1 2. Option pricing in a nite market model (February 14, 2012) 1 Introduction The main purpose of this chapter is to prove the rst and second fundamental theorem of asset pricing in a so called nite market model. 2.1 A nite market Throughout this chapter (; F; P ) is a probability space with a non-empty nite set. The time is discrete and consists of the points 0; 1; :::; T, where T is a positive integer. Moreover, there is given a ltration F 0 = f;, g F 1 F 2 ::: F T = F and n + 1 traded asset price processes S 1 = (S 1 (t)) T t=0; :::; S n+1 = (S n+1 (t)) T t=0: The corresponding vector price process is denoted by S; that is S = (S 1 (t); :::; S n+1 (t)) T t=0 and it is assumed that the random vector S(t) = (S 1 (t); :::; S n+1 (t)) is F t - measurable at each point of time t: It is appropriate to think of F t as the known information at time t: Throughout, it will be assumed that S n+1 (t) > 0; t = 0; :::; T: In particular, S n+1 can be used as a numéraire. Since is nite the -algebra F is generated by a partition of, and without loss of generality assume each member of this partition is of positive probability. Hence, we may assume = f! 1 ; :::;! d g ;
2 and F = 2 P [f! k g] > 0; k = 1; :::; d: In particular, if X is an F-measurable random variable, X = 0 a.s. if and only if X(!) = 0 for every! 2 : Finally, we assume there are no transaction costs or other frictions on the market. The above assumoptions de ne our nite market model. 2.2 Self- nancing strategies An R m -valued stochastic process (X(t)) T t=1 is predictable if X(t) is F t 1 - measurable for t = 1; :::; T: A (portfolio) strategy h = (h(t)) T t=0 is an R n+1 -valued stochastic process with (h(t)) T t=1 predictable and h(0) = h(1): The corresponding value process is given by Xn+1 V h (t) = h(t) S(t) = h i (t)s i (t); t = 0; :::; T: i=1 The strategy h is said to be self- nancing when V h (t) = h(t + 1) S(t); t = 1; :::; T 1: Since h(0) = h(1) this relation also holds for t = 0. If (a(t)) T t=0 is a sequence in a vector space, let 4a(t) = a(t) a(t 1); t = 1; :::; T: Theorem 2.2.1. If h is a strategy h is self- nancing if and only if 4V h (t) = h(t) 4S(t); t = 1; :::; T:
3 PROOF. The defnitions yield 4V h (t) = h(t) S(t) h(t 1) S(t 1) and and, hence if and only if h(t) 4S(t) = h(t) S(t) h(t) S(t 1) 4V h (t) = h(t) 4S(t); t = 1; :::; T which proves the theorem. h(t 1) S(t 1) = h(t) S(t 1); t = 1; :::; T The so called gain process G h = (G h (t)) T t=0 corresponding to a strategy h is given by G h (0) = 0 and G h (t) = h(1) 4S(1) + ::: + h(t) 4S(t); t = 1; :::; T: Theorem 2.2.2. A strategy h is self- nancing if and only if V h (t) = V h (0) + G h (t); t = 0; 1; :::; T: PROOF. Suppose t 2 f1; :::; T g and note that V h (t) V h (0) = tx 4V h (u): u=1 Now by Theorem 2.2.1, if h is self- nancing V h (t) V h (0) = tx h(u) 4S(u) = G h (t): u=1
4 Conversely, if the last identity holds 4V h (t) = 4(V h (t) V h (0)) = h(t) 4S(t) and Theorem 2.2.1 shows that h is self- nancing. This completes the proof of the theorem. 2.3 The fundamental theorems of asset pricing An arbitrage (opportunity or strategy) is a self- nancing strategy h such that or, equivalently, V h (0) = 0; V h (T ) 0; and E [V h (T )] > 0 V h (0) = 0; V h (T ) 0; and max 1kd V h(t;! k ) > 0; Next we want to give conditions on the nite market model model so that it is free of arbitrage. To this end choose S n+1 as a numéraire and introduce the price processes S(t) = ( S 1 (t); :::; S n+1 (t)) = S(t) S n+1 (t) : Clearly, each S(t) is F t -measurable. The corresponding nite market model is called the normalized market. If h is a strategy, the value process and gain process in the normalized market are denoted by V h and G h, respectively. The equation is equivalent to the equation h(t) S(t) = h(t + 1) S(t) h(t) S(t) = h(t + 1) S(t)
and we conclude that h is self- nancing in the original market if and only if h is self- nancing in the normalized market. Furthermore, note that From S n+1 (t) = 0; t = 1; :::; T: V h (t) = V h(t) S n+1 (t) follows that our original nite market model is free of arbitrage if and only if the normalized market is free of arbitrage. A probability measure P ~ de ned on the -algebra F is said to be an equivalent martingale measure if the following two conditions hold: (i) P ~ and P are equivalent, that is P ~ and P have the same null sets, (ii) under P ~ ; ( S(t); F t ) T t=0 is martingale, that is ( S i (t); F t ) T t=0 is a martingale for every i = 1; :::; n + 1: If Q is a probability measure on F and X a random variable, we write Z XdQ = E Q [X]. Moreover, if ~ P is an equivalent martingale measure E ~P [X] is written ~ E [X] : 5 Theorem 2.3.1. (First fundamental theorem of asset pricing) The market model is free of arbitrage if and only if there exists an equivalent martingale measure. To prove the rst fundamental theorem of asset pricing we need the following lemma with a = 0 (the general case is needed later to prove the second fundamental theorem of asset pricing). Lemma 2.3.1. Suppose a is a real number and g = (g(t)) T t=1 a predictable process in R n : Then there exists a unique self- nancing strategy h such that V h (0) = a and (h 1 (t); :::; h n (t)) = g(t); t = 1; :::; T:
6 PROOF. De ne (h 1 (t); :::; h n (t)) = g(t); t = 1; :::; T; h i (0) = h i (1); i = 1; :::; n; and, moreover, h n+1 (0) such that nx h i (0) S i (0) + h n+1 (0) = a: i=1 Observe that S n+1 (t) = 1 for every t and (h 1 (0); :::; h n+1 (0)) is a constant vector. Put h n+1 (1) = h n+1 (0) so that h(0) = h(1): Next we use induction to de ne h n+1 (t + 1); t = 1; :::; T 1; such that nx h i (t) S nx i (t) + h n+1 (t) = h i (t + 1) S i (t) + h n+1 (t + 1); t = 1; :::; T 1: i=1 i=1 It follows that h = (h(t)) T t=1 is predictable and self- nancing. The uniqueness is obvious from the construction of h n+1 ; which completes the proof of the theorem. PROOF THEOREM 2.3.1 (: Suppose P ~ is an equivalent martingale measure and let h be a self- nancing strategy satisfying V h (0) = 0 and V h (T ) 0: We must show that V h (T ) = 0 or, stated otherwise V h (T ) = V h (T )=S n+1 (T ) = 0. Now given t 2 f1; ::; T g ; ~E h(t) 4S(t) = E ~ h ~E h(t) 4S(t) i j Ft 1 = E ~ h h(t) E ~ 4S(t) i j F t 1 = E ~ [h(t) 0] = 0: Since it follows that V h (T ) = V h (0) + G h (T ) = 0 + ~E Vh (T ) = 0: TX h(t) 4S(t) But V h (T ) 0 and we conclude that V h (T ) = 0 as ~ P [f! k g] > 0; k = 1; :::; d: t=1 ): Let S n (t) = ( S 1 (t); :::; S n (t)); t = 0; :::; T: If g = (g(t)) T t=1 is an R n -valued predictable process we de ne Hg = g(1) 4 S n (1) + ::: + g(t ) 4 S n (T )
and use Lemma 2.3.1 to get a self- nancing strategy such that V h (0) = 0 and = 0 + V h (T ) = V h (0) + G h (T ) TX h(t) 4S(t) = Hg: t=1 Here recall that 4 S n+1 (t) = 0; t = 1; :::; T: Since the market model is free of arbitrage Hg =2 C, where C = Y 2 L 2 (; F; P ); Y 0 och Y (! k ) > 0 for some k = 1; :::; d : Next introduce the vector space L = Hg; g = (g(t)) T t=1 predictable process in R n and the compact convex set Then K = fy 2 C; E [Y ] = 1g L 2 (; F; P ): K \ L = ;: and from the Theorem 1.2.5 we obtain Z 2 L 2 (; F; P ) such that and Here E [(Hg)Z] = 0 if g = (g(t)) T t=1 predictable process in R n E [Y Z] > 0 if Y 2 K: 1f!k g Z(! k ) = E P [f! k g] Z > 0; k = 1; :::; d and we de ne a probability measure ~ P on F by setting 7 d ~ P = Clearly, P and ~ P are equivalent and Z E [Z] dp: ~E [Hg] = 0 if g = (g(t)) T t=1 predictable process in R n :
8 Finally, to prove that( S(t); F t ) T t=0 is a ~ P -martingale, choose u 2 f1; :::; T g and pick a predictable process g = (g(t)) T t=1 in R n such that g(t) = 0 if t 6= u: Then 0 = ~ E [Hg] = ~ E g(u) 4 S n (u) and it follows that ( S n (t); F t ) T t=0 is a ~ P -martingal. Clearly, then the process ( S(t); F t ) T t=0 is a ~ P -martingal. This completes the proof of the rst fundamental theorem of asset pricing. A contingent claim is an F T -measurable random variable Y, representing a contract that pays out the amount Y (!) at time T if! occurs: In the following Y and the contract are identi ed. We will say the contract is replicable if there exists a self- nancing strategy h such that V h (T ) = Y: The model is called complete if every contingent claim is replicable. In a nite market the class of all contingent claims is equal to L 2 (; F T ; Q) for every probability measure Q on F T : Theorem 2.3.2. Suppose the market model is free of arbitrage. (a) If P ~ is an equivalent martingale measure and h a self- nancing strategy, then under P ~ T ( V h (t)) T Vh (t) t=0 = S n+1 (t) t=0 is a martingale. (b) If h is a self- nancing strategy which replicates the contengent claim Y; then V h (t) = S n+1 (t) E ~ Y S n+1 (T ) j F t ; t = 0; 1; ::; T for every equivalent martingale measure ~ P : PROOF. (a) By Theorem 2.2.2 applied to the normalized market, we have V h (t) = V h (0) + G h (t)
9 and it follows that V h (T ) = V h (t) + TX u=t+1 h(u) 4 S(u): Now suppose ~ P is an equivalent martingale measure and let t+1 u T: Since S = ( S(t); F t ) T t=0 is a ~ P -martingale ~E h(u) 4 S(u) j F t = ~ E h ~E h(u) 4 S(u) j Fu 1 j Ft i = E ~ hh(u) E ~ 4S(u) i j F u 1 j Ft = E ~ [h(u) 0 j F t ] = 0 and thus ~E Vh (T ) j F t = ~ E Vh (t) j F t = Vh (t) that is ~E Vh (T ) S n+1 (T ) j F t = V h(t) S n+1 (t) : This proves Part (a). Part (b) is an immediate consequence of Part (a), which completes the proof of Theorem 2.3.2. Theorem 2.3.3. (Second fundamental theorem of asset pricing) Suppose the market model is free of arbitrage. Then the model is complete if and only if there is a unique martingale measure. PROOF. ): Suppose X is an F T -measurable random variable and P ~ 0 and ~P 1 are two equivalent martingale measures. Set Y = S n+1 (T )X and let h be a self- nancing strategy which replicates Y: By applying Theorem 2.3.2, we get P V h (0) = S n+1 (0)E ~ i Y ; i = 0; 1 S n+1 (T ) and, hence P E ~ 0 P [X] = E ~ 0 Y P = E ~ 1 Y P = E ~ 1 [X] : S n+1 (T ) S n+1 (T )
10 Thus ~ P 0 = ~ P 1. (: Let ~ P denote the equivalent martingale measure: As above for each R n - valued predictable process g = (g(t)) T t=1 set and introduce the vector space Hg = g(1) 4 S n (1) + ::: + g(t ) 4 S n (T ) L = a + Hg; a 2 R and g = (g(t)) T t=0 a predictable process in R n : We now claim that L = L 2 (; F; ~ P ): In fact, if this is wrong by Theorem 1.2.5 there is a Z 2 L 2 (; F; ~ P ) such that Z 6= 0 and ~E [Y Z] = 0 if Y 2 L: The choice Y = 1 yields E ~ [Z] = 0: Moreover, choosing a positive real number M > Z we get a probability measure P ~ 0 on F by setting d ~ P 0 = (1 + Z M )d ~ P : Clearly, P ~ 0 and P ~ are equivalent and if Y 2 L; P E ~ 0 [Y ] = E Y ~ + 1M Y Z In particular, = ~ E [Y ] + 1 M ~ E [Y Z] = ~ E [Y ] : E ~ P 0 [Hg] = 0; if g = (g(t)) T t=1 is a predictable process in R n. From this, as in the proof of the rst fundamental theorem of asset pricing, ( S(t); F t ) T t=0 is a ~ P 0 -martingale and we have got an equivalent martingale measure ~ P 0 di erent from ~ P. From this contradiction we conclude that L = L 2 (; F; ~ P ): Next suppose Y 2 L 2 (; F; ~ P ). To show that Y is replicable we introduce X = Y S n+1 (T ) :
Since L = L 2 (; F; ~ P ) there is a real number a and a predictable process g = (g(t)) T t=1 in R n such that X = a + Hg: Now by Lemma 2.3.1 there exists a self- nancing strategy h satisfying V h (0) = a and (h 1 (t); :::; h n (t)) = g(t); t = 1; :::; T: But then X = V h (0) + G h (T ) = V h (T ) = h(t ) S(T ) and it follows that Y = h(t ) S(T ) = V h (T ): 11 Exercises 1. Suppose the nite market is free of arbitrage. Show that the restricted market with with time set f0; :::; T 1g is free of arbitrage. 2. Suppose the nite market is free of arbitrage and complete. Show that the restricted market with time set f0; :::; T 1g is complete. 2.4 The price of a contingent claim To de ne a price of a contingent claim Y in our nite market model we will require that there are no arbitrages or, stated otherwise, there is at least one equivalent martingale measure. If, in addition, Y is replicable by Theorem 2.3.3 the expression S n+1 (t) E ~ Y S n+1 (T ) j F t is independent of the underlying equivalent martingale measure ~ P and equals V h (t) for every strategy h which replicates Y: This leads us to the following De nition 2.4.1. Suppose the market is free of arbitrage. A replicable contingent claim Y has the price Y (t) = S n+1 (t) E ~ Y S n+1 (T ) j F t
12 at time t; where the expectation is with respect to an arbitrary equivalent martingale measure. The next example shows that the quantity S n+1 (t) E ~ Y S n+1 (T ) j F t may depend on the martingale measure ~ P even if the market is free from arbitrage. Example 2.4.1. Consider a nite market model with one stock and one bond in a single period from t = 0 to t = 1: The stock (bond) price at time t equals S(t) (B(t)): Furthermore, suppose x 1 ; x 2 ; x 3 2R; x 1 < x 2 < x 3 ; and S(1) = S(0)e X where S(0) is a positive constant and X:! fx 1 ; x 2 ; x 3 g a random variable such that p i = P [X = x i ] > 0; i = 1; 2; 3: The bond price is deterministic and B(1) = B(0)e r where B(0) and r are positive constants. Finally, suppose x 1 < r < x 3 : Next assume the strategy h is an arbitrage and h(1) = (h S ; h B ); that is and h S S(0) + h B B(0) = 0 h S S(0)e x i + h B B(0)e r 0; i = 1; 2; 3 where strict inequality occurs for some i. Setting a = h S S(0), a(e x i e r ) 0; i = 1; 2; 3 where strict inequality occurs for some i. Hence a 6= 0 and we have got a contradiction. Thus by the rst fundamental theorem of asset pricing there is at least one equivalent martingale measure. In the following, for simplicity, choose X as the identity map in = fx 1 ; x 2 ; x 3 g so that P is a linear combination of Dirac measures P = p 1 x1 + p 2 x2 + p 3 x3 :
13 If P and ~ P are equivalent probability measures where ~p i > 0; i = 1; 2; 3, Moreover, if and only if Here ~P = ~p 1 x1 + ~p 2 x2 + ~p 3 x3 ~p 1 + ~p 2 + ~p 3 = 1: S(0) = e r ~ E [S(1)] e x 1 r ~p 1 + e x 2 r ~p 2 + e x 3 r ~p 3 = 1: e x 1 r < 1 < e x 3 r and it is evident that there exists in nitely many martingale measures. In the following discussion assume r < x 2. Drawing a gure it is simple to see that the map ~ P! ~p 3 is injective. De ning Y such that Y (x 1 ) = Y (x 0 ) = 0; and Y (x 3 ) = 1 it follows that the quantity e r E ~P [Y ] = e r ~p 3 is di erent for di erent equivalent martingale measures. Exercises 1. Suppose the market is free of arbitrage and let Y be a replicable contingent claim. Add the price process ( Y (t)) T t=0 to the other price processes and show that the augmented market model does not admit arbitrage. 2.5 American options An American derivative may be exercised at each time point. If the exercise takes place at time t the holder of the contract gets the amount X(t) and
14 after that point of time the contract expires. Here X(t) is an F t -measurable random variable for each t = 0; 1; :::; T. Next we want to de ne the price V (t) of the derivative at an arbitrary time t given that the contract is alive at this time. Clearly, V (T ) = X(T ): To de ne the price at a time point before maturity T from now on in this section we assume that the market is complete with the equivalent martingale measure P ~. Under this assumption de ne V (t) = max(x(t); S n+1 (t) E ~ V (t + 1) S n+1 (t + 1) j F t ); t = T 1; T 2; :::; 1; 0: It is optimal to exercise the derivative at time t T X(t) > S n+1 (t) E ~ V (t + 1) S n+1 (t + 1) j F t since it costs the amount S n+1 (t) ~ E V (t + 1) S n+1 (t + 1) j F t 1; if to buy a portfolio at time t which replicates V (t + 1) at time t + 1: Note that V (t) S n+1 (t) E ~ V (t + 1) S n+1 (t + 1) j F t and we conclude that the sequence T V (t) S n+1 (t) ; F t t=0 is a ~ P -supermartingale. If the same sequence is a ~ P -martingale there is no extra gain to exercise the American derivative before maturity. Furthermore note that if (U(t); F t ) T t=0 is a ~ P -supermartingale such that then U(t) X(t) ; t = 0; :::; T S n+1 (t) U(t) max( X(t) S n+1 (t) ; ~ E [U(t + 1) j F t ]); t = 0; :::; T 1:
15 Hence the price process V (t) T S n+1 (t) t=0 is equal to the smallest P ~ -supermartingale, which dominates the process T X(t) S n+1 (t) : t=0 The aim in this section is to describe the price of the derivative in terms of so called stopping times. Let t 0 2 f0; 1; :::; T g : A map :! ft 0 ; :::; T g is called a stopping time with respect to the ltration (F t ) T t=0 if the event [ t] 2 F t for every t 2 ft 0 ; :::; T g : Note that a constant map from into ft 0 ; ; :::; T gis a stopping time. Moreover, if 0 ; 1 :! ft 0 ; :::; T g are stopping times and 0 _ 1 = max( 0 ; 1 ) 0 ^ 1 = min( 0 ; 1 ) are stopping times. The following result is a special case of a martingale theorem by J. L. Snell (see e.g. J. Neveu, Discrete-Parameter Martingales, North-Holland Publishing Company, 1972). Theorem 2.5.1. If :! ft 0 ; :::; T g is a stopping time with respect to the ltration (F t ) T t=0; then V (t 0 ) S n+1 (t 0 ) E ~ X() S n+1 () j F t 0 where equality occurs if = min t 2 ft 0 ; :::; T 1g ; X(t) > S n+1 (t) ~ E V (t + 1) S n+1 (t + 1) j F t with the convention that the minimum over the empty set equals T:
16 PROOF. To simplify notation let S n+1 (t) = 1 for every t: We rst prove that V (t 0 ) ~ E [X() j F t0 ] where :! ft 0 ; :::; T g is a stopping time with respect to the ltration (F t ) N t=0: The case t 0 = T yields = T and as V (T ) = X(T ) this case is obvious. Next assume t 0 < T and V (t 0 + 1) ~ E [X() j F t0 +1] for every stopping time :! ft 0 + 1; :::; T g with respect to the ltration (F t ) N t=0: Now if :! ft 0 ; :::; T g is a stopping time with respect to the ltration (F t ) N t=0we have ~E [X() j F t0 ] = ~ E X()1 [=t0 ] j F t0 + ~ E X()1[>t0 ] j F t0 = ~ E X(t 0 )1 [=t0 ] j F t0 + ~ E X_(t0 +1)1 [>t0 ] j F t0 = X(t 0 )1 [=t0 ] + 1 ~ [>t0 ] E X _(t0 +1) j F t0 = X(t 0 )1 [=t0 ] + 1 ~ h i [>t0 ] E ~E X_(t0 +1) j F t0 +1 j Ft0 Now by induction we get V (t 0 )1 [=t0 ] + 1 [>t0 ] ~ E [V (t 0 + 1) j F t0 ] V (t 0 )1 [=t0 ] + 1 [>t0 ]V (t 0 ) = V (t 0 ): V (t 0 ) E [X() j F t0 ] if :! ft 0 ; :::; T g is a stopping time and t 0 2 f0; :::; T g : Finally, if n = min t 2 ft 0 ; :::; T 1g ; X(t) > E ~ o [V (t + 1) j F t ] it is readily seen that there is equality in each step in the induction proof above. This concludes the proof of the theorem. Exercises
1. Let (V (t)) T t=0 denote the price process of an American styled derivative. Show that V (t) ~E E S n+1 (t) ~ V (t + 1) ; t = 0; :::; T 1: S n+1 (t) 17