Chapter 15: Jump Processes and Incomplete Markets. 1 Jumps as One Explanation of Incomplete Markets

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1 Chapter 5: Jump Processes and Incomplete Markets Jumps as One Explanation of Incomplete Markets It is easy to argue that Brownian motion paths cannot model actual stock price movements properly in reality, as sudden price changes due to political and economical events are not naturally incorporated. The presence of jumps has another aspect: hedging strategies will be difficult to achieve their goals, which leads us to incomplete markets from one perspective. Black-Scholes model assumes that stock prices have continuous paths, which suggests Var [S(t + t) S(t)] 0 as t 0. The effectiveness of delta hedging hinges on the fact that changes in the underlying and derivatives can be made sufficiently small if the time period is short enough. Namely, we assume that over this short period of time, the change to the portfolio (V δs) V S δ S = S ( V S δ ) S so we just need to choose δ = V/ S and the hedge is quite effective. If S can no longer be assumed to be small, there is no reason to expect that this can still work. This observation explains the difficulty in portfolio managements once a market crash happens. This phenomenon is just one aspect of the general condition called incomplete market, namely the market risk cannot be eliminated by using various derivatives. The fact that we do not expect derivatives to serve that perfect hedging role suggests that we no longer have a unique no-arbitrage price for derivatives in an incomplete market. We will need to explore other pricing methodologies, such as utility function pricing, or a risk-preference pricing. A risk-preference pricing consideration can partially explain the volatility smile phenomenon. For example, investors tend to overhedge, buying more into those slightly out-of-the-money puts (K < S 0 ), causing ultimately the implied vol for puts to have down-sloping skew. This is a demonstration of the risk-averse behavior from many investors. 2 Modeling Jumps with a Tree In this simple tree model, we want to illustrate the fact that no unique no-arbitrage price would emerge. However, prices of different derivatives on the same asset should satisfy some deterministic relation, even though the price of each individual derivative cannot be determined uniquely from the no-arbitrage consideration.

2 In the numerical example used in the text, a stock with a current price of $00 will have three possible outcomes after one time period: two normal moves 0, 00, and a sudden jump to 50, with probabilities p, p 2 and p 3 respectively. We assume a zero interest rate for this time period. Suppose there is a risk-neutral probability measure: 0p + 00p p 3 = 00 () with the constraint of probability measure p + p 2 + p 3 =, 0 p, p 2, p 3. The solutions can be characterized by p = 5p 3, p 2 = 6p 3, and 0 p 3 /6, so they are not unique. Suppose there is a call (C () ) with strike K = 00 which provides a payoff of (0, 0, 0) in these three scenarios, and another cal (C (2) ) with strike K = 90 which provides a payoff of (20, 0, 0). Any price for C () between 0 and 25/3 and any price for C (2) between 0 and 6.67 will not cause an arbitrage, if considered separately. However, when we consider these two calls simultaneously, we find they must satisfy a consistency condition in order to avoid arbitrage. Specifically, we use a vector to denote possible outcomes for each one asset: C () = 0 0 0, C (2) = , S = We can see that they are linearly dependent, in particular, B =. S 50B = 5C () 4C (2) (2) Let S 0, B 0, C () 0 and C (2) 0 be the prices of the assets at time zero, because the payoffs satisfy the above relation in vector form, which means the matching at every possible outcome, no-arbitrage condition will enforce the same relation for the prices. Since S and B have well observed prices (00 and in this example), we must have a relation between the prices of these two calls 5C () 0 4C (2) 0 = 50, (3) even though we cannot determine either of the prices individually. In this example, what we learn is that given only two assets on the market with reliable prices, if there are three possible outcomes, we will not be able to determine the price of any payoff (derivative). This can be described in a linear algebra language: you have two linearly independent vectors in R 3 so they cannot span R 3, meaning that some payoffs cannot be replicated by a linear combination of S and B. This is what incomplete refers to. Once you bring in another security with a payoff linearly independent of those two and if this new security also has a market price, then the market is so-called completed, or we say that the new set (now containing three linearly independent vectors in R 3 ) is a basis of R 3. In 2

3 financial terms, every risk scenario corresponds to one vector in R 3 and it can be replicated by a portfolio consisting of those three assets, which is saying that every vector can be expressed as a linear combination of those three vectors. If this is the case, then the market is complete. In the real world, there are infinitely many risk scenarios, so a complete market would require a basis for an infinitely dimensional vector space, which is necessarily infinite. There is no way that we can find a set of securities like this on the market, so the idea of complete market is an idealistic fantasy and it is natural to criticize the use of the complete market assumption in practice. 3 Modeling Jumps in a Continuous Framework How should we model jumps in a stock price path? The following two features are to be carefully included: unexpected timing; when a jump occurs, the amplitude is not foreseeable. Merton pioneered the use of Poisson process in jump models (973), in which two crucial assumptions are made: ) a parameter called the intensity λ controls how likely jumps can occur; 2) the process is memoryless, that is the previous jumps will not increase/decrease the likelihood of future jumps. Let us define N t to be the number of jumps before time t, and model it as a Poisson random variable, with distribution λt (λt)j P [N t = j] = e j! In particular, for small t, we have an estimate P [N t = ] λt To model the jump amplitude, we imagine that a jump modifies a stock price by multiplying a factor J: S : S t S t = S t J, or S = S t S t = (J )S t. This suggests that we can add one term to our previous modeling of the instantaneous return ds t = µ dt + σ dw t + (J )dn t (5) S t In order to obtain a risk-neutral probability measure, we must have [ ] dst E = µ dt + E[J ]λ dt = r dt S t Let us assume deterministic J for a while, the above relation requires (4) µ = r (J )λ (6) 3

4 4 Merton s Option Pricing Formula A no-arbitrage price for a derivative of S with payoff F (S T ) can be written as e rt E [F (S T )] for the risk-neutral probability measure resulting from one choice of λ. What is the distribution of S T? We can use Ito s formula to get d log S t = (µ 2 ) σ2 dt + σ dw t + (log J)dN t (7) Integrating t from 0 to T, log S T log S 0 = (µ 2 σ2 ) T + σ T Z + (log J)N T, or S T = S 0 e (µ r)t J N T exp ((r 2 ) σ2 T + σ ) T Z (8) Here Z N(0, ) and N T is a Poisson random variable with parameter λt. As we can see, the expression for S T is similar to that of the Black-Scholes model, with S 0 replaced with S 0 e (µ r)t J N T. If we fix N T for a while, the expectation of a call option payoff can lead to a Black-Scholes like formula E [ e rt (S T K) +] = C BS ( S0 e (µ r)t J N T, K, σ, T ) To take into consideration of the random variable N, we use a conditional expectation formula [ [ E 2 E e rt (S T K) +]] ( = C BS S0 e (µ r)t J j, K, σ, T ) P[N T = j] = j=0 j=0 ( C BS S0 e (µ r)t J j, K, σ, T ) λt (λt )j e j! Finally we allow J to have a distribution so another expectation calculation is expected. The consideration for J is that it should be positive so the stock price can never drop below zero. One obvious example is the lognormal distribution and this is the original Merton s choice: so J = m exp( 2 ν2 + νz), and E[J] = m. In this way log S T = log S 0 + log J N(log m 2 ν2, ν) (9) (µ 2 σ2 ) T + N T (log m 2 ν2 ) + σ T Z 0 + ν 4 N T k= Z k, (0)

5 which involves standard normal random variables Z k, k = 0,,..., and a Poisson random variable N T. Since all the normal random variables here are independent, the equation can be simplified to log S T = log S 0 + (µ 2 ) σ2 T + N T (log m 2 ) ν2 + σ 2 T + N T ν 2 W () where W is a standard normal. Taking another expectation of E 2 above, we have the Merton s jump diffusion price of a call option with C M = e λ T (λ T ) n C BS (S 0, K, T, σ n, r n ) (2) n! n=0 σ n = σ 2 + nν2 T (3) r n = r λ(m ) + n log m T (4) λ = λm (5) As we learned from the Black-Scholes pricing methodology, an asset price should be validated through certain procedure. This option price from Merton s model is different from our previous prices with the following issues:. It is a no-arbitrage price, but not an unique no-arbitrage price. We argue from the following reasons: (a) The equivalent martingale measure depends on the choice of λ, and the estimate/choice of λ is a subjective matter; (b) In choosing λ for the model, investor s risk preference is reflected; (c) For vanilla options with convex payoff, the jump model price is always higher than the no-jump model price; (d) The volatility in the model is unchanged. 2. If the price is not a unique no-arbitrage price, how should we hedge an option in this model? Can we construct and execute a perfect hedge? The answer will be no, but we could still establish some super and sub replications. Assume that an option O with payoff f(s T ), if we can find a self-financing portfolio Q such that O(T ) = f(s T ) Q(T ) in all scenarios, then we must have O(t) Q(t) for any t < T. Similarly if we have a lower bound R(T ) O(T ) then we must have R(t) O(t). Putting together we have R(t) O(t) Q(t) 5

6 So in order to find bounds to sandwich the price of the option, we should look for portfolios that cover the payoff of the option from above and from below in all of the scenarios. 3. Hedging consideration We want to demonstrate that in the jump models the ability to hedge depends on whether J is random or not. (a) If J is deterministic, we can use α shares of the one option O, shares of the underlying S to hedge one option C: C S + α O S + = 0 (6) C(S, t) + αo(s, t) + S = C(SJ, t) + αo(sj, t) + SJ (7) As we can see, once J is determined, we can solve this system for α and to obtain the hedge. (b) In case of random J, there is no other equation to allow us to determine the hedge ratios. If we follow Black-Scholes delta hedging, for a convex payoff we can see that the value of the option is always above it s hedge. In the case of Brownian paths the underlying moves are supposed to be small for a short period of time so the error of hedge is quite small. After the short period of time a new hedge will kick in. However, in the jump model, these underlying moves are not small so any occurrence of jump can lead to a substantial hedge error. (c) So the question is: given that no perfect hedge (in the sense of eliminating risks) is possible, what should we use instead? There are several possibilities: i. minimum variance: this is to minimize the difference between the hedge and the option value. ii. counting on no occurrence of jumps, want to maintain a perfect hedge. As it turns out, the idea of super and sub replications will play an important role in generating certain hedge. 5 Asset Pricing in Incomplete Markets As we see earlier, an incomplete market for a discrete model can be explained in a linear algebra language. We now formalize this approach with a definition: A market where some payoffs cannot be replicated by trading in marketed securities is called an incomplete market. Consider this one-step model: we have N traded securities (meaning a market value is available) with market prices S j, j =,..., N. After one time period, 6

7 there are M possible scenarios k =,..., M. In scenario k, security S j will have a value S j (k). Now let us consider an arbitrary payoff function P (k), k =,..., M and ask if it is possible to construct a portfolio to have this payoff. This is to ask to find c, c 2,..., c N such that c S (k) + c 2 S 2 (k) + + c N S N (k) = P (k), for k =,..., M. We can also express this in the vector notation c S + c 2 S 2 (k) + + c N S N (k) = P R M If N = M, and S, S 2,..., S N are linearly independent, then the answer is always yes. In case N < M, P / Range(S,..., S N ), then we cannot have replication, which leads to an incomplete market. 7