American call options
It will now become clear why our leading example above was a put. Specif-ically, we show that the prices of European and American call options writ-ten on a stock not paying dividends are the same.
Denote by CE(n) the price at time n of a European call option with ex-ercise price K and exex-ercise time N. The prices of the American version of the same option will be denoted by CA(n) so CA(n) = Y(n) in the general notation.
Proposition 5.14
For all n we have CA(n)= CE(n).
Proof First, note that the inequality CE(n) ≤ CA(n) holds as a result of the No Arbitrage Principle. If CE(n) > CA(n) for some n, at this time we buy the American and sell the European call creating a positive cash flow which we immediately invest in the risk-free asset. We do not exercise the American option until time N, when both options are (or are not) exercised, with zero balance in each case.
We thus need to prove that CE(n) ≥ CA(n). We know that CE(n) is the value at n of a replicating strategy and so after discounting we have a martingale. In particular, CE(n) is a supermartingale and to show that CE(n) ≥ CA(n) it is now sufficient to show that CE(n) dominates the pay-off I(n) = Z(n). Indeed, the Snell envelope CA(n) = Y(n) is the minimal supermartingale dominating the payoff so it cannot exceed any other such supermartingale, in particular CA(n)≤ CE(n).
So, it remains to see that CE(n)≥ I(n) and using call-put parity we have CE(n)= S (n) − K(1 + R)−(N−n)− PE(n)
≥ S (n) − K(1 + R)−(N−n)
> S (n) − K
which implies CE(n) ≥ S (n) − K. Of course, CE(n) ≥ 0 so CE(n) ≥ max{0, S (n) − K} = I(n) and the proof is complete. The inequality CE(n)≥ I(n) gives a model-independent argument since it shows that it is never optimal to exercise the American call option before expiry as the option value exceeds the available payoff at time n.
It may seem that the holder of the American call option has an advan-tage over the holder of the European option since the latter cannot exercise before expiry. However the European option can be sold at any time for the same amount as can be obtained from exercising the American call.
Finally, we give a simple example showing that if the stock pays divi-dends, it may be profitable to exercise the American call prior to the expiry date.
Example 5.15
Suppose S (0) = 120, U = 20%, D = −10%, and consider a call with N = 2, K = 120. Assume that R = 10% and just before time 1 a dividend of 14 is announced and will be paid to the current shareholders soon after time 1 (there is always a gap between these events). This will move the prices at time 1 down by this amount since if the share is sold, the new owner will not receive the dividend, and the resulting tree is shown below
The call gives us some money at time 2 in just one scenario, namely uu. For European call at time 1 in the up state the value is
1 1.1×2
3× 36 = 21.18, so
CE(0)= 1 1.1×2
3× 21.18 = 12.84, CA(0)= 1
1.1×2
3× 24 = 14.54.
144 156
130
120 117
108 112.8
94
84.6
In this state the payoff of the American call is 24 so it is higher, and the option should be exercised in this state. The initial prices of the options are of course different; nonetheless, the American call, if correctly exercised at time 1, is more profitable.
Exercise 5.10 Investigate the dependence of American call price on a dividend expressed as a percentage of the stock price.
Exercise 5.11 Show that European and American derivative securi-ties not paying dividends have the same prices if the payoff is of the form h(S (n)) for convex h: [0, ∞) → [0, ∞) with h(0) = 0. Find a counterexample if the last condition is violated.
Bounds on option prices
Here we consider some simple bounds on the initial prices of American options, similar to those proved earlier for European options. Arbitrage arguments again provide the principal tool, and a good deal of the detail is left to the reader, as these methods should be very familiar by now. We begin with the relationship between calls and puts.
Recall the call-put parity relation, proved as Theorem 4.41
CE− PE = S (0) − K(1 + R)−N. (5.6) For American options we cannot obtain an equality, but nonetheless the following inequalities provide useful bounds for the prices of American puts and calls on a stock that does not pay dividends.
Proposition 5.16
Let CA, PAdenote the initial prices of, respectively, an American call and an American put option, each with strike K, on the stock S (which pays no dividends). Then
S (0)− K ≤ CA− PA≤ S (0) − K(1 + R)−N.
Proof Suppose that the first inequality fails, so that CA − PA − S (0) + K < 0. We construct an arbitrage as follows: sell a put, short a share and
buy a call. Invest the balance S (0)+PA−CA, which is positive, since the call costs less than S (0), in the money market. Suppose the put is exercised at time n. We can borrow K in the money market to settle the put and use the share we receive to close the short position. Our net position then consists of the call and cash of
(S (0)+ PA− CA)(1+ R)n− K > (1 + R)nK− K > 0.
On the other hand, if the put is not exercised, we exercise the call at time N and obtain a share for K, enabling us to close the short position in the stock. Closing the money market position we then have
(S (0)+ PA− CA)(1+ R)N− K > K(1 + R)N− K > 0.
In both cases we have a riskless profit, which implies that CA− PA−S (0)+
K≥ 0, proving our first inequality.
By Proposition 5.14 the American and European call prices are the same, so CE = CA, while the American put price always dominates that of the European put, PA≥ PE. Using (5.6) this proves the second inequality.
Exercise 5.12 Suppose the underlying stock S pays a dividend D at time n< N. Show that
S (0)− D(1 + R)−n− K ≤ CA− PA≤ S (0) − K(1 + R)−N.
Bounds on the individual option prices can also be found quite simply.
For the American call we obtain the same bounds as in the European case.
By Propositions 2.45 and 5.14,
max{0, S (0) − K(1 + R)−N} ≤ CA≤ S (0).
For the put price PA, note that we must have K−S (0) ≤ PA, since otherwise we would buy the put and exercise it immediately, gaining a risk-free profit of K− S (0) − PA> 0. A strict upper bound for PAis given by K: if PA≥ K we can sell the put and invest the proceeds in the money market. If the put is exercised at time n we obtain the stock for K and sell it at once for S (n). This leaves a balance of PA(1+ R)n− K + S (n) > 0, while if the put is not exercised we retain PA(1+ R)N > 0. We have therefore proved that
(K− S (0))+≤ PA< K.
Exercise 5.13 Show that the following hold for a stock that pays a dividend D at time n≤ N:
max{0, S (0) − D(1 + R)−n− K(1 + R)−N, S (0) − K} ≤ CA, max{0, K(1 + R)−N+ D(1 + R)−n− S (0), K − S (0)} ≤ PA.
Dependence on the underlying
We make use of the various bounds on option prices obtained above to investigate the dependence of these prices on various parameters. The first results are obvious, although it would be instructive for you to provide a rigorous arbitrage proof. We already know that the call CE(S ) is a non-decreasing function of S, so that, since CA(S ) = CE(S ), the same holds for the American call price. We denote these prices by CA(S ), PA(S ) to emphasise the variable being examined.
Exercise 5.14 Show that the American put option price PA(S ) is a non-increasing function of the underlying S.
We can use our option price bounds to say rather more:
Proposition 5.17 If S < S then
CA(S )− CA(S )< S − S , PA(S )− PA(S )< S − S . Proof Using the inequalities in Proposition 5.16 we have
PA(S )− CA(S )≤ K − S , CA(S )− PA(S )≤ S − 1
(1+ R)NK. Add, then
[CA(S )− CA(S )]+ [PA(S )− PA(S )]
≤ S − S + (1 − 1 (1+ R)N)K
≤ S − S .
Both terms in square brackets on the left are non-negative (by the above exercise) so neither can be greater than the term on the right. In other words, the (positive) slope of S → CA(S ) is less than 1, while the (negative) slope of S → PA(S ) is greater than−1. As the differences on the left-hand side are non-negative, these two functions in fact satisfy Lipschitz conditions with Lipschitz constant 1.
Our final result is left as an exercise.
Exercise 5.15 Show that American call and put prices are both con-vex functions of the underlying.
Dependence on the strike price
Consider the American call and put prices as functions of the strike K. We denote them by CA(K), PA(K) for this purpose. Trivially, the call price is non-increasing (it coincides with CE(K) so Proposition 2.46 applies) and the put price is non-decreasing since, if K < K , the right to sell a share for K is more valuable than the right to sell it for K . We again have Lipschitz constant 1 for both functions, as the next exercise implies.
Exercise 5.16 Provide arbitrage arguments to verify the following inequalities when K < K :
CA(K )− CA(K )< K − K , PA(K )− PA(K )< K − K .
Note that Exercise 2.31 provides a stronger result for European options, with Lipschitz constant (1+ R)−1 < 1 in each case. Finally, we note that CA(K) = CE(K) ensures that K → CA(K) is a convex function, and prove that K→ PA(K) is also convex. (For simplicity we again just consider the midpoint.)
Proposition 5.18 If K < K then
PA(K + K 2 )≤ 1
2PA(K )+ 1
2PA(K ).
Proof Suppose 2PA(K +K2 )> PA(K )+ PA(K ). Sell two puts with strike
K +K
2 , buy two puts with strikes K , K and invest the difference risk free.
For any n≤ N we have 2(K +K2 − S (n))+ ≤ (K − S (n))++ (K − S (n))+, since the function f (x) = (x − S (n))+ is convex. If the holder of the puts sold wishes to exercise, we do the same and the inequality makes sure that we have sufficient funds. The risk-free investment gives the arbitrage
profit.
5.5 Proofs
Theorem 5.10
A stopping time τ is optimal if and only if Z(τ) = Y(τ) and Yτ(n) is a martingale.
Proof Suppose that Z(τ) = Y(τ) and Yτ(n) is a martingale, which implies (by the constant expectation property)
Y(0)= E(Yτ(N))= E(Y(τ)),
the last equality holding by the definition of the stopped process. The last number is equal to E(Z(τ)) by our hypothesis. Proposition 5.8 tells us that Y(0) = max E(Z(ν)), so E(Y(τ)) = max E(Z(ν)) and τ is optimal as claimed.
Conversely, if τ is optimal then by (5.3) and (5.5) we have Y(0) = E(Z(τ)). By the definition of the Snell envelope we have E(Z(τ)) ≤ E(Y(τ)).
But we know that Yτ is a supermartingale soE(Y(τ)) = E(Yτ(N)) ≤ Y(0).
Putting these inequalities together we get
Y(0)= E(Z(τ)) ≤ E(Y(τ)) = E(Yτ(N))≤ Y(0)
soE(Z(τ)) = E(Y(τ)). In addition, we know that Z(τ) ≤ Y(τ), so Z(τ) = Y(τ).
It remains to prove that Yτis a martingale and to this end it is sufficient to show that Yτ(n) = E(Y(τ)|Fn). We know that Yτ is a supermartingale hence Yτ(n) ≥ E(Yτ(N)|Fn) = E(Y(τ)|Fn). Then to show that the random variables Yτ(n) andE(Y(τ)|Fn) are equal, it is sufficient to show that their expectations are equal:
E(Yτ(n))= E(E(Y(τ)|Fn)).
The right-hand side isE(Y(τ)) by the tower property The left-hand side is the same as the following chain of inequalities shows
Y(0)≥ E(Yτ(n)) (Yτis a supermartingale)
≥ E(E(Yτ(N)|Fn)) (as above)
= E(Y(τ))
= E(Z(τ)) (as established above)
= Y(0) (since τ is optimal).
Theorem 5.11Doob decomposition
If Y(n) is a supermartingale then there exist a martingale, M(n), and a predictable, non-decreasing sequence, A(n), given by a recursive formula
A(n)= A(n − 1) − (E(Y(n)|Fn−1)− Y(n − 1)) with A(0)= 0, such that
Y(n)= Y(0) + M(n) − A(n).
This decomposition is unique in the above classes of processes M and A. Proof To find A(1), which, as an F0-measurable random variable, must be a constant, we take the expectation of the target relation:
A(1)= −Y(1) + Y(0) + M(1) to find
A(1)= −E(Y(1)) + Y(0) sinceE(M(1)) = M(0) = 0. We must then put
M(1)= A(1) + Y(1) − Y(0)
= Y(1) − E(Y(1)).
Next, to find A(2) satisfying
A(2)= −Y(2) + Y(0) + M(2)
we take the conditional expectation of this target identity with respect to F1, recalling that A is to be predictable, to get
A(2)= −E(Y(2)|F1)+ Y(0) + E(M(2)|F1)
= −E(Y(2)|F1)+ Y(0) + M(1) (if M is going to be a martingale)
= −E(Y(2)|F1)+ Y(0) + A(1) + [Y(1) − Y(0)]
having inserted the form of M(1) obtained above. Rearranged, this gives a formula for A(2):
A(2)= A(1) − E([Y(2) − Y(1)]|F1). This suggests a general recursive formula: for n≥ 1 write
A(n)= A(n − 1) − E([Y(n) − Y(n − 1)]|Fn−1) (5.7) with the form of M implied by our goal
M(n)= A(n) + Y(n) − Y(0).
This was an educated guess based on desired properties, and it is time for a rigorous argument. First note that A is increasing since the supermartingale property of Y givesE(Y(n) − Y(n − 1)|Fn−1)≤ 0.
Next, we use induction to prove predictability: A(1) is constant, so F0-measurable, and if A(n − 1) is Fn−2-measurable, it automatically is Fn−1-measurable and with the second term on the right of (5.7) also being Fn−1-measurable we find that A(n) isFn−1-measurable.
It remains to see that M is a martingale:
E(M(n)|Fn−1)
= E([A(n) + Y(n) − Y(0)]|Fn−1)
= A(n) + E(Y(n)|Fn−1)− Y(0)
= A(n − 1) − E([Y(n) − Y(n − 1)]|Fn−1)+ E(Y(n)|Fn−1)− Y(0)
= A(n − 1) + Y(n − 1) − Y(0)
= M(n − 1).
For uniqueness, suppose on the contrary that we have another such de-composition, with Y(n)= Y(0) + M (n)− A (n), so that
Y(n)− Y(0) = M(n) − A(n) = M (n)− A (n). Then
M(n)− M (n)= A(n) − A (n)
and note that for n= 0 we have M(0) = M (0) since A and A start from 0.
Next, take the conditional expectation on both sides with respect toFn−1
E([M(n) − M (n)]|Fn−1)= E([A(n) − A (n)]|Fn−1) which gives
M(n− 1) − M (n− 1) = A(n) − A (n).
So, the equality of M and M at time n− 1 implies the equality of A and A at time n, which yields equality of M and M at n. By induction on n this
proves the uniqueness.
Theorem 5.13
The stopping timeτmaxis optimal.
Proof First, directly from the definition,τmaxis a stopping time since the set{τmax= n} is determined by A(n + 1), which is Fn-measurable.
Next, again by definition, A(τmax)= 0, so Yτmax(n) = Mτmax(n), and Yτmax is a martingale.
To complete the proof it is sufficient to show that Y(τmax)= Z(τmax). Fix arbitraryω and denote by j the instant τmax(ω). Insert the Doob decompo-sition into the definition of the Snell envelope
Y( j)= max(Z( j), E(Y( j + 1)|Fj))
= max(Z( j), E(M( j + 1) − A( j + 1)|Fj))
= max(Z( j), M( j) − A( j + 1)) since M is a martingale and A is predictable.
Next, by the definition ofτmax, A( j) = 0, A( j+1) > 0. Since M( j) = Y( j), we have M( j)− A( j + 1) < Y( j). So Y( j) = max(Z( j), M( j) − A( j + 1)) =
Z( j).