Elements of Stochastic Analysis with application to Finance (52579) Pavel Chigansky

Transcription

1 Elements of Stochastic Analysis with application to Finance Pavel Chigansky Department of Statistics, The Hebrew University, Mount Scopus, Jerusalem 9195, Israel address:

2 These are the lecture notes for the graduate course I taught at the Statistics Department of the Hebrew University during the spring semester of 211. They are an adapted adoption of the textbooks [1], [2] and [3]. The reader is assumed to be familiar with mathematical probability at the level of [4]. As always the author will be delighted to get any comments, suggestions, etc. P.Ch., Jerusalem, June, 211

3 Contents Chapter 1. Option pricing in discrete time 5 1. The one period binomial market model 5 2. The multi period binomial market model Multi-asset one-period model 17 Chapter 2. Stochastic processes in continuous time Elements of the general theory The Brownian motion and its existence Some properties of the Brownian motion Itô s stochastic integral The Itô chain rule Stochastic differential equations The Feynman-Kac formula 1 8. Girsanov s change of measure 17 Chapter 3. Option pricing in continuous time 117 Bibliography 133 3

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5 CHAPTER 1 Option pricing in discrete time An Israeli company orders equipment from its US partner, to be shipped in a year from now, and will pay 1 million US $ at the moment of delivery. Today s exchange rate of the US $ is 3.4 NIS, but in a year it may change in either direction. The products of the Israeli firm are sold at the national market and it s budget is managed in the local currency. Hence it needs to secure the future payment against increase in the exchange rate. One obvious way to do this is to buy 1 million US $ today and put them in a bank account perhaps, enjoying a modest interest rate. However, this possibility would require holding back the funds, which could have been used for development or more profitable investments. Alternatively, the Israeli company can buy a contract from e.g. a bank, according to which the bank will sell 1 million US $ at the rate 3.4 in a year from now and the company will be obliged to buy it. Such contract allows the company to buy the US $ at at the rate, not higher than 3.4, but if the actual exchange rate drops below 3.4, it will overpay. To address this issue, a different contract is possible: the bank will sell 1 million US $ at the rate 3.4 in a year from now, and the company will have the right, but not the obligation, to buy it. Hence if the actual exchange rate drops below 3.4, the company will be able to buy at the lower rate. This type of contract is called European call option. Of course, the option writer the bank will charge a premium for the service usually at the time of signing the contract. But what would be a fair price for the option? We shall see that this question leads to an elegant theory of option pricing, which will be the main subject of our course. In this chapter we shall address the problem, assuming that the prices evolve randomly between discrete time epochs. In the next chapter we shall review some basic theory of stochastic processes and will use it to model the prices, evolving continuously in time. While much more technically involved, the continuous time formulation admits some explicit simple pricing formulas, widely used by practitioners. 1. The one period binomial market model The market is the process B t, S t, t =, 1, where B = 1 B 1 = 1 + r 5

6 6 1. OPTION PRICING IN DISCRETE TIME is the risk free asset e.g. bank account with the interest rate r > and S = s S 1 = S Z 1 is the risky asset stocks, etc. with the share price s > and random yield Z 1 {1 + u, 1 + d} with p u := PZ 1 = 1 + u >, p d := PZ 1 = 1 + d > Definition 1.1. A portfolio strategy h is a vector x, y R 2. value process of the portfolio h is V h t = xb t + ys t. The Remark 1.2. For the one-period model, the portfolio is the same for both t = and t = 1: it is chosen at time t = and its value changes from V h to V 1 h due to the random change in stock s price. We shall assume that negative and fractional values of x and y are allowed. Negative x stands for borrowing money from the bank and negative y means a short position in stocks. Also we shall assume liquid market, i.e. any amount of money can be borrowed or invested in bank and any amount of shares can be bought or sold. Definition 1.3. h is an arbitrage portfolio if V h = PV h 1 > = 1 The market is arbitrage free if no arbitrage portfolio exists.. Since V h 1 takes only two values, the latter equality is equivalent to V h 1 > Proposition 1.4. The market as above is arbitrage free if and only if Proof. direct calculation d r u. Definition 1.5. A probability Q is a martingale measure if S = r E QS 1, where E Q is the expectation with respect to Q. Proposition 1.6. The one period binomial model is arbitrage free if and only if there exists a martingale measure. and Proof. Note that Q is defined by its values at 1 + u and 1 + d: q u = QZ 1 = 1 + u, q d = QZ 1 = 1 + d, E Q S 1 = s1 + dq d + s1 + uq u.

7 1. THE ONE PERIOD BINOMIAL MARKET MODEL 7 Suppose there exists a martingale measure, i.e. there are q u and q d, q u + q d = 1 such that s = 1 s1 + dq d + s1 + uq u. 1 + r This implies r = dq d + uq u, i.e. d r u and by Proposition 1.4 the market is arbitrage free. The converse claim is obtained by reversing the implications. Corollary 1.7. If the market is arbitrage free, the martingale measure is unique and is given by q u = r d u d, q d = u r u d. Proof. Solve the equations r = dq d + uq u and 1 = q u + q d. Definition 1.8. A simple contingent claim is a random variable X of the form X = ΦZ 1, where Φ is a contract function. Example 1.9. For the European call option, { Φz = sz K + sz < K = sz K sz K where K is the strike price part of the contract. Definition 1.1. Πt; X, t =, 1 is the arbitrage free price process of the contingent claim X, if the extended market B, S, Π is arbitrage free. The arbitrage free price of the option is fair in the sense that any deviation from this price creates arbitrage on the market, where the option is traded as an asset. In particular, the fair price of the option premium is Π, X. Definition A contingent claim X is reachable if there is a portfolio h, called hedging or replicating portfolio, such that V h 1 = X, P a.s. Definition The market is complete if any contingent claim is reachable. Proposition If X is hedged by h, the price process Πt; X := V h t is arbitrage free. Proof. exercise Proposition The binomial market is complete and a contingent claim X = ΦZ 1 is hedged by the portfolio 1 + uφ1 + d 1 + dφ1 + u x = 1 + ru d Φ1 + u Φ1 + d y = su d

8 8 1. OPTION PRICING IN DISCRETE TIME Proof. Since V h 1 = x1+r+ys 1, the requirement V h 1 = ΦZ 1, implies x1 + r + ys1 + u = Φ1 + u x1 + r + ys1 + d = Φ1 + d. The claimed answer is obtained by solving these equations. Corollary Risk Neutral valuation formula If the market is arbitrage free, then Π; X = r E QX, where Q is the martingale measure. Note that the valuation formula suggests that the fair price premium of the contingent claim is the discounted expected value of X with respect to the martingale measure, rather that the objective measure P Exercises. Problem Find the arbitrage free price and and the replicating portfolio of the European call option with the strike price K = 3.6, for the following one-period binomial market with the bond price dynamics: B = 1 B 1 = 1 + r, where r =.1 is the interest rate, and the stock price dynamics: S = s, S 1 = S Z 1, where s = 3.5 NIS/US$ and Z 1 is a random variable taking values in {u, d} = {.2,.2} with probabilities PZ 1 = 1 + u = p u =.4 PZ 1 = 1 + d = p d =.6. Problem This problem recalls some of the basic notions of mathematical probability and exercises them in the simple setting with the finite probability space. 1 Let Ω = {ω 1,..., ω n } for some fixed n 2. Recall that the power set of Ω, denoted 2 Ω, is the set of all subsets of Ω including. Argue that 2 Ω is finite with cardinality 2 Ω = 2 n, i.e. it contains 2 n elements 2 Show that F := 2 Ω is an algebra of sets, i.e. it is closed under taking intersections and complements and thus also unions.

9 1. THE ONE PERIOD BINOMIAL MARKET MODEL 9 3 Recall that a probability measure on Ω, F is a function P : F [, 1] from sets events to numbers in the interval [, 1], such that PΩ = 1 and PA B = PA + PB whenever A B = additivity property. Using these axioms, show that P is determined by its values on the singletons {ω}, ω Ω Consider a probability space Ω, F with two probability measures P and Q, defined on it. The probability Q is absolutely continuous with respect to P, denoted Q P, if PA = = QA =, A F. If both Q P and P Q, then Q and P are said to be equivalent, denoted Q P. The measures P and Q are orthogonal, denoted by Q P, if there exists a set A, such that PA = and QA = 1. 4 Let P be the uniform measure on Ω, i.e. P{ω} = 1/n and Q be defined by { 1/2 ω {ω 1, ω 2 } Q{ω} = otherwise Is P Q? Is Q P? Is Q P? Is Q P? 5 For P and Q from the previous question, characterize the sets of measures, which are a absolutely continuous w.r.t. P; b equivalent to P; c absolutely continuous w.r.t. Q; d equivalent to Q e orthogonal to P; f orthogonal to Q. 6 Do any two measures have to be either orthogonal or equivalent? If not, give a counterexample. Recall that expectation of a random variable X : Ω R with respect to P is 1 E P X = Xω i P{ω i } i 7 Calculate E P X and E Q X of Xω = 1 {ω {ω1,ω n }} 8 Show that if ξ is a nonnegative random variable with E P ξ = 1, then the set function QA := E P ξ1 {ω A} is a probability measure. 9 Show that if Q P, then there exists a unique non-negative random variable ξ : Ω R, called Radon-Nikodym derivative of Q with respect to P and denoted dq dp ω, such that QA = E P ξ1 {ω A}. Find ξ for P and Q as above. 1 the expectations for uncountable Ω are defined as Lebesgue integrals with respect to a probability measure

10 1 1. OPTION PRICING IN DISCRETE TIME 1 Show that if Q P, then dq dp ω = dp dq ω 1 ω Ω. 11 Prove the change of measure formula for expectations: for a bounded variable X and Q P, E Q X = E P dq dp X. Calculate E Q X for X from 7, using this formula 12 Recall that the conditional expectation of a r.v. X given the r.v Y under probability P is the random variable 2 E P X Y ω = y:py =y> E P X1 {Y =y} PY = y 1 {Y ω=y}. 13 Calculate E P X Y and E Q X Y for X as in 7 and Y = 1 {ω {ω1,ω 2 }}. 14 Show that E P X E P X Y 1 {Y =y} =, y R and consequently 2 E P X E P X Y EP X ϕy for any function ϕ. In other words, the conditional expectation of X given Y minimizes the mean square error in the problem of predicting the value of X, given the realization of Y. 15 Prove the following Bayes formula or change of measure for conditional expectations E P X dq Y E Q X Y = dp dq E P Y dp and apply it for X and Y as above. 16 Explain why all the definitions above can be extended to infinitely countable Ω. Which of the definitions/questions above are meaningless for Ω = [, 1]? We shall see later how the notions introduced above can be defined in much more general setting, than countable Ω ultimately, for Ω which is a set of functions on the time interval [, T ]. 2 2 {Y = y} is the usual abbreviation for {ω : Y ω = y}

11 2. THE MULTI PERIOD BINOMIAL MARKET MODEL 11 Problem Consider the following definition of arbitrage portfolio: V h = V h 1 PV h 1 > > 1 Explain why the above definition of arbitrage is weaker than the one defined in class 2 Show that the market is arbitrage free in this sense if and only if d < r < u 3 Show that the market is arbitrage free if and only if there exist an equivalent martingale measure, i.e. a probability measure Q, such that Q P and S = r E QS 1 S 2. The multi period binomial market model In this section we consider the market model B, S = B t, S t t {,...,T }, where B t = B t r, B = 1 t = 1,..., T is the price process of a risk free asset e.g. bank account, with the interest rate 3 r and S t = S t 1 Z t, S = s t = 1,..., T is stock price process, with initial price s > and Z 1,..., Z T are i.i.d. r.v. p u = PZ 1 = 1 + u >, p d = PZ 1 = 1 + d >. Definition 2.1. A portfolio h is a stochastic process h = h t = x t, y t, t =,..., T, which is predictable with respect to the price process, i.e. h t is measurable with respect to F S t 1 = σ{s,..., S t 1 } h = h 1 is set by convention. The corresponding value process is V h t = x t 1 + r + y t S t, t =,..., T Remark 2.2. x t and y t are the amounts of money in the bank and stocks respectively. At time t =, one chooses the portfolio h = x, y, whose value is V h = x + y s = x 1 + y 1 s 3 r is assumed, since otherwise arbitrage is possible think why. If, however, one must hold her money either in the bank or in stocks and not at home with zero interest rate, then r > 1 makes sense as well.

12 12 1. OPTION PRICING IN DISCRETE TIME This portfolio is kept till time t = 1, when the price is updated to S 1 = sz 1. Now the value of the portfolio is V 1 h = x 11 + r + y 1 S 1. Since more information on the market is available at this moment, it makes sense to update the portfolio to h 2. This portfolio is kept till time t = 2, when the new price S 2 is revealed and the value of the portfolio becomes V h 2 = x r + y 2 S 2. Again, the portfolio is updated to rely on more data, namely {S, S 1, S 2 }, etc. Definition 2.3. A portfolio h is self-financing if x t 1 + r + y t S t = x t+1 + y t+1 S t, t = 1,..., T 1. The latter means that the rebalancing of the portfolio is made without bringing in or out any external funds. Definition 2.4. an arbitrage possibility is a self-financing portfolio h with the property u. V h = PV h T = 1 PV h T > > Proposition 2.5. If the binomial market is arbitrage free, then d < r < Proof. exercise. Definition 2.6. Q is an equivalent martingale measure EMM of the market, if Q P and the process S t /1 + r t is a martingale under Q, i.e. S t 1 = r E Q St S t 1, Q a.s. t = 1,..., T. Proposition 2.7. For the arbitrage free binomial market, there exists the unique EMM Q under which Z 1,..., Z T are i.i.d. and QZ 1 = 1 + u = r d u d >, QZ 1 = 1 + d = u r u d >. Proof. direct calculation Theorem 2.8 the First Fundamental Theorem. The binomial market is arbitrage free if and only if there exists at least one EMM. Proof. One direction follows from the Propositions 2.5 and 2.7. Conversely, assume that there exists an EMM Q and suppose that h is an arbitrage possibility. Since h is an arbitrage possibility, V h = and thus recall the first home assignment E Q V h = E P dq dp V h =.

13 2. THE MULTI PERIOD BINOMIAL MARKET MODEL 13 Since Q P, PV T h > > implies QV T h > > and hence E QV h T > why? On the other hand, for any self-financing portfolio h exercise: E Q V h T = r T E QV h, which is a contradiction. Hence the market is arbitrage free. Definition 2.9. A simple contingent claim is a random variable of the form X = ΦS T, where Φ is the contract function. An option bought at the time t = guarantees possible profit at the future time T and hence at any time t = 1,..., T 1 has a value and can be considered as an asset on its own 4. Consequently, the option can be traded on the market, which raises the question of how to choose a fair price for it at each time t =,..., T. If the underlying market is arbitrage free, it makes sense to price the option so that the extended market, consisting of the original assets and the option, remains arbitrage free. One simple consequence of this principle is that the only possible choice of the price at time T is ΠT ; X := X. Indeed, if we trade the option at e.g. a smaller price ΠT ; X < X, one can easily construct an arbitrage portfolio: do nothing till time T i.e. choose x t = and y t = for t =,..., T and at time buy the option for ΠT ; X and sell it for X, gaining X ΠT ; X >. The question remains as to how the arbitrage free price is chosen at any other time t =,..., T 1. A handy tool to calculate this price is the following notion: Definition 2.1. A contingent claim X is reachable if there exists a selffinancing portfolio h, called hedge or replicating portfolio, such that V h T = X, P-a.s. If a hedge exists for the given contingent claim X, its value process is an arbitrage free price for the option: Πt; X := V h t, t =,..., T. Indeed, suppose that at some t T, the option is offered on the market at a price Πt; X < Vt h, then we wait till time t x s = and y s =, s =,..., t 1, buy at time t the option at the price Πt; X and sell it as the hedge for X at the price Vt h, investing the profit Vt h Πt; X > in the bank account, thus creating arbitrage. Hence the problem of pricing of a contingent claim on an arbitrage free market is reduced to finding a hedge and calculating its value process. As we shall see shortly, this is indeed possible for the binomial market at hand. In fact, the binomial market is complete, i.e. a hedge can be found for all simple contingent claims, which gives the ultimate solution to the arbitrage free pricing problem. 4 indeed, options are routinely traded on the stock markets

14 14 1. OPTION PRICING IN DISCRETE TIME Remark If a market is arbitrage free, but incomplete and a contingent claim X cannot be hedged, the pricing principle outlined above is not applicable. We shall encounter such examples in a more general market model, considered in the next section. Proposition The arbitrage free binomial model is complete and any simple contingent claim X = ΦS T can be replicated by the self-financing portfolio h = x t, y t, t =,..., T with x t k = uv t k 1 + dv t k + 1, k =,..., t 1 + r u d y t k = 1 V t k + 1 V t k, S t 1 u d where V t k is the value of the portfolio at the t, k-node of the corresponding binomial tree, satisfying the backward recursion: V t k = 1 q u V t+1 k q d V t+1 k, k =,..., t 1 + r V T k = Φ s1 + u k 1 + d T k with q u = r d u d, q d = u r u d. In particular, the arbitrage free premium for the contingent claim X is given by: Π; X = V = r T T k= T k q k uq T k d Φ s1 + u k 1 + d T k = r T E QΦS T, 2.1 where Q is the unique EMM guaranteed for the arbitrage free binomial market by Theorem 2.8. Proof. the binomial algorithm, demonstrated in class through a numerical example Remark suggests the neutral valuation pricing formula for the contingent claim X: Π; X = r T E QX. Note that the hedge h does not appear in this formula anymore. Remarkably, this formula remains valid even in incomplete arbitrage free markets for the claims, which cannot be replicated c.f. Remark 2.11.

15 2.1. Exercises. 2. THE MULTI PERIOD BINOMIAL MARKET MODEL 15 Problem Let Ω = {ω 1,..., ω m } and F be the power set 5 of Ω. Let X be a random variable X : Ω R d, d 1 and let R X be the range of X, i.e. R X = {Xω : ω Ω}. 1 Argue that R X is a finite set with cardinality 6 less than m 2 Let F X be the following collection of subsets of Ω: { F X := {ω Ω : Xω Γ} : Γ R X}. Show that F X is an algebra of subsets 7 of Ω and F X F. 3 Suppose ω the outcome of the experiment is not known to us and we see only Xω. An event A occurs if ω A. Explain how the value of Xω tells whether any event in F X occurred or not. Conversely, suppose that we know whether each event in F X occurred or not. Explain why this reveals the value of Xω. 4 Let X and Y be random variables on Ω. Show that F Y F X if and only if there exists a function ϕ : R X R Y, such that Y ω = ϕ Xω for all ω Ω 5 Argue that F Y = F X if and only if there exists a one-to-one function ϕ, such that Y ω = ϕ Xω for all ω Ω. 6 Give an example of a pair of random variables, so that neither F X F Y nor F Y F X holds. 7 Show that F X F Y is an algebra of subsets of Ω. Given an example of F X and F Y so that F X F Y is not an algebra. 8 Let F X F Y be the minimal algebra which contains F X F Y. Let F X,Y be the algebra generated by X, Y, i.e. { } F X,Y = {ω Ω : Xω Γ 1, Y ω Γ 2 } : Γ 1 R X, Γ 2 R Y. Show that F X,Y = F X F Y 9 Show that F X F X,Y Problem Let ξ be a real valued random variable 8 on a probability space Ω, F, P. Show that Pξ > > implies Eξ >. Hint: explain why {ω : ξω > } = n 1 {ω : ξω 1/n} and use the Chebyshev inequality to show that Eξ = implies Pξ > =. 5 the set of all subsets of Ω 6 cardinality of a finite set is the number of its points 7 i.e. it is closed under intersections and complements and thus unions. Note that the is automatically added to F X. 8 do not assume that it takes values in a finite or countable set

16 16 1. OPTION PRICING IN DISCRETE TIME Problem Show that F S t = F Z t for all t = 1,..., T. Hint: use the results from Problem Problem Find the arbitrage free price of the European call option with the strike price K = 8 and maturity time T = 3 for the binomial market with the bank account interest rate r =.1, B = 1 and the stock price dynamics: B t = 1 + rb t 1, t = 1, 2, 3 S = s S t = S t 1 Z t, t = 1, 2, 3, where s = 8, and Z t s are i.i.d. random variables with values in {u, d} = {.5,.5} and PZ 1 = u = p u =.6 PZ 1 = d = p d =.4 Problem Let B, S = B t, S t t {,...,T } be a multi-period binomial market. Prove that the arbitrage free price of a simple contingent claim X = ΦS T is given by Π; X = r T E QX = r T T T q t uq t T t d Φ s1 + u t 1 + d T t, t= where Q is the EMM, under which the stock yields Z t, t = 1,..., T are i.i.d. with q u = QZ 1 = 1 + u and q d = QZ 1 = 1 + d. Problem Prove the claim of Proposition 2.5. Hint: construct an concrete arbitrage possibility h Problem Assume that d < r < u and let Q be the equivalent martingale measure we saw that there is one and it is unique. Show that the value process Vt h, t =,..., T of a self-financing portfolio satisfies r E Q V h t F S t 1 = V h t 1, t = 1,..., T.

17 3. MULTI-ASSET ONE-PERIOD MODEL 17 Problem Show that the binomial market is arbitrage free if there exists an EMM. Hint: assume that there exists an EMM Q and an arbitrage portfolio h and get a contradiction Problems and can be useful. 3. Multi-asset one-period model Consider the market model S t, t =, 1, consisting of N assets: S 1 t S t =. S N t, t =, 1 where S = s R N is a deterministic vector and S 1 ω is a random vector on the probability space Ω = {ω 1,..., ω M } of M outcomes with p i := Pω i > : { } S 1 ω S 1 ω 1,..., S 1 ω M. We shall make the following Assumption 3.1. S 1 > w.l.o.g. S 1 := 1 S 1 1ω >, ω Ω and will refer the first asset St 1 as numeraire. In particular, the case S1 1 = 1 + r is the familiar risk free asset bank account with the interest rate r >. The corresponding notions of the portfolio, etc. are adjusted appropriately: Definition 3.2. The value of the portfolio h = h 1,..., h N is N Vt h = h i St i =: h S t. i=1 Definition 3.3. The portfolio h is an arbitrage possibility if V h = PV h 1 ω = 1 PV h 1 ω > > For technical convenience define the normalized market S t := S t /S 1 t. The asset corresponding to the normalizing price is called numeraire see Assumption 3.1 above and S t are the assets prices under numeraire S 1 t. Note that S t corresponds to the market, which includes a zero interest rate bank account. Proposition 3.4. Let Vt h Then = h S t as before and V h t := h S t, t =, 1.

18 18 1. OPTION PRICING IN DISCRETE TIME 1 V t h = Vt h /St 1 2 a portfolio h is an arbitrage on the S-market if and only if it is an arbitrage on the S-market 3 S t 1 ω = 1 for all ω Ω. Proof. by inspection Definition 3.5. A probability Q is an EMM if Q P and S t is a martingale under Q: S 1 S = E Q S1 1. Remark 3.6. For S1 1 := 1+r, the familiar definition from the previous models is recovered. Theorem 3.7 the First Fundamental theorem of Finance. The market is arbitrage free if and only if there exists an EMM Q Proof. We shall exhibit a vector of strictly positive probabilities q, such that M S = S 1q i i. 3.1 To this end, define an R N M matrices S1 1 ω 1... S1 1ω M D :=....., D := S1 Nω 1... S1 Nω M S 1 Nω 1... SN 1 ω M i=1 By Proposition 3.4, the market has an arbitrage if and only if there exists a portfolio h, such that h S = h D h Du >, where u is an arbitrary vector with strictly positive entries e.g. u := p can be taken. The latter can be rewritten in the form to fit the Farkas lemma 9 h S h S h D h Du <. 9 Lemma [Farkas] For any d,..., d k vectors in R n, exactly one of two problems have a solution: 1 find the numbers λ i, i = 1,..., k, such that d = k i=1 λ id i 2 find a row vector h R n such that h d < and h d i for all i = 1,..., k.

19 3. MULTI-ASSET ONE-PERIOD MODEL 19 Hence, by the Farkas lemma, the market is arbitrage free if and only if the problem Du = [ D S S ] λ }{{} R N M+2 has a solution λ with λ i. Set β i := λ i, i = 1,..., M and α := λ M+2 λ M+1, then the latter reads: α S = Dβ + u. The first line of the this set of equalities yields α = M i=1 β i + i u i > and thus 3.1 holds with q := β + u/α as required. Now we approach the question of pricing the contingent claims. Definition 3.8. A simple contingent claim X is the random variable of the form ΦS 1. Let Πt; X denote the price of the contingent claim at times t =, 1. As in the previous models, the arbitrage free price is obviously Π1; X := X. Our goal is to choose the price Π; X i.e. the premium for the contingent claim, so that the extended market S, Π is arbitrage free. To this end, note that Πt; X can be viewed as an asset of the type, which our model supports, once we fix a deterministic price for it at time t = i.e. it has a fixed price at t = and a random price, depending on S 1, at t = 1. Hence the extended market S, Π fits the framework of the model under consideration. If the original market S is arbitrage free, then there exists possibly nonunique EMM Q and if we set Π; X := E Q ΦS 1 /S 1 1, 3.2 the extended market S, Π remains arbitrage free by Theorem 3.7. Thus we proved the Proposition 3.9 risk-neutral valuating formula. Let S be an arbitrage free market and Q be a corresponding EMM, then 3.2 is the arbitrage free price of the contingent claim X under the numeraire S 1 1. Remark 3.1. Note that we didn t appeal to replicating portfolios to find the arbitrage free price of X, unlike in the case of the binomial model. In the binomial market the risk-neutral pricing formula is an outcome of completeness of the market: once we know that a contingent claim can be hedged, its arbitrage free price is constructed using the value process of the hedge. The essentially new element of the multi-asset market is that it may be incomplete: Proposition The multi-asset one-period market is complete, i.e. any contingent claim can be hedged by a replicating portfolio, if and only if rankd = M.

20 2 1. OPTION PRICING IN DISCRETE TIME i.e. Proof. The portfolio h is replicating if V1 h ω = Xω for all ω Ω, hd = Xω 1,..., Xω M. The latter system of linear equations have a solution for any X if and only if the rows of the matrix D are linearly independent vectors in R M, which is equivalent to rankd = M recall why. Remark If e.g. N < M, i.e. the number of assets is less than the number of possible prices of the assets, there will be non-reachable contingent claims. Our pricing principle is based on the assumption that the underlying market is arbitrage-free, which raises the question how the algebraic condition found in the latter proposition can be expressed in this case. The answer is given in the following Theorem 3.13 the Second Fundamental theorem of Finance. The arbitrage-free market is complete if and only if the EMM is unique. Proof. If the market is arbitrage free, the equation S = Dq does have a solution q with strictly positive entries. This solution is unique if and only if D has rank M: indeed, if the rank of D is less than M, then Dv = for some v R M and q := q + εv with ε > small enough has strictly positive entries and solves S = Dq. Note that D and D has the same rank why? and hence the EMM is unique if and only if rankd = M, i.e. if and only if the market is complete by Proposition Let s explore the connection between the risk-neutral formula 3.2 and the pricing by hedging. Suppose the market is complete and let Q be the unique EMM. It is still possible that X can be hedged by two different hedges h and h look back at Proposition 3.11 for N > M. In this case we shall expect that the two hedges will give the same price for the contingent claim, i.e. their values at t = coincide, since otherwise arbitrage is possible. Indeed, this is the case: V h = hs = he Q S 1 /S 1 1 = E Q V h 1 /S 1 1 = E Q X/S 1 1 = Π; X 3.3 regardless of the particular hedge h. Suppose now that the market is not complete, i.e. there exists at least two EMM s Q and Q. Does the risk-neutral valuating formula 3.2 yield the same price for Q and Q? If X is reachable, say, by a hedge h, then the answer is positive, since V h = E QX/S1 1 and V h = E QX/S 1 1 by 3.3. What if X does not admit a hedge? Note that in this case, the option seller cannot find a replicating portfolio, but the prices given by 3.2 are still such that the market S, Π is arbitrage free, i.e no portfolio yields

21 3. MULTI-ASSET ONE-PERIOD MODEL 21 arbitrage. As the following numerical example shows, many arbitrage free prices are possible in this latter case 1. Example Consider the market with two assets N = 2 and three final prices of the stocks M = 3: B t := S 1 t = 1, t =, 1, i.e. the numeraire is zero interest rate bank account and S := S 2 = 1 S 1 := S 2 1 {.8, 1, 1.2}, the latter taking values with positive probabilities the values are of no importance. Both Q := 1/4, 1/2, 1/4 and Q = 1/3, 1/3, 1/3 are martingale measures: and E Q S 1 = = 1 = S E QS 1 = = 1 = S. Consider the Europian call option with strike price K = 1, i.e. Φs = s K + =.2 1 {s=1.2}. This contingent claim is not reachable, since the vector Xω1, Xω 2, Xω 3 =,,.2 is clearly not in the span of the rows of D = The neutral risk valuation formula gives two different arbitrage free prices: and Π; X = E Q X = 1/4.2 =.5 Π; X = E QX = 1/3.2 = Check e.g. that the silly contingent claim with the contract function Φs = s is reachable and check that its premium is the same under both Q and Q Exercises. Problem Separating hyperplane theorem. Let C and D be nonintersecting closed convex subsets of R d and assume that D is bounded. Then there a separating hyperplane exists, i.e. there exist a nonzero vector a R d and a constant b such that a x b for all x C and a x b for all x D the set {x R d : a x = b} is the separating hyperplane. Prove this statement, following the steps below. 1 Note that existence of more than one arbitrage free price is not a contradiction: one is not allowed to trade two copies of the option with different prices

22 22 1. OPTION PRICING IN DISCRETE TIME Consider the distance between the sets C and D, defined by: where x 2 2 = d i=1 x2 i dc, D = inf{ c d 2 : c C, d D}, is the Euclidian norm. 1 Give an example of sets C and D in R 2 such that dc, D =, but C D =. Hint: try open sets 2 Show that if C and D are closed and D is bounded, then C D = implies dc, D >. Hint: prove by contradiction: assume that dc, D =, i.e. there exist sequences u n C and v n D, such that u n v n 2 ; use the Bolzano-Weierstrass theorem 11 to extract a convergent subsequence v nk v and argue that v D; show that u nk v and argue that v C, i.e. v C D 3 Show that if C and D are closed and D is bounded, then dc, D = c d 2 for some c C and d D. Hint: use the Bolzano-Weierstrass theorem again By 2, dc, D > and by 3, dc, D = c d 2 for some c C and d D. Consider the hyperplane with a := d c and b := 1 2 d 2 2 c 2 2. This is our candidate for the separating plane, i.e. fx := a x b for x D and fx for x C 4 Check that fx = d c x d d c Prove that fx for x D by contradiction. Assume that fx < for some x D and conclude that d d + tx d c 2 dt 2 = 2d c x d <, t = and, consequently, d + tx d c 2 2 < d c 2 2 for all t small enough. Argue that d + tx d D, i.e. that dc, D < c d 2, which contradicts the choice of c and d. 6 Prove that fx for x C, similarly to look up for it, if not familiar

23 3. MULTI-ASSET ONE-PERIOD MODEL 23 Problem strict separation of a point and a closed convex set. If C is a closed convex set and x C, then there exist a nonzero vector a and a number b, such that a x < b for x C and a x > b. Follow the steps, to deduce this statement from the Separating Hyperplane theorem: 1 Argue that for some 12 ε >, Bε x C = 2 By the Separating Hyperplane theorem there exist a nonzero a and b such that a x b for x C and a x b for x B ε x. Show that the latter is equivalent to a x + a u b, u such that u 2 ε. 3 Show that min u: u 2 ε a u = ε a 2 and the minimum is attained at u := εa/ a 2. In particular, a x ε a 2 b. 4 Let b := b + ε 2 a 2 and show that a x < b for x C and a x > b Problem The Farkas lemma states that for the vectors d,..., d K R N, exactly one of the following problems has a solution: a find numbers λ j such that d = K j=1 λ jd j b find a row vector h R N such that 13 hd < hd j, j = 1,..., M Follow the following steps to prove the lemma 1 Show that a and {b cannot hold simultaneously K } 2 Define a set C := i=1 β id i : β i R + {} R n. Show that 14 a C is a cone, i.e. v C implies av C for any number a > b C contains the origin of R N c C is a convex set d C is a closed subset of R N 3 Argue that the claim of the Farkas lemma follows, if we assume that for d C and show that b holds. 4 Assume that d C. Use the Problem to show that there exist an α R and an h R N so that hd < α < h x, x C. 5 Argue that C implies α < and hence hd <. 12 Bε x := {x R d : x x 2 ε} is the closed ball around x with radius ε > 13 for a row vector h and a column vector v, hv := h v = i h iv i 14 if not familiar, look up for the definitions of convex and closed subsets

24 24 1. OPTION PRICING IN DISCRETE TIME 6 Check that for any ξ i, i = 1,..., K K K α < h ξ i d i = ξ i h d i, i=1 i=1 and argue that the latter implies h d i. Hint: the inequality is impossible for any ξ i if h d i <. Why? Problem Let D be an n m matrix of real numbers and ξ a row vector in R n. 1 Show that the problem hd = ξ has the unique solution h for all ξ, if and only if rankd = m. 2 Recall the definitions KerD = {x R m : Dx = } and ImD = {D y : y R N } R m. Show that the orthogonal compliment of ImD in R m : ImD { = z R m : z v = v ImD } coincides with KerD.

25 CHAPTER 2 Stochastic processes in continuous time 1. Elements of the general theory Definition 1.1. A random process X with continuous time parameter t [, is a collection of random variables X t ω, t [,, defined on a probability space Ω, F, P. Definition 1.2. The distributions of the random vectors X t1,..., X tn for t 1,..., t n [, and n N are called finite dimensional distributions f.d.d of X Existence of stochastic processes. The fundamental question is whether a stochastic process with given finite dimensional distributions exists? An affirmative answer would be a construction of a probability space Ω, F, P and a stochastic process X t ω, t [, with the given f.d.d. To this end consider the following: Definition 1.3. A family of distributions Q t1,...,t n A 1... A n, t 1,..., t n [,, A 1,..., A n BR, n N is consistent if i for any permutation t i1,..., t in ii of t 1,..., t n Q ti1,...,t in A i1... A in = Q t1,...,t n A 1... A n Q t1,...,t n A 1... A n 1 R = Q t1,...,t n 1 A 1... A n 1 Example 1.4. The family of distributions n Q t1,...,t n A 1... A n = fx i dx i 1.1 A i where f is a probability density on R, is consistent, while n Q t1,...,t n A 1... A n = fx i t n dx i, A i is not why? Obviously, the family of f.d.d. of a random process if exists is consistent why? Hence consistency of f.d.d. is the necessary condition for existence 25 i=1 i=1

26 26 2. STOCHASTIC PROCESSES IN CONTINUOUS TIME of a random process with these f.d.d. Remarkably, consistency turns to be also sufficient. Definition 1.5. C is a cylinder set of the space of real valued functions on [,, denoted by R [,, if there is a finite set of times t 1,..., t n and Borel subsets A 1,..., A n of R, such that C = {ω R [, : ω t1 A 1,..., ω tn A n }. Denote by B [, the σ-algebra generated by the cylinder sets, i.e. the minimal σ-algebra of subsets of R [,, containing the cylinder sets. A set Γ belongs to B [, if and only if there is a sequence of times t n and Borel sets A n, such that see a guided exercise below Γ = {ω R [, : ω t1 A 1, ω t2 A 2,...}. This implies that many sets of interest, such as e.g. {ω R [, : sup ω t c}, t the set of all continuous functions, the set of all functions continuous at a point, etc. are not measurable w.r.t. B [, see a guided exercise below. Roughly speaking, this means that the cylinder σ-algebra is not rich enough for such a big space as R [,. One of the reasons to consider R [, with the cylinder σ-algebra and not e.g. the Borel σ-algebra with respect to some metric is the following theorem, which is the main tool for constructing random processes with given f.d.d.: Theorem 1.6 Daniel-Kolmogorov. Let Q t1,...,t n be a consistent family of distributions, then there is a unique probability measure on the measurable space R [,, B [,, such that the extension of P to any finite set of times coincides with the corresponding f.d.d.: P t1,...,t n = Q t1,...,t n. Remark 1.7. The proof applies Caratheodory s extension theorem, stating that a σ-additive measure on an algebra can be uniquely extended to the σ-algebra generated by this algebra. In the context of D-K theorem, the algebra is the collection of the cylinder sets check, that it is indeed an algebra!. The family of the given consistent distributions define an additive measure on this algebra. The nontrivial part is to check that this measure is in fact σ-additive. Once the latter is established, the Caratheodory theorem is applied. Given a consistent family of distributions, the D-K theorem gives a probability space Ω, F, P with Ω = R [, and F = B [,, referred to as canonical, such that the coordinate process X t ω = ω t has the required f.d.d. Note that nothing is said about the properties of the obtained process: in particular, it may have quite bizarre trajectories. For example, the process

27 1. ELEMENTS OF THE GENERAL THEORY 27 with f.d.d. 1.1 exists, but has discontinuous paths at all points; in fact, the map ω, t X t ω is not even measurable with respect to F B[, and hence it is not even clear whether the Lebesgue integral X sωds can be defined for each ω see an exercise below Types of measurability. Viewed as a map of both arguments ω, t X t ω, various types of measurability can be defined. Definition 1.8. The process X is measurable, if the map ω, t X t ω is F B[, -measurable 1. Since jointly measurable function is measurable with respect to each one of its coordinates, a measurable bounded process is Lebesgue integrable w.r.t. t for each ω, i.e. t X sωds is again a measurable process. In many applications, the probability space comes with an increasing family of σ-algebras F t, t, i.e. F s F t F for all t s. The family F t is called filtration of σ-algebras and the quadruple Ω, F, F t, P is referred to as the filtered probability space or stochastic basis. The filtration is said to satisfy the usual conditions if it is right continuous: F t+ := ε> F t+ε = F t, t and F contains all the P-null sets of F. These are technical conditions, without which certain desirable properties do not hold beyond the scope of our course. Definition 1.9. The process X is F t -adapted, if the map ω X t ω is F t -measurable for each t. The practical meaning of X being F t -adapted is that the trajectory of X can be reconstructed precisely if one knows which events from F t occurred. For example, if S is a stock price process and X t is the portfolio to be chosen at each t on the basis of the prices observed up to time t, X is to be adapted to the natural filtration of S, i.e. Ft S := σ{s s, s t}. Note that an adapted process may not be Lebesgue integrable w.r.t. time variable t for each ω, hence the processes are usually required to be both measurable and adapted. Definition 1.1. The process X is progressively measurable if the map ω, t X t ω is F t B[, t]-measurable for each t. Clearly, a progressively measurable process is both adapted and measurable. Counterexamples show that the converse is false, unless X has sufficiently regular paths. It can be shown that if X is adapted and measurable and has cadlag 2 paths, then it has a progressively measurable modification 1 recall that for a pair of measurable spaces X, X and Y, Y, the function ψ : X Y is X/Y-measurable if ψ 1 B := {x X : ψx B} X for all B Y. When the range σ-algebra Y is obvious from the context, the map is briefly said to be X-measurable. 2 a function f : [, R is cadlag if it is right continuous and has limits from the left

28 28 2. STOCHASTIC PROCESSES IN CONTINUOUS TIME refer to the next section for the definition of modification. The space of cadlag functions plays the central role in the general theory of stochastic processes. Once again, let s stress that the D-K theorem does not impose much structure on the trajectories of the emerging process Equality of random processes. Definition The cadlag processes X and Y are equal in distribution or identically distributed if they have the same f.d.d., i.e. X t1,..., X tn d = Y t1,..., Y tn, t 1,..., t n, n Note that identically distributed processes must not even be defined on the same probability space. Definition X is a modification version of Y if P {ω Ω : X t ω = Y t ω} = 1, t. Clearly, modifications are identically distributed why?. Definition X and Y are indistinguishable if P {ω Ω : sup X t ω Y t ω = } = 1. t Clearly, indistinguishable processes are modifications of each other. The following example demonstrates that modifications must not be indistinguishable. Example Let Ω, F, P = [, 1], B[, 1], λ, where λ is the Lebesgue measure and define X t and Y t = 1 {t=ω}, t [, 1]. Then but sup t [,1] X t ω Y t ω = 1. PX t = Y t = Pt ω = 1, Lemma If X and Y are versions with cadlag paths, they are indistinguishable prove. Note that the process Y in Example 1.14 violates the conditions of this lemma Exercises. Problem Recall that a collection A of subsets of a set Ω is an algebra if Ω A and A is closed under finite intersections and compliments, i.e. A, B A = A B A and A c A. 1 Show that algebra is closed under unions as well 2 Argue that the collection of all open intervals in R is not an algebra 3 Argue that finite unions of nonintersecting open intervals do not form an algebra

29 1. ELEMENTS OF THE GENERAL THEORY 29 4 Prove that the collection of unions of nonintersecting intervals of the form a, b], a < b together with R itself is an algebra call it A Recall that a collection S of subsets of a set Ω is a σ-algebra if Ω S and S is closed under compliments and countable number of intersections, i.e. whenever A n is a sequence in S, n 1 A n S and A c n S. 5 Argue that the algebra A from 4 is not a σ-algebra 6 Show that a σ-algebra is closed under countable unions 7 Let σa be the smallest σ-algebra which contains the algebra A. Argue that σa exists. Hint: prove that if S u, u U is an arbitrary collection of σ- algebras not necessarily countable, then u U S u is a σ-algebra. Use this and the fact that the collection of all subsets of R is a σ-algebra 8 Show that singletons, open and closed intervals are contained in σa. 9 Show that the set of all rational numbers Q is contained in σa 1 The Borel σ-algebra of the subsets of R, denoted BR, is the smallest σ-algebra, which contains the open intervals the existence is established as in 7 above. Prove that BR = σa. Hint: by 8, the open intervals are included in σa and hence BR σa. Give a similar argument for the other inclusion. Problem Recall the basic setting of the probability theory. The probability space Ω, F, P consists of a set Ω of points or elementary events, a σ-algebra of its subsets events F see the definition above and a probability measure P, i.e. a σ-additive 3 function F [, 1]. A random variable X with values in R is a F/BR-measurable function, i.e. {ω : Xω B} F for all B BR. Note that P X B = P{ω : Xω B}, B BR is well defined. It is not hard to show that P X is a probability measure on BR, induced by X. The function x P X, x] =: F X x is called the cumulative distribution function c.d.f of X. 1 Construct a probability space i.e. specify Ω, F, P so that Xω = ω is a Bernoulli random variable 2 Construct a probability space so that Xω = ω is a Poisson random variable 3 Presuming that the Lebesgue probability space Ω, F, P = [, 1], B[, 1], λ 3 i.e. for any sequence of pairwise disjoint sets An n N, P n=1a n = n=1 PA n

30 3 2. STOCHASTIC PROCESSES IN CONTINUOUS TIME is already constructed 4, construct a random variable Xω on it with the given continuous strictly increasing c.d.f. F. Hint: argue that Xω = F 1 ω is a random variable and show that it has the right c.d.f 4 Construct a Poisson r.v. on the Lebesgue probability space Remark Construction of the Lebesgue space is one of the pearls of the measure theory. The preceding problem hints that many random variables can be constructed starting from the Lebesgue space in fact, even such a complex random process as the Brownian motion can be constructed from a single instance of the Lebesgue space!. How do we construct a random vector with values in R n and given joint c.d.f? What is the appropriate σ-algebra for R n? How can we construct a probability space and an infinite sequence of random variables a discrete time random process with prescribed probability distribution? Can we construct a family of random variables, indexed by a continuous parameter the random process in continuous time? The following problem aims to stimulate curiosity. Problem Consider the Lebesgue space [, 1], B, P as in the previous problem we changed the name P := λ. 1 Show that P{ω} = for all ω R. Find the mistake in the following absurd: 1 = P ω [,1] {ω} = P{ω} =. ω [,1] 2 Argue that any countable subset of Ω is Borel measurable and has zero Lebesgue measure Hint: by definition a countable set can be enumerated and is a countable union of its points 3 Show that any number ω [, 1] can be expanded into series ω = 2 i ω i, where ω i {, 1} i=1 4 Show that the set of numbers with two distinct expansions coincides with the set of dyadic rationas and thus has Lebesgue measure zero e.g. 1 2 =.1... = , etc. 4 i.e. for any set A B[, 1], the probability measure λa is well defined and λa, b = b a

31 1. ELEMENTS OF THE GENERAL THEORY 31 5 Define an infinite sequence of random variables X n ω = ω n, n N argue that these are indeed r.v. s Prove that X 1,..., X n are i.i.d. Ber1/2 for any n 1. 6 Define Y 1 ω = i odd 2 i+1/2 ω i and Y 2 ω = i even 2 i/2 ω i. Show that Y 1 and Y 2 are i.i.d. random variables with uniform distribution on [, 1] 7 Suggest a way to generate an infinite sequence of i.i.d. r.v. s Y n, n 1 with Y 1 U[, 1]. 8 Suggest a way to generate an infinite sequence of i.i.d. N, 1 random variables 9 Consider the set { A := ω Ω : lim n 1 n } n ω i = 1/2. i=1 Give examples of ω s from A and from A c. The strong law of large numbers asserts that PA = 1. In particular A is uncountable. Does this imply that A c is countable? Problem expectation = the Lebesgue integral. Let Ω, F, P be a probability space and X be a nonnegative random variable. Define a sequence of simple 5 random variables X n ω = n2 n k=1 k 1 2 n 1 {k 1/2 n Xω k/2 n } + n1 {Xω n} and define the Lebesgue integral of X n w.r.t. P EX := X n dp := n2 n k=1 k 1 2 n P ω : {k 1/2 n Xω k/2 n } + npxω n. 1 Explain why measurability of X is crucial for the definition of EX n 2 Draw X 1, X 2 and X 3 for Xω = ω 1/2 and calculate their Lebesgue integrals. Do they coincide with the Riemann integrals? 3 Show that X n ω Xω for all n 1 and lim n X n ω = Xω for any ω Ω. 5 simple random variable takes a finite number of values

32 32 2. STOCHASTIC PROCESSES IN CONTINUOUS TIME 4 Use Lebesgue s Monotone Convergence theorem 6 to show that the limit lim n EX n exists. This limit is the Lebesgue integral of X, denoted XdP or EX. 5 Extend the definition of EX for an arbitrary random variable X, with E X < Hint: note that X = X + X and X = X + + X, where X + = X and X = X 6 Prove Xω = 1 {ω Q}, ω [, 1] is Lebesgue integrable on the Lebesgue space and calculate the integral. Remark A theorem of Lebesgue tells that a function on [, 1] is Riemann integrable if and only if the set of its discontinuity points has the Lebesgue measure zero, in which case the Riemann and the Lebesgue integrals coincide. However, Lebesgue integrability conditions are much weaker: only measurability is required. Moreover, the construction of the Lebesgue integral does not depend on the particular geometry of the space unlike the construction of the Riemann integral on R. This makes possible integration in very abstract infinite dimensional spaces such as the space of sequences or space of continuous functions, etc.: all one needs is a measurable structure and a measure on the space! Problem This problem explores the cylinder sets and the σ- algebra they generate. A subset C R [, is a cylinder if there exists a finite number of distinct times t 1,..., t n and a set A BR n, such that C = { ω R [, : ω t1,..., ω tn A }. 1.2 Depending on the structure of the set A, three distinct collections of cylinder sets can be considered: a C 1 is the collection of the cylinder sets as in 1.2 with A of the form A := I 1... I n, where I i are open intervals of R; b C 2 is the collection of the cylinder sets as in 1.2 with A of the form A := A 1... A n, where A i BR; c C 3 is the collection of the cylinder sets as in 1.2 with A BR n Clearly, C 1 C 2 C 3 and hence σ{c 1 } σ{c 2 } σ{c 3 } why? 1 Show that C 3 is an algebra 6 Theorem: if Xn is an increasing sequence of nonnegative random variables, i.e. X n ω and X n ω X n+1 ω P-a.s., then the function Xω := lim n X n ω is a random variable and EX = lim EX n n