K 1 < K 2 = P (K 1 ) P (K 2 ) (6) This holds for both American and European Options.
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1 Slope and Convexity Restrictions and How to implement Arbitrage Opportunities 1 These notes will show how to implement arbitrage opportunities when either the slope or the convexity restriction is violated. Even though I will focus on the slope and convexity restrictions, the idea is the same for the violation of any arbitrage bound. 1 Notation: I will use the following notation throughout these notes: K Strike price T Maturity r Risk-free rate C (K) The price of an American call at time t < T c (K) The price of a European call at time t < T P (K) The price of an American put at time t < T p (K) The price of a European put at time t < T r(t t) B B (t, T ) e Also, I will assume that there are options trading at three strike prices K 1, K 2 and K 3 such that: K 1 < K 2 < K 3 2 The slope restriction: All that you need to know to check the slope restriction (and to implement arbitrage opportunities, when it doesn t hold) are 4 simple but powerful inequalities: C (K 2 ) C (K 1 ) + (K 2 K 1 ) 0 (for both American and European calls) (1) P (K 1 ) P (K 2 ) + (K 2 K 1 ) 0 (for both American and European puts) (2) c (K 2 ) c (K 1 ) + (K 2 K 1 ) B 0 (for European calls only) (3) p (K 1 ) p (K 2 ) + (K 2 K 1 ) B 0 (for European puts only) (4) 1 Please send any comments, suggestions and corrections to breno.schmidt@usc.edu 1
2 In this section, I will show why the inequalities above should hold and how they lead to the so called slope restrictions. I start by showing that the first two inequalities are easily derived from the following result: Result 1 The price of a call (put) option is an decreasing (increasing) function of the strike price. In other words, K 1 < K 2 = C (K 1 ) C (K 2 ) (5) This holds for both American and European Options. K 1 < K 2 = P (K 1 ) P (K 2 ) (6) In words, calls with a larger strike price are worth less than otherwise identical calls, and puts with a larger strike price are worth more than otherwise identical puts To show that Result 1 is true, consider the payoff of two (otherwise identical) European calls with different strikes. A quick look at graph 1 is enough to conclude that the investor is always better off (at maturity) holding the call with the lowest strike. Now, consider an American call option: at any time before maturity exercising the option with the lowest strike will be more profitable than exercising the option with the highest strike (the payoff graph is basically the same). A similar argument shows that puts with the highest strike is worth more than the put with the lowest price. Put Options with Different Strike Prices Call Options with Different Strike Prices K 2 At any given stock price, we are better off holding the put with the largest strike price K 2 At any given stock price, we are better off holding the call with the smallest strike price K 1 K 1 This simple observation about the sensitivity of calls and puts leads to a powerful result: Result 2 The sensitivity of calls and puts to the strike price is bounded (from above for calls and from below for puts). Specifically, dc dk 0 dp dk This holds for both American and European options. 2
3 As you probably know, increasing (decreasing) functions have the property that the derivative (the slope of the graph) evaluated at any point is positive (negative) 2. An easy way to see that is the following: note that Result 1 implies: C (K 2 ) C (K 1 ) K 2 K 1 0 for every values of K 2 and K 1, such that K 2 K 1 > 0 (if you are still not convinced, take another look at graph 1). Loosely speaking, making the difference (K 2 K 1 ) smaller and smaller in the inequality above leads to dc dk 0. A similar argument can be made for the puts. Thus, the slope of the graph of a call (as a function of the strike price) is never negative and the slope of the graph of a put (as a function of the strike price) is never positive. But we can do better than that: not only we can tell the sign of these slopes, but also we can find lower and upper bounds for calls and puts, respectively. The following result makes this idea precise. Result 3 The sensitivity of the calls and puts to their strike prices is bounded from above and from below, respectively. Specifically, 1 dc dk 0 dp dk 1 This holds for both American and European options// In particular, for European Options only, we have that: B dc dk 0 dp dk B To show that the above result is true, consider the following strategy: Buy a call with strike price K 2 and sell an otherwise identical call with strike price K 1. What would be the payoff of such strategy if the options were European? What would be the payoff of such strategy if the options were American and we exercise them both today? Graph 3 answers both questions. 2 Actually, this is not quite right. The correct theorem is that strictly increasing (decreasing) continuous functions have positive (negative) derivatives. But as mentioned before, we will abstain from such formalisms 3
4 Payoff of the Strategy: C(K2) - C(K1) C(K2) The payoff of this strategy is never positive! 0 C(K2) - C(K1) -C(K1) Since the price of the call with the highest strike is worth more (Result 1), there will be an outflow of money at maturity. In the case of an American call, this outflow will occur at the time of exercise. Of course, there is no way someone would follow that strategy since it always leads to a negative payoff. But, suppose there was a way to assure the holder of such strategy an additional amount of money (K 2 K 1 ) > 0 at the time of exercise. We could then guarantee a positive payoff at maturity for any stock price. Graph 4 illustrates that idea. Payoff of the Strategy: C(K2) - C(K1) + (K2 - K1) x B This is the payoff of the strategy after summing (K2 - K1) to C(K2) - C(K1) C(K2) K 2 - K 1 0 C(K2) - C(K1) -C(K1) But how could someone guarantee a positive inflow of money of (K 2 K 1 ) at time of exercise? In the case of Europeans the time of exercise corresponds to the maturity of the option. So, in this case, you just have to lend the present value of (K 2 K 1 ), i.e. (K 2 K 1 ) B (t, T ). In the case of American options, you 4
5 would require that amount of money ( K 2 K 1 ) every period, since there is always the possibility of early exercise. Now, since the payoff of such strategy cannot be negative, its value today has to be positive. Thus: C (K 2 ) C (K 1 ) + (K 2 K 1 ) 0 c (K 2 ) c (K 1 ) + (K 2 K 1 ) B 0 (for both American and European calls) (for European calls only) or, equivalently: C (K 2 ) C (K 1 ) (K 2 K 1 ) c (K 2 ) c (K 1 ) (K 2 K 1 ) 1 B (for both American and European calls) (for European calls only) If we make the difference K 2 K 1 smaller and smaller, in the limit we will have slope restrictions 1 dc dc 0 for American and European and B 0 for European options dk dk only3.// As an exercise, do the same for the put options. 3 The Convex Restriction Again, all that you have to know to derive the convex restriction is that: C (K 3 ) C (K 2 ) K 3 K 2 C (K 2) C (K 1 ) K 2 K 1 0 As before, lets show that the above inequality is true and then how it leads to a convex restriction.// By definition, the inequality above says that the price of a call is a convex function of the strike price. Loosely speaking, we say that a function y = y (x) is convex when the sensitivity of y to variations in x increases as we increase x. In sum, the following result holds: Result 4 The price of a call or put option is a convex function of the strike price. In other words, for K 1 < K 2 : K 1 < K 2 = dc dk < dc K1 dk (7) K2 K 1 < K 2 = dp dk < dp K1 dk (8) K2 where dx dk K1 represents the derivative (slope or sensitivity) of the function X with respect to K, when evaluated at the point K 1. This holds for both American and European Options. 3 Note that we are making use of Result 1 here. 5
6 In words, as the difference between the strike prices increases, the sensitivity to the strike price of two otherwise identical options also increases. To show that this is true, consider a butterfly spread using calls (i.e. buy one call with the highest strike, buy one call with the lowest strike and sell two calls with the intermediary strike). For K 1 < K 2 < K 3, the payoff of such strategy is depicted below: Payoff of a Butterfly Strategy - c(k 1 ) The Payoff is never negative! K 1 - c(k 3 ) K 3 - c(k 2 ) It is clear that the payoff of such strategy is positive (actually, non-negative) everywhere. Thus, its price today have to be positive. Thus, it must be true that: C (K 3 ) C (K 2 ) [C (K 2 ) C (K 1 )] 0 (9) Now, assume that K 2 = K 1+K 3 2, which implies that (K 3 K 2 ) = (K 2 K 1 ). We can divide both sides of (9) above by (K 3 K 2 ) = (K 2 K 1 ) and get: or, equivalently, C (K 3 ) C (K 2 ) K 3 K 2 C (K 2) C (K 1 ) K 2 K 1 0 (10) C (K 3 ) C (K 2 ) C (K 2) C (K 1 ) K 3 K 2 K 2 K 1 /bigskip Because this restriction actually defines a convex function, it is called convexity restriction.// Now, if we make the difference (K 3 K 2 ) = (K 2 K 1 ) go to zero, we have: K 3 K 1 = dc dk dc K3 dk This holds for both American and European calls.// As an exercise, do the same for puts. K1 6
7 4 How to Implement Arbitrage Opportunities Arbitrages opportunities are very easy to construct when some inequality restriction is violated. All that you have to do is to use the inequality that describes the violation to construct a portfolio that gives you some money today. Then, you just have show that there is no way to lose money anytime in the future. We will look at two examples corresponding to violations of the slope and convexity restrictions for European options. 4.1 Violation of the slope restriction for European Options As mentioned above, all that we need to check for a violation of the slope restriction are the two slope inequalities : c (K 2 ) c (K 1 ) + (K 2 K 1 ) B 0 p (K 1 ) p (K 2 ) + (K 2 K 1 ) B 0 (for European calls only) (for European puts only) Now, say that the slope restriction is violated for calls. In this case, we have that: or, equivalently, c (K 2 ) c (K 1 ) + (K 2 K 1 ) B < 0 c (K 2 ) + c (K 1 ) (K 2 K 1 ) B > 0 (11) Here is how you have to interpret (11): since the elements in this inequality are amounts of money, I can make money today by replicating what is happening on the left hand side of (11). More specifically, just copy the numbers above to the column of the table that represents the transactions today: Transaction Today Maturity c (K 2 ) +c (K 1 ) (K 2 K 1 ) B Total > 0 Now, figure out what you have to do to get the right signs (which will lead to a positive amount of money today) Transaction Today Maturity Buy call with strike price K 2 c (K 2 ) Sell call with strike price K 1 +c (K 1 ) Lend (K 2 K 1 ) B dollars (K 2 K 1 ) B Total > 0 7
8 The final step is to check that we do not lose money at maturity. Obviously, this will always be the case if some arbitrage bound is being violated: Transaction Today Maturity Buy call with strike price K 2 c (K 2 ) max (0, S (T ) K 2 ) Sell call with strike price K 1 +c (K 1 ) max (0, S (T ) K 1 ) Lend (K 2 K 1 ) B dollars (K 2 K 1 ) B K 2 K 1 Total > 0? Since there is uncertainty about the behavior of the asset at maturity, we need to check each case: Transaction Maturity: If S (T ) < K 1 < K 2 If K 1 < S (T ) < K 2 If K 1 < K 2 < S (T ) Buy call with strike price K 2 0 S (T ) K 2 S (T ) K 2 Sell call with strike price K (S (T ) K 1 ) Lend (K 2 K 1 ) B dollars K 2 K 1 Total K 2 K 1 > 0 S (T ) K 1 > 0 0 And we are done! Note that the approach taken here is quite general: whenever you have an arbitrage bound being violated, you have an inequality. Play around with this inequality to make it look like X + Y + Z > 0. Then, put what is on the left side in the column describing the transactions (remember, X, Y and Z are the values today). By construction, you will always make money today. All that is left is to show is that you will not lose money in the future. Once again, this will always be the case (by definition) if an arbitrage restriction is being violated. 4.2 Violation of the convex restriction for European calls: Lets do the same thing for the convex restriction for calls. Since the convex restriction can be translated into the inequality (9), a violation of the convex restriction is the same as 4 : or, equivalently: C (K 3 ) C (K 2 ) [C (K 2 ) C (K 1 )] < 0 [C (K 3 ) C (K 2 )] + [C (K 2 ) C (K 1 )] > 0 But what is on the left side? A butterfly spread, right? So you know you will never lose money at maturity (take another look at the graph above). But even if you didn t know that, the same conclusion could be reached by just repeating the argument above: Write down the inflows and outflows of money today and the corresponding transactions. 4 There is a catch here: I am assuming that (K 3 K 2 ) = (K 2 K 1 ). If that is not the case you have to work with inequality (10). The argument is pretty much the same, though. 8
9 Transaction Today Maturity Buy call with strike price K 1 c (K 1 ) Sell call with strike price K 2 c (K 2 ) Buy call with strike price K 3 c (K 3 ) Sell call with strike price K 2 c (K 2 ) Total: > 0 Since this is just a butterfly, its payoff is always positive, so that I am guaranteed not to lose money in the future. You can actually show that by plugging in the payoffs: Transaction Today Maturity Buy call with strike price K 1 c (K 1 ) max (0, S (T ) K 1 ) Sell call with strike price K 2 c (K 2 ) -max (0, S (T ) K 2 ) Buy call with strike price K 3 c (K 3 ) max (0, S (T ) K 3 ) Sell call with strike price K 2 c (K 2 ) -max (0, S (T ) K 2 ) Total: > 0? Once again, we have to check each possible case. A quick tip: if you are unsure about all the cases you have to consider, plot the payoff of the entire strategy on a graph. Transaction Maturity: (note: K 2 = K 1+K 3 ) 2 If S (T ) < K 1 If K 1 < S (T ) < K 2 If K 2 < S (T ) < K 3 If K 3 < S (T ) Buy call (strike K 1 ) 0 S (T ) K 1 S (T ) K 1 S (T ) K 1 Sell call (strike K 2 ) 0 0 (S (T ) K 2 ) (S (T ) K 2 ) Buy call (strike K 3 ) S (T ) K 3 Sell call (strike K 2 ) 0 0 (S (T ) K 2 ) (S (T ) K 2 ) Total: 0 S (T ) K 1 > 0 2K 2 K 1 S (T ) > 0 0 And we are done. What if you use puts instead of calls to construct the butterfly? Before redoing all the tables for the put case, draw a graph and compare to the one above... 9
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