n+ 1) − Yτmin(n)= 1{n+1≤τmin}(Y(n+ 1) − Y(n)).
If{τmin ≥ n + 1} then Y(n) > Z(n). By the definition of the Snell envelope, Y(n) as the greater of the two numbers Z(n) andE(Y(n + 1)|Fn) must equal the latter: Y(n)= E(Y(n+1)|Fn). We insert this in the right-hand side above and compute the conditional expectation
E(Yτmin(n+ 1) − Yτmin(n)|Fn)
= 1{n+1≤τmin}E(Y(n + 1) − E(Y(n + 1)|Fn)|Fn)= 0 which proves that Yτminis a martingale. Martingales have constant expecta-tion so for all n,
E(Yτmin(n))= E(Yτmin(0))= Yτmin(0)
the initial value being non-random. In particular, E(Yτmin(N))= Yτmin(0) but by the definition ofτmin, E(Yτmin(N))= E(Zτmin(N)) and also Y(0)= Yτmin(0). Putting all these together we get Y(0)= E(Zτmin(N)) as claimed in (5.4).
Finally we show thatτminis optimal. Take any stopping timeν. We know that Yνis a supermartingale so the expectation decreases with time and so
Y(0) ≥ E(Yν(N)). By the definition of the stopped process, E(Yν(N)) = E(Y(ν)) which is greater than or equal to E(Z(ν)) by the definition of the Snell envelope. So
Y(0)≥ max{E(Z(ν)) : ν is a stopping time}.
But (5.4) holds so Y(0) is equal to one of the expectations on the right.
Therefore, we must have equality, i.e. (5.3) holds, which completes the
proof.
Remark 5.9
Optimality of the early exercise strategy follows from simple economic considerations. If at a certain stage the option holder chooses not to exer-cise while the payoff is higher than the expectation of the future possibil-ities, then she effectively replaces the security held by an asset which is worth less, so suffers a loss (being the profit to the other party, the seller of the option). This will be seen clearly in the hedging argument presented below.
Putting together (5.4) and optimality ofτmin(that is, (5.3) holds) we have the following important conclusion concerning the price of the option:
Y(0)= max{E(Z(ν) : ν is a stopping time} (5.5) (Y(0) is the price P(0) if we consider the Snell envelope of a sequence of put option payoffs).
Exercise 5.4 In a binomial tree with three steps consider all possible stopping times and find numerically the one that maximises the ex-pected discounted payoff (expectation with respect to the risk-neutral probability).
Exercise 5.5 Find the optimal exercise time for the straddle de-scribed in Exercise 5.1.
Analysing the optimal exercise strategy in our leading example, path by path, we can see that before we exercise, the prices follow a martingale scheme since in the Snell envelope, the maximum of the two is the dis-counted martingale expectation. After we exercise (i.e. stop), the sequence becomes constant and so it is obviously a martingale. Thus we have found a simple and elegant criterion for optimality, however with little practical
meaning since verification of the martingale condition of the stopped pro-cess is not straightforward.
Theorem 5.10
A stopping time τ is optimal if and only if Z(τ) = Y(τ) and Yτ(n) is a martingale.
Proof See page 133.
5.3 Hedging
We keep working within the framework of our 5-step binomial example.
Suppose we have written and sold the American put, where S (0) = 100, U = 15%, D = −10%, K = 100, R = 5%. Having cashed the price, 4.51, we are exposed to some risk. To make our position secure we construct a replicating strategy which, we recall, is based on taking a position in the underlying and completing the portfolio with a position in the money market account. Our position in stock is determined by the delta computed by
Δ(0) = Pu− Pd
Su− Sd = 1.22 − 10
115− 90 = −0.350644
where superscripts denote the prices resulting from the up and down move-ments of the stock in the first step. This number is negative which means short-selling the stock. The money market position is then
4.51 + 0.350644 × 100 = 39.579 39.58 so this amount is invested risk free for one period.
Time n= 1
Consider the case where the stock goes down. There are two cases: the holder of the option either exercises or not.
• Suppose the option is exercised.
We owe the payoff 10 and we have to repurchase the fraction of the stock to close the short position the good news being that the stock is cheap since we only have to pay 0.35 × 90. The risk-free investment exactly covers this cost since R= 5% and
−10 − 0.35 × 90 = −41.56 = −39.58 × (1.05).
• Suppose the option is not exercised.
We compute the new delta
Δ(1) = 2.94 − 19
103.50 − 81 = −0.71.
This means that we have to increase our short position by short-selling additional 0.71 − 0.35 = 0.36 shares, which will generate some money to be added to our risk-free investment:
0.36 × 90 + 39.58 × (1 + 5%) = 74.23.
However, we do not need all this money for further hedging since the ‘value of waiting’ is (with q= 0.6)
1
1.05(0.6 × 2.94 + 0.4 × 19) = 8.92
so for replication such a value of our strategy is needed. Therefore to cover our liabilities: short stock position worth−0.71 × 90 = −64.4, and the op-tion:−8.92, we need 73.15 so we could consume 74.23−73.15 = 1.08. This would mean that our strategy would not be self-financing so we prefer not to withdraw any funds, which leads us to suspect that we are superhedging strictly, as will be confirmed below.
Time n= 2
Consider both cases of subsequent stock movements following the first step, which is down, and assume that the option is exercised or repurchased in each scenario to close the position.
• Stock goes up to Sdu = 103.50. We pay 2.94 for the option and we have to buy back 0.71 of a share and to cash our savings:
−2.94 − 0.71 × 103.50 + 74.23 × (1.05) = 1.135.
• Stock goes down to Sdd = 81. We pay 19 as the exercise pay-off. We buy back 0.71 of a share and we clear our money market account:
−19 − 0.71 × 81 + 74.23 × (1.05) = 1.135.
In each case as a result of the sub-optimal policy of the option holder (the option should have been exercised at time n= 1) we have a surplus (which is obviously the extra money 1.08 from the previous step increased by the risk-free return).
This is an example of a general fact which we will discuss below in the general setting.
The Doob decomposition
The sequence of prices Y(n) is a supermartingale, as we know. In partic-ular, their expected values decrease. To make such a sequence a martin-gale one should somehow compensate this ‘leakage of mass’ by adding some suitable quantities. We will show that such a compensation is al-ways possible, and moreover, that it is unique within a certain family of processes.