Introduction Univ. Prof. Dr. Helmut Strasser. Advanced Financial Mathematics and Structured Derivatives Module 1 Munich, June 11th - 14th, 2008

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1 Introduction Univ. Prof. Dr. Helmut Strasser Advanced Financial Mathematics and Structured Derivatives Module 1 Munich, June 11th - 14th, 2008 Module 1 Slide 1 - Title c Helmut Strasser, May 30, 2009

2 Single period models Basic concepts Module 1 Slide 2 - Title c Helmut Strasser, May 30, 2009

3 The simplest case: Single period A market with two assets: M = (B, S), period [t, T ]. Spot prices: B t, S t Terminal values: B T, S T (Extension to markets with m > 2 assets is straightforward.) Example: Binomial model r(t t) B bank account with fixed interest rate r: B t = 1, B T = e S risky asset: P (S T = S t u) = p or P (S T = S t d) = 1 p with d < u and 0 < p < 1. Module 1 Slide 3 - Single period model c Helmut Strasser, May 30, 2009

4 Retrieving empirical stock prices y=getstocks("^dji",from=c(1,1,2004),to=c(31,12,2004)) print(y) pplot(y) pplot(returns(y)) y=getstocks(c("msft","nok"),from=c(1,1,2004),to=c(31,12,2004)) pplot(y) pplot(returns(y)) Module 1 Slide 4 - R: Empirical stock prices c Helmut Strasser, May 30, 2009

5 Statistics of empirical stock prices y=getstocks(c("ko"),from=c(1,1,2000),to=c(31,12,2001)) Scatter(y) Scatter(Returns(y)) summary(returns(y)) Hist(Returns(y)) Normalplot(Returns(y)) Boxplot(Returns(y)) Shape(as.vector(Returns(y))) Module 1 Slide 5 - R: Descriptive statistics c Helmut Strasser, May 30, 2009

6 Example: Lognormal model r(t t) B bank account with fixed interest rate: B t = 1, B T = e Assume that the log-returns of the risky asset are normally distributed: log S T S t N(a(T t), σ 2 (T t)) Volatility: Standard deviation σ of the log-returns (per time unit). ( Average asset price: E(S T S t ) = S t exp a(t t) + σ2 (T t) ) 2 Usual parametrization: µ = a + σ 2 /2 growth rate of average prices µ(t t) = S t e Lognormal model: where Z N(0, 1). ) S T = S t exp [(µ σ2 (T t) + σ ] T t Z 2 Module 1 Slide 6 - Lognormal model c Helmut Strasser, May 30, 2009

7 Lognormal distribution ff=function(meanlog=0,sdlog=1) { fun=function(x) dlnorm(x,meanlog,sdlog) } plot(ff(),0,5);grid() plot(ff(meanlog=0.04,sdlog=0.5),0,5);grid() sd=0.5 plot(ff(meanlog=-sd^2/2,sdlog=sd),0,5);grid() x=rlnorm(500,meanlog=-sd^2/2,sdlog=sd) Scatter(x) Hist(x,breaks=20) plot(density(x),xlim=c(0,5)) Module 1 Slide 7 - R: Lognormal distribution c Helmut Strasser, May 30, 2009

8 Lognormal model X=rpBS(rate=0.04,vol=0.2) pplot(x(1)) pplot(x(20)) x=apply(x(500),1,mean) plot(attr(x(),"time"),x,type="l") Module 1 Slide 8 - R: Lognormal model c Helmut Strasser, May 30, 2009

9 Portfolio: H = (H B, H S ) (number of units in the portfolio) Value: V t = H B B t + H S S t : initial value, price of the portfolio V T = H B B T + H S S T : terminal value, payoff of the portfolio Admissible portfolio values: V = {V = H B B + H S S : (H B, H S ) R 2 } (Assumptions: divisibility of assets, short selling allowed) Theorem: Every linear combination of admissible portfolio values is an admissible portfolio value. Module 1 Slide 9 - Portfolios c Helmut Strasser, May 30, 2009

10 Definition: The market M has the no-arbitrage property (NA) if for every V V V t 0 and V T 0 V t = V T Example: Binomial case NA holds iff d < e r(t t) < u. Consequences: Let M be a market with NA. Then: (1) There is no free lunch (NFL): V t = 0 and V T 0 V T = 0 whenever V V. (2) Prices are uniquely determined, i.e. V T = V T V t = V t whenever V, V V (This means: If NA holds then pricing makes sense.) (3) There is at most one admissible portfolio value V V which is riskless. Module 1 Slide 10 - No arbitrage property c Helmut Strasser, May 30, 2009

11 Forward price Period [t, T ], market model: B = (B t, B T ) = (1, e r(t t) ), S = (S t, S T ). V = S S t B is an admissible portfolio: V t = 0, V T = S T S t e r(t t). Spot price: S t r(t t) Forward price: F := S t e Forward contract: Exchange at time T of the asset and the forward price. This contract has price zero. Module 1 Slide 11 - Forward price c Helmut Strasser, May 30, 2009

12 Multinomial case: Characterize the NA-property! Bet and Win: A simple example A bet is offered on a soccer game with teams A and B (remis excluded). A company offers bets with quotes (a, b). Characterize the quotes where arbitrage (a riskless profit) is possible! Assume that two companies offer bets on the same game, but with different quotes. Discuss the NA property! Extend your findings to bets with k alternatives (e.g. horse races) and m companies offering quotes! A theoretical problem: Let (B, S) be a single period market. Assume that there is a numeraire (a strictly positive portfolio N V). Show that NA is equivalent to NFL! Module 1 Slide 12 - Problems c Helmut Strasser, May 30, 2009

13 Single period models Claims Module 1 Slide 13 - Title c Helmut Strasser, May 30, 2009

14 Definition: A claim is a payoff at time T depending on the history of the market. Examples: Call C T = (S T K) +, Put P T = (K S T ) +, (K strike price). Binary Call C b T = 1 (S T K), Binary Put P b T = 1 (S T <K) Vanilla claims: Claims with liquid prices. The claims are traded at an exchange, and market efficiency provides prices which avoid arbitrage. Exotic claims: Claims which are not traded at an exchange but sold over the counter. Pricing problem: How to find prices for exotic claims? Module 1 Slide 14 - Claims c Helmut Strasser, May 30, 2009

15 Put-Call parity Let C and P be a call and a put for the same strike price K with arbitrage free prices. Put-Call parity: C T P T = (S T K) + (K S T ) + r(t t) = S T K C t P t = S t Ke Implications: r(t t) C t S t Ke which Early exercise of a call is of no value. P t Ke r(t t) S t is > S t K which is < K S t Early exercise of a put can be valuable. Module 1 Slide 15 - Put-Call parity c Helmut Strasser, May 30, 2009

16 Pricing by replication: Find a portfolio (H B, H S ) such that the portfolio value satisfies V T = C T. Then take V t as the price C t of the claim C at time t. Theorem: If the market satisfies NA, then this rule provides the one and only price which avoids arbitrage possibilities. Question: When is it possible to replicate a claim? Module 1 Slide 16 - Replication c Helmut Strasser, May 30, 2009

17 Example: Binomial model. Let C T := { C u T whenever S T = S u T = S tu C d T whenever S T = S d T = S td Then a replicating portfolio necessarily satisfies: We see: C T H S S T = H B e r is riskless! H B e r + H S S u T = Cu T H B e r + H S S d T = Cd T C H S S T has only a single outcome: C u T HS S u T = Cd T HS S d T This implies H S = Cu T Cd T S u T Sd T =:, H B = e r (C u T S u T) The number of units of S in the portfolio equals the Delta of the claim. Module 1 Slide 17 - Binomial case c Helmut Strasser, May 30, 2009

18 Problem: A stock price is currently 20$, and it is known that at the end of three months the stock price will be either 18$ or 22$. We are interested in pricing a European call option to buy the stock for 21$ at the end of three months. Suppose that the risk-free rate is 12% per annum. Module 1 Slide 18 - Problem c Helmut Strasser, May 30, 2009

19 In a binomial model every claim can be replicated. The single period binomial model is complete. Definition: A market model is complete if every claim can be replicated by an admissible portfolio value. Problem: Discuss completeness in the multinomial case! Module 1 Slide 19 - Market completeness c Helmut Strasser, May 30, 2009

20 Pricing based on liquid options Assume that the market provides prices for calls C 1, C 2,..., C m with strikes K 1 < K 2 <... < K m. Let C T = f(s T ) be a claim whose payoff f(x) is continuous and piecewise linear payoff function with edges at K 1 < K 2 <... < K m. This claim is replicable: Let k 1, k 2,..., k m+1 be the slopes of the payoff function f(x). Then m f(x) = f(0) + k 1 x + (k i+1 k i )(x K i ) + Therefore C T = f(s T ) = f(0) + k 1 S T + m i=1 (k i+1 k i )C i T and C t = f(0)e r(t t) + k 1 S t + m i=1 (k i+1 k i )C i t i=1 Module 1 Slide 20 - Using liquid options c Helmut Strasser, May 30, 2009

21 Single period models Risk neutral pricing Module 1 Slide 21 - Title c Helmut Strasser, May 30, 2009

22 M = (B, S): a market, B a bank account. P, Q: probability models such that the expectation E(S T ) exists. Definition: A probability model Q is called risk neutral if the expectation of S T equals the forward price of S t : E Q (S T S t ) = e r(t t) S t. Risk neutrality of Q means: (1) The growth rate of expected prices equals the riskless rate. (2) Forward contracts are risk neutral, the payoff of forward contracts equals the expectation of the underlying. Important: Typical statistical probability models P need not be risk neutral! Module 1 Slide 22 - Risk neutral models c Helmut Strasser, May 30, 2009

23 Example: Binomial model. Let Q(S T = S t u) = q and Q(S T = S t d) = 1 q. Then Q is risk neutral iff q = er(t t) d. u d Note: For the binomial case there exists a risk neutral model iff the NA-property holds. Problem: Discuss risk neutral models for the multinomial case! Problem: Discuss uniqueness of risk neutral models both for the binomial and for the multinomial case! Module 1 Slide 23 - Examples c Helmut Strasser, May 30, 2009

24 Pricing rule: Let Q be a risk neutral probability model. Then the prices of admissible portfolios equal the expectations under Q of the discounted payoffs. Proof: Risk neutrality means ( E Q (S T S t ) = e r(t t) ST ) S t E Q S e r(t t) t = S t This carries over to all attainable portfolio values: E Q ( VT ) S e r(t t) t = E Q ( H B B T + H S S T e r(t t) S t ) = H B B t + H S S t = V t Question: How to find prices, if the statistical model P is not risk neutral? Answer: Find some (equivalent) risk neutral probability model Q! Problem: Solve binomial pricing via risk neutral pricing! Module 1 Slide 24 - Pricing with risk neutral models c Helmut Strasser, May 30, 2009

25 Question: Can we be sure to be able finding risk neutral probability model? Fundamental theorem: For any (single period) market there exists a risk neutral model iff the NA-property holds. Proof: The proof consists of two parts. Part 1 (easy): If there is an equivalent risk neutral model then the NA-property holds. Part 2 (difficult): If the NA-property holds, then there is some equivalent risk neutral model. Module 1 Slide 25 - Fundamental theorem c Helmut Strasser, May 30, 2009

26 Single period models Lognormal pricing Module 1 Slide 26 - Title c Helmut Strasser, May 30, 2009

27 Risk neutral lognormal model: where Z N(0, 1) (P ). ) S T = S t exp [(µ σ2 (T t) + σ ] T t Z 2 µ(t t) This model satisfies: E P (S T S t ) = S t e It follows: A lognormal model Q is risk-neutral iff µ = r, i.e. ) S T = S t exp [(r σ2 (T t) + σ ] T t Z 2 where Z N(0, 1) (Q). Risk neutral pricing: If C T is some claim, we may calculate and call it the price of the claim C T at time t. C t = E Q(C T S t ) er(t t) Module 1 Slide 27 - Risk neutrality c Helmut Strasser, May 30, 2009

28 Theoretical background: Can such a procedure be justified? (1) The single period lognormal model is not complete. Usually, claims cannot be replicated within a single period model which is not binomial. (2) We did not see so far how a risk neutral model should be able to render the initial capital of hedging portfolios in a multiperiod situation. (3) We cannot understand so far how a change of the model can be performed in such a way that it stays equivalent to the original statistical model (whatever that means). These are serious questions. We will answer them in the forthcoming chapters. At present: We will perform the simple calculations which render prices of claims under a risk neutral lognormal model. Module 1 Slide 28 - Theory c Helmut Strasser, May 30, 2009

29 Calculation strategy Let C T = f(s T ), i.e. assume that the payoff depends on the terminal value only. ( S ) T C T = f S t where S T is independent of the known initial asset price S t. S t S t Then E Q (C T S t ) = h(s t ) where [ h(x) := E Q (f x S ]) T = E Q (f S t [ x exp The main calculation work consists in evaluating Then the result is ((r σ2 2 ) (T t) + σ )]) (T t) Z u(x, a, b) := E(f(x exp(a + bz))) where Z N(0, 1) [ ) E Q (C T F t ) = h(s t ) = u S t, (r σ2 (T t), σ ] T t 2 Module 1 Slide 29 - Calculation strategy c Helmut Strasser, May 30, 2009

30 Price formula for a European call: C T = (S T K) + Let F t := S t e r(t t) and σ t = σ T t. Then where C t e r(t t) = F t N(d 1 ) KN(d 2 ) d 1 = log(f t/k) + σ 2 t /2 σ t, d 2 = log(f t/k) σ 2 t /2 σ t Price formula: C t = S t N(d 1 ) Ke r(t t) N(d 2 ) Module 1 Slide 30 - Call prices c Helmut Strasser, May 30, 2009

31 Monte Carlo Simulation: Value r=0.04 S=rBS(vol=0.2,rate=r,maturity=1,spot=1) mean(call(1)(s( )))*exp(-r) Call.BS(vol=0.2,rate=r)$Value Module 1 Slide 31 - R: MC Call Value c Helmut Strasser, May 30, 2009

32 MC: Call value Linear interpolation S=rBS(vol=0.2,rate=0.04) Call(1)(S(10)) Call(c(1.2,1,0.8))(S(10)) sp=seq(from=0.5,to=1.5,by=0.1) Call(1/sp)(S(15)) pr=sp*apply(sp*call(1/sp)(s(15)),2,mean)*exp(-0.04) plot(sp,pr,type="l") S=rBS(vol=0.2,rate=0.04) pr=sp*apply(call(1/sp)(s(10000)),2,mean)*exp(-0.04) plot(sp,pr,type="l") pr0=call.bs(spot=sp,vol=0.2,rate=0.04)$value lines(sp,pr0,col="red") Module 1 Slide 32 - R: MC Call Value c Helmut Strasser, May 30, 2009

33 MC Call value Smoothing spline interpolation S=rBS(vol=0.2,rate=0.04) sp=seq(from=0.5,to=1.5,length=50) pr=sp*apply(call(1/sp)(s(50000)),2,mean)*exp(-0.04) sm=smooth.spline(sp,pr,spar=0.5) price=function(x) predict(sm,x)$y plot(price,0.5,1.5) pr0=call.bs(spot=sp,vol=0.2,rate=0.04)$value lines(sp,pr0,col="red") Module 1 Slide 33 - R: MC Call Value c Helmut Strasser, May 30, 2009

34 MC BinCall value S=rBS(vol=0.2,rate=0.04) BinCall(1)(S(10)) BinCall(c(1.2,1,0.8))(S(10)) S=rBS(vol=0.2,rate=0.04) sp=seq(from=0.5,to=1.5,length=50) pr=apply(bincall(1/sp)(s(50000)),2,mean)*exp(-0.04) sm=smooth.spline(sp,pr,spar=0.5) price=function(x) predict(sm,x)$y plot(price,0.5,1.5) Module 1 Slide 34 - R: MC BinCall Value c Helmut Strasser, May 30, 2009

35 Diagrams call.price.diagram(rate=0.05) put.price.diagram(rate=0.05) pf=function(x,y) Call.BS(spot=x,vol=y)$Value Persp("pf(x,y)",xlim=c(0,2),ylim=c(0,1),xlab="spot", ylab="vol",zlab="price") Module 1 Slide 35 - R: Diagrams c Helmut Strasser, May 30, 2009

36 price Call price surface vol spot Module 1 Slide 36 - Diagram c Helmut Strasser, May 30, 2009

37 Problem: Draw the graph of call price as a function of the spot price. Problem: Describe the behaviour of call price under variation of volatility and of maturity. What is the influence of the riskless rate? Problem: Give a rough estimate of the call price if the call is deep in the money or deep out of the money. Problem: Assume that the riskless rate is zero and that volatility is known. estimate the call price of a call at the money? How can we Module 1 Slide 37 - Problems c Helmut Strasser, May 30, 2009

38 Hedging interpretation: Delta Delta: S t C t = N(d 1 ) (has to be proved!) The price formula contains the information about the Hedge portfolio: C t = S t N(d 1 ) }{{} asset part Scenario 1: Option is deep in the money. C t S t Ke r(t t), = N(d 1 ) 1. Scenario 2: Option is deep out of the money. C t 0, = N(d 1 ) 0. Scenario 3: Option is at the money S t Ke r(t t). C t S t σ t 2π, = N(d 1 ) 1 2 σ t 2 2π. Ke r(t t) N(d 2 ) }{{} cash part Module 1 Slide 38 - Delta c Helmut Strasser, May 30, 2009

39 Fluctuation of Delta: The asset part of the hedge portfolio is determined by option s Delta. The sensitivity of the option s Delta with respect to the underlying is the option s Gamma: Γ = S t = 2 S 2 t Scenario 1: Option is deep in the money. C t S t Ke r(t t), = N(d 1 ) 1, Γ 0. Scenario 2: Option is deep out of the money. C t 0, = N(d 1 ) 0, Γ 0. Scenario 3: Option is at the money S t Ke r(t t). C t S t σ t C t = 1 2π exp(d 2 1/2) S t σ t 2π, = N(d 1 ) 1 2 σ t 2 2π, Γ c σ t (large!) Module 1 Slide 39 - Gamma c Helmut Strasser, May 30, 2009

40 MC: Call Delta and Gamma S=rBS(vol=0.2,rate=0.04) sp=seq(from=0.5,to=1.5,length=50) pr=sp*apply(call(1/sp)(s(10000)),2,mean)*exp(-0.04) sm=smooth.spline(sp,pr,spar=0.5) delta=function(x) predict(sm,x,deriv=1)$y plot(delta,0.5,1.5) dl0=call.bs(spot=sp,vol=0.2,rate=0.04)$delta lines(sp,dl0,col="red") gamma=function(x) predict(sm,x,deriv=2)$y plot(gamma,0.5,1.5) gm0=call.bs(spot=sp,vol=0.2,rate=0.04)$gamma lines(sp,gm0,col="red") Module 1 Slide 40 - R: MC Call Delta and Gamma c Helmut Strasser, May 30, 2009

41 MC: BinCall Delta and Gamma S=rBS(vol=0.2,rate=0.04) sp=seq(from=0.5,to=1.5,length=50) pr=apply(bincall(1/sp)(s(50000)),2,mean)*exp(-0.04) sm=smooth.spline(sp,pr,spar=0.5) delta=function(x) predict(sm,x,deriv=1)$y plot(delta,0.5,1.5) gamma=function(x) predict(sm,x,deriv=2)$y plot(gamma,0.5,1.5) Module 1 Slide 41 - R: MC BinCall Delta and Gamma c Helmut Strasser, May 30, 2009

42 Diagrams pf=function(x,y) Call.BS(spot=x,vol=y)$Delta Persp("pf(x,y)",xlim=c(0,2),ylim=c(0,1),xlab="spot", ylab="vol",zlab="delta") pf=function(x,y) Call.BS(spot=x,vol=y)$Gamma Persp("pf(x,y)",xlim=c(0,2),ylim=c(0,1),xlab="spot", ylab="vol",zlab="gamma") Module 1 Slide 42 - R: Diagrams c Helmut Strasser, May 30, 2009

43 Delta surface price vol spot Module 1 Slide 43 - Diagram c Helmut Strasser, May 30, 2009

44 price Gamma surface vol spot Module 1 Slide 44 - Diagram c Helmut Strasser, May 30, 2009

45 Problem: Draw the graph of Delta and Gamma as a function of the spot price. Problem: Describe the behaviour of Delta and Gamma under variation of volatility and of maturity. Problem: Give a rough estimate of Delta and Gamma if the call is deep in the money or deep out of the money. Problem: Assume that the riskless rate is zero and that volatility is known. estimate the Delta of a call at the money? How can we Module 1 Slide 45 - Problems c Helmut Strasser, May 30, 2009

46 Multiperiod models Basic concepts Module 1 Slide 46 - Title c Helmut Strasser, May 30, 2009

47 A market with two assets: M = (B, S) n (equidistant) periods: t 0 = t, t 1, t 2,..., t n = T (trading times) Asset prices: (B 0, S 0 ), (B 1, S 1 ),..., (B n, S n ) Trading strategy: (H B k, HS k ) number of units in the portfolio between times t k 1 and t k Value process: V = (V 0, V 1,..., V n ) Definition: A trading strategy (H B k, HS k ) is called self-financing if at every trading time t k, k = 1, 2,..., n 1 it satisfies the rebalancing condition H B k B k + H S k S k = H B k+1b k + H S k+1s k The value process of a self-financing trading strategy is called an admissible value process. Value at time t k : V k = H B k B k + H S k S k = H B k+1 B k + H S k+1 S k Module 1 Slide 47 - Self-financing trading c Helmut Strasser, May 30, 2009

48 Theorem: An admissible value process depends only on the value changes of the underlying assets. Proof: V k V k 1 = (H B k B k + H S k S k ) (H B k B k 1 + H S k S k 1 ) = H B k (B k B k 1 ) + H S k (S k S k 1 ) In other words: V n = V 0 + n Hk B (B k B k 1 ) + k=1 n Hk S (S k S k 1 ) k=1 Short notation: V is admissible iff V = H B B + H S S and dv = H B db + H S ds Module 1 Slide 48 - Self-financing trading c Helmut Strasser, May 30, 2009

49 Let V be the set of all admissible value processes. Theorem: Every linear combination of admissible value processes is an admissible value process. Definition: The market M has the no-arbitrage property (NA) if for every V V V 0 0 and V n 0 V 0 = V n Note: The same consequences are true as for the single period case. Question: How can we check whether a multiperiod model satisfies the NA-property? Answer: Later. Module 1 Slide 49 - No Arbitrage c Helmut Strasser, May 30, 2009

50 Goal: Pricing of claims Definition: An attainable claim is a random variable C which can be written as the terminal value of an admissible value process (is replicated by a self-financing trading strategy). Terminology: Replicating claims by self-financing trading strategies is called hedging. Theorem: If the market satisfies the NA property then for every attainable claim the one and only price which avoids arbitrage is the initial value of a hedging strategy. Problem: How to obtain the price of an attainable claim? Module 1 Slide 50 - Pricing of claims c Helmut Strasser, May 30, 2009

51 Example: Binomial case At every trading time t k we have: V k = H B k B k + H S k S k Therefore V k H S S k = H B k B k is riskless: Given S k 1, the right hand side is a known constant, since H B k depends only on S k 1. This implies that (given S k 1 ) V u k H S k S u k = V d k H S k S d k Hk S = V k u V k d Sk u Sd k =: k Thus, we may reconstruct the value process backwards: H B k = V k H S k S k B k, V k 1 = H B k B k 1 + H S k S k 1 Module 1 Slide 51 - Binomial case, Delta c Helmut Strasser, May 30, 2009

52 Multiperiod models Binomial trees Module 1 Slide 52 - Title c Helmut Strasser, May 30, 2009

53 Binomial tree: Stock values s k,i = u i d k i s 0 Bond values b k = e kr (Reasonable choices of u, d, r are discussed later.) Time S V 0 s 0 v 0 1 s 1,0 s 1,1 v 1,0 v 1,1 2 s 2,0 s 2,1 s 2,2 v 2, v 2,1 v 2,2 n s n,0 s n,1 s n,2... s n,n v n,0 v n,1 v n,2... v n,n If C = f(s) is a claim let v n,i := f(s n,i ) and calculate backwards: H S k := v k,i+1 v k,i s k,i+1 s k,i, H B k = v k,i H S k s k,i b k, v k 1,i = H B k b k 1 + H S k s k 1,i The initial value v 0 is the price of the claim. Module 1 Slide 53 - Binomial tree algorithm c Helmut Strasser, May 30, 2009

54 In a multiperiod binomial model every claim can be replicated. The multiperiod period binomial model is complete. Module 1 Slide 54 - Market completeness c Helmut Strasser, May 30, 2009

55 Numerical example: u = up 1/ n, d = 1/u, r = rate/n t=binomtree(n=5,up=1.06,rate=0.04,claim=call(1)) lapply(t$stock,round,3) lapply(t$value,round,3) [[1]] [[1]] [1] 1 [1] [[2]] [[2]] [1] [1] [[3]] [[3]] [1] [1] [[4]] [[4]] [1] [1] [[5]] [[5]] [1] [1] [[6]] [[6]] [1] [1] Module 1 Slide 55 - R: Binomial tree example c Helmut Strasser, May 30, 2009

56 Numerical examples: system.time(binomtree(n=100,up=1.06,rate=0.04,claim=call(1))) system.time(binomtree(n=500,up=1.06,rate=0.04,claim=call(1))) system.time(binomtree(n=1000,up=1.06,rate=0.04,claim=call(1))) Module 1 Slide 56 - R: Binomial tree efficiency c Helmut Strasser, May 30, 2009

57 Early exercise: European claim: The right to exercise at maturity American claim: The right to exercise before maturity (early exercise) 1. The value of an American claim is never smaller than the value of the corresponding European claim. 2. Early exercise is preferable whenever the intrinsic value of the claim is bigger than the option value. 3. Eary Problem is worthless for a call on a stock without dividends. The reason is that the value of the call is always larger than the intrinsic value. 4. Eary exercise can be valuable for a put on a stock without dividends. The reason is that the value of the put is sometimes smaller than the intrinsic value. Module 1 Slide 57 - Early exercise c Helmut Strasser, May 30, 2009

58 Early exercise in a binomial tree: Let v k,i be the values of the American claim. Beginning with v n,i = f(s n,i ) calculate backwards: but H S k := v k,i+1 v k,i s k,i+1 s k,i, H B k = v k,i H S k s k,i b k v k 1,i = max(f(s k 1,i ), H B k b k 1 + H S k s k 1,i ) The initial value v 0 is the price of the American claim. Module 1 Slide 58 - Binomial algorithm with early exercise c Helmut Strasser, May 30, 2009

59 Example: Early exercise in a binomial tree t=binomtree(n=30,up=1.06,rate=0.04,claim=call(1)) t=binomtree(n=30,up=1.06,rate=0.04,claim=call(1),mode="am") t=binomtree(n=30,up=1.06,rate=0.04,claim=put(1)) t=binomtree(n=30,up=1.06,rate=0.04,claim=put(1),mode="am") Module 1 Slide 59 - R: Early exercise example c Helmut Strasser, May 30, 2009

60 Binomial model as discretization: How to calibrate a binomial model to given statistical properties? Suppose that the multiperiod binomial model arises through discretization of a time interval [t, T ] such that µ(t t) E(S T S t ) = S t e growth V (log S T /S t ) = σ 2 (T t) rate of expected prices volatility How to choose u, d and p when n periods are used? (large n) Solution: u = 1/d = e σ (T t)/n, p = e µ(t t)/n d u d Module 1 Slide 60 - Discretization c Helmut Strasser, May 30, 2009

61 Details: Let u = e a, d = 1/u = e a. S k = s 0 X 1 X 2 X k where X i = E(S n ) = s 0 E(X) n = s 0 (pe a + (1 p)e a ) n { P (Xi = e a ) = p P (X i = e a ) = 1 p (X i ) independent Calibration of the growth rate of expected prices: µ(t t)/n E(X) = e p = eµ(t t)/n e a e a e a Module 1 Slide 61 - Details c Helmut Strasser, May 30, 2009

62 Calibration of volatility: Choose a such that L n := i log X i satisfies V (L n ) = σ 2 (T t). log X i = a(2y i 1) where Y i = { P (Yi = 1) = p P (Y i = 0) = 1 p) L n = a(2z n n) where Z n = i Y i is binomial distributed E(L n ) = an(2p 1) V (L n ) = 4a 2 np(1 p) Now we can adjust the jump height a in order to match volatility. In this discrete time picture the jump height seems to depend on the parameter µ. However, this is not the whole story! Module 1 Slide 62 - Details c Helmut Strasser, May 30, 2009

63 The case of large n: (n ) Since a 2 n const, we should have a c/ n. This implies (by Taylor expansion e x 1 + x + x 2 /2) p n ( µ(t t) 2c c 4 ), E(L n ) µ(t t) c2 2, V (L n) c 2 Choosing c 2 = σ 2 (T t) (i.e. a = σ (T t)/n) gives E(L n ) µ(t t) σ2 (T t), V (L n ) σ 2 (T t) 2 Note: This is the lognormal model, since L n is approximately normally distributed by the CLT. Module 1 Slide 63 - Details c Helmut Strasser, May 30, 2009

64 Solution: u = 1/d = e σ (T t)/n, p = e µ(t t)/n d u d Interpretation: Volatility determines the size of the jumps in the binomial model. Expected returns can be varied by the choice of probabilities without changing volatility or the jumps. The set of available paths of the binomial model depends only on the volatility but not on the expected returns. Hedging requires only the set of available paths but no probabilities. For the calibration of a model for pricing reasons volatility only has to be taken into account. Module 1 Slide 64 - Interpretation c Helmut Strasser, May 30, 2009

65 Numerical example: Call.BS(vol=0.2,strike=1,rate=0.04,spot=1,maturity=1)$Value CRR(n=100,vol=0.2,rate=0.04,spot=1,maturity=1,claim=Call(1)) CRR(n=1000,vol=0.2,rate=0.04,spot=1,maturity=1,claim=Call(1)) Pricing under a risk neutral lognormal model and binomial hedging render the same results. Module 1 Slide 65 - R: Example c Helmut Strasser, May 30, 2009

66 Numerical example: Monte Carlo simulation S=rBS(vol=0.2,rate=0.04,maturity=1,spot=1) mean(call(1)(s( )))*exp(-0.04) CRR(n=1000,vol=0.2,rate=0.04,spot=1,maturity=1,claim=Call(1)) Call.BS(vol=0.2,rate=0.04,maturity=1,spot=1,strike=1)$Value Module 1 Slide 66 - R: Example c Helmut Strasser, May 30, 2009

67 Multiperiod models Risk neutral models Module 1 Slide 67 - Title c Helmut Strasser, May 30, 2009

68 Question: Can we extend the concept of risk neutral models to the multiperiod case? Definition: An n-period model Q defined on [t, T ] is called risk neutral if E Q (S k+1 S 0, S 1,..., S k ) = S k e r n (r n := r(t t)/n being the riskless rate for a single period.) In words: Given the past of the asset, the expected return for the next period equals the riskless rate. E Q (S k+1 S 0 = s 0, S 1 = s 1,..., S k = s k ) = s k e r n Notation: F k := F(S 0, S 1,..., S k ) past of time k. Risk neutrality means: E Q (S k+1 F k ) = S k e r n Module 1 Slide 68 - Risk neutral models c Helmut Strasser, May 30, 2009

69 Question: Can we be sure to be able finding risk neutral probability model? Fundamental theorem: For any (multiperiod) market there exists a risk neutral model iff the NA-property holds. Proof: The proof consists of two parts. Part 1 (easy): If there is a risk neutral model then the NA-property holds. Part 2 (difficult): If the NA-property holds, then there is some risk neutral model. Module 1 Slide 69 - Fundamental theorem c Helmut Strasser, May 30, 2009

70 Are risk neutral probability models useful for pricing claims in the multiperiod case? Change of Numeraire: Use B k = e kr n = er(t k t) as numeraire Discounting: S k := S k B k, V k := V k B k, B k := B k B k = 1 Theorem: If Q is risk-neutral and (H B k, HS k ) is self-financing then E Q (V n ) = V 0 E Q (V n ) = V 0 e nr n = V r(t t) 0e Under a risk-neutral probability model the following statements are true: The expected return of a self-financing trading strategy equals the riskless rate. The initial value of an admissible value process equals the expected value of the discounted terminal value. The arbitrage-free price of an attainable claim equals the expected value of the discounted claim. Module 1 Slide 70 - Pricing rule c Helmut Strasser, May 30, 2009

71 Proof of the theorem: Since (H B k, HS k ) is self-financing we have V n = V 0 + n Hk S (S k S k 1 ) k=1 Since Q is a risk-neutral probability model the discounted prices satisfy the martingale property E Q (S k F k 1 ) = S k 1 By the redundancy rule we obtain E Q (H S k (S k S k 1 ) F k 1 ) = H S k E Q (S k S k 1 F k 1 ) = 0 By the expectation rule it follows that Problem: Fill in the details of the arguments! E Q (H S k (S k S k 1 )) = 0 Module 1 Slide 71 - Proof c Helmut Strasser, May 30, 2009

72 The multiperiod binomial model: Let d < u. Then (X 1,..., X n are independent) S k = s 0 X 1 X 2 X k where X i = { u with probability q d with probability 1 q The model satisfies the NA property (for each single period) iff d < e r < u. Risk neutral probabilities: q = er n d u d Risk neutrality: E Q (S k F k 1 ) = E Q (S k 1 X k F k 1 ) = S k 1 e r n Module 1 Slide 72 - Binomial case c Helmut Strasser, May 30, 2009

73 Log-Returns: log X i = log(u)(2y i 1) where Y i = { 1 with probability q 0 with probability 1 q S n = s 0 u 2Z n n where Z n = i Y i is a binomial random variable. Closed form pricing: Let C = f(s n ) und (V k ) V such that V n = C. Then n ( ) V 0 = e nr n E Q(f(S n )) = e nr n n f(s 0 u 2k n ) q k (1 q) n k k k=0 Module 1 Slide 73 - Log Returns c Helmut Strasser, May 30, 2009

74 Binomial model as discretization: How to calibrate a risk neutral binomial model to given statistical properties? Suppose that the multiperiod binomial model arises through discretization of a time interval [t, T ] such that µ(t t) E P (S T S t ) = s 0 e µ: V P (log S T /S t ) = σ 2 (T t) r(t t) Solution: u = 1/d = e σ (T t)/n, q = e r(t t)/n d u d Interpretation: Volatility under P determines the size of jumps. The probability models P and Q are equivalent, i.e. they are defined on the same set of available paths. growth rate of expected prices σ: volatility r: riskless rate Module 1 Slide 74 - Discretization c Helmut Strasser, May 30, 2009

75 Risk neutral lognormal model: For risk neutral binomial models the option prices obtained by hedging (binomial trees) are equal to those obtained by risk neutral pricing. The risk-neutral lognormal model is the limit of risk neutral binomial models if the number of discretization intervals goes to infinity. Therefore, prices obtained by risk neutral pricing with a lognormal model are consistent with pricing by binomial trees. Module 1 Slide 75 - Lognormal model c Helmut Strasser, May 30, 2009

76 Continuous time models Introduction Module 1 Slide 76 - Title c Helmut Strasser, May 30, 2009

77 From discrete to continuous time Binomial model: S k = S 0 exp(l k ) where σ 2 (T t) k L k = (2Y i 1) = E(L k ) + σ n i=1 ) (µ σ2 k(t t) + σz k 2 n ( Lk E(L k ) ) }{{ σ } =: Z k The increments Z k+h Z k are approximately normally distributed and centered, are independent of the past F k, have variances V (Z k+h Z k ) = h(t t)/n equal to elapsed time. Module 1 Slide 77 - From discrete to continuous c Helmut Strasser, May 30, 2009

78 The Black-Scholes model For t [0, T ] let ] S t = S 0 exp [(µ )t σ2 + σw t 2 where the increments W t W s are normally distributed and centered, are independent of the past F s, have variances V (W t W s ) = t s equal to elapsed time. A market model M = (B, S) where B t = e rt and where S t satisfies the conditions stated above is called a Black-Scholes model. Module 1 Slide 78 - Black-Scholes model c Helmut Strasser, May 30, 2009

79 Properties of the Black-Scholes model: For each fixed time t the asset prices S t follow a lognormal model. The log-returns have independent and stationary increments. Module 1 Slide 79 - Properties c Helmut Strasser, May 30, 2009

80 Continuous time models The Wiener process Module 1 Slide 80 - Title c Helmut Strasser, May 30, 2009

81 Stochastic processes Definition: A stochastic process is a family (X t ) t [0,T ] of random variables defined on a time interval [0, T ]. Path: A path of a process (X t ) t [0,T ] is a particular realization (x t ) considered as a function t x t of time. Past: The past of a process (X t ) t [0,T ] at time s is the information set F s := F(X u : u s) corresponding to all paths up to time s. Increments: The random variable X t X s, s < t, is called the increment of the process (X t ) t [0,T ] on the subinterval [s, t]. Martingale: A process (X t ) is a martingale if E(X t F s ) = X s whenever s < t. Module 1 Slide 81 - Stochastic processes c Helmut Strasser, May 30, 2009

82 Continuity and jumps: Let (x t ) t [0,T ] be a particular path of a process (X t ) t [0,T ]. If at a particular time point t the difference x t := x t x t is zero then the path is continuous at t. Otherwise it has a jump at t with jump height x t. A process is called continuous if all paths are continuous everywhere. There are no jumps. Module 1 Slide 82 - Path properties c Helmut Strasser, May 30, 2009

83 Wiener process Definition: A stochastic process (W t ) t [0,T ] is a Wiener process (Brownian motion process) if 1. it is continuous, 2. the increments satisfy W t W s N(0, t s), 3. the increments are independent of the past. Mathematical probability theory: There are Wiener processes. Problem: Show that the Wiener process is a martingale. Problem: Find E(W 2 t F s ). Problem: Find A t such that (W 2 t A t ) is a martingale. Problem: Find lim t W t t. Module 1 Slide 83 - Wiener process c Helmut Strasser, May 30, 2009

84 pplot(rpwiener()(1)) pplot(rpwiener()(20)) Module 1 Slide 84 - Computer experiments c Helmut Strasser, May 30, 2009

85 Generalized Brownian motion: X t = µt + σw t Drift: The drift of a continuous process is the average forward velocity ( lim E Xt+h X ) t F t := µ(t) h 0 h Problem: Show that the drift of a generalized Brownian motion is constant and equal to µ. What about the Wiener process? Diffusion: The diffusion coefficient of a continuous process is the average forward quadratic increment rate ( lim E (Xt+h X t ) 2 ) F t := σ 2 (t) h 0 h Problem: Show that the diffusion coefficient of a generalized Brownian motion is constant and equal to σ 2. What about the Wiener process? Module 1 Slide 85 - Drift and Diffusion c Helmut Strasser, May 30, 2009

86 X=rpWiener(mean=1,cov=0.1) pplot(x(1)) Module 1 Slide 86 - Computer experiment c Helmut Strasser, May 30, 2009

87 Black-Scholes Model ] Let (W t ) be a Wiener process and S t = S 0 exp [(µ )t σ2 + σw t. 2 The logarithms of a Black-Scholes model have drift µ σ 2 /2 and diffusion coefficient σ 2. Problem: Find E(S t ). Problem: Find E(S t F s ), s < t. Definition: A Black Scholes market model M = (B, S) is a risk neutral model if B t = e rt and µ = r. Problem: Let (B, S) be a risk neutral Black Scholes model. Show that (S t ) is a martingale. Module 1 Slide 87 - Black Scholes c Helmut Strasser, May 30, 2009

88 pplot(rpbs()(1)) pplot(rpbs()(20)) pplot(rpbs(vol=0.2,rate=0.08)(20)) Module 1 Slide 88 - Computer experiments c Helmut Strasser, May 30, 2009

89 Continuous time models Variation of paths Module 1 Slide 89 - Title c Helmut Strasser, May 30, 2009

90 Size of increments: V (W t W s ) = t s V ( Wt W s t s ) = 1 It follows: As (t s) 0 the difference W t W s shrinks of the order t s. This is remarkable: As (t s) 0 the differences (t s) are much smaller than t s. f(t) f(s) For smooth functions f(t) the ratios remain stable as (t s) 0 t s (actually tend to f (s)). For the Wiener process the ratios W t W s t s explode as (t s) 0. Module 1 Slide 90 - Size of increments c Helmut Strasser, May 30, 2009

91 Subdivisions: Subdivision of an interval [a, b]: a = t 0 < t 1 < t 2 <... < t n = b Sequence of subdivisions: t 10 < t 11 t 20 < t 21 < t 22 t 30 < t 31 < t 32 < t t n0 < t n1 < t n2 <... < t n,n 1 < t nn... Riemannian sequence of subdivisions: max i t ni t n,i 1 0 as n Module 1 Slide 91 - Subdivisions c Helmut Strasser, May 30, 2009

92 Variation of paths: Let f(t) be any function defined on an interval [a, b]. Total variation (first order): Va b (f) := lim n for any Riemannian sequence of subdivisions of [a, b]. i f(t ni ) f(t n,i 1 ) Quadratic variation: QVa b (f) := lim n for any Riemannian sequence of subdivisions of [a, b]. i f(t ni ) f(t n,i 1 ) 2 Module 1 Slide 92 - Variation c Helmut Strasser, May 30, 2009

93 pplot(as.path("sin(2*pi*t)",ngrid=5000),ylim=c(-1,1)) pplot(pvariation(as.path("sin(2*pi*t)",ngrid=5000),order=1)) pplot(pbracket(as.path("sin(2*pi*t)",ngrid=5000)),ylim=c(-1,1)) W=rpWiener(ngrid=5000) pplot(w(1)) pplot(pvariation(w(1),order=1)) pplot(pbracket(w(1))) pplot(papply("x+sin(2*pi*t)",w(1))) pplot(pbracket(papply("x+sin(2*pi*t)",w(1)))) Module 1 Slide 93 - Computer experiments c Helmut Strasser, May 30, 2009

94 Smooth functions Theorem: Let f(t) be a smooth function defined on an interval [a, b]. (1) The the total variation is V t a (f) = t a f (s) ds. (2) The quadratic variation is [f] t 0. A smooth function has finite variation and zero quadratic variation. Problem: Explain part (1) in an informal way by means of subdivisions. Wiener paths Theorem: Let (W t ) be a Wiener path. (1) The the total variation is Va t (W ). (2) The quadratic variation is [W ] t = t. A Wiener path has infinite variation, but finite quadratic variation. Problem: Explain part (2). Module 1 Slide 94 - Summary c Helmut Strasser, May 30, 2009

95 Stochastic calculus Orientation Module 1 Slide 95 - Title c Helmut Strasser, May 30, 2009

96 The problem of integration: The time value of a hedge portfolio with multiperiod trading is V n = V 0 + n Hk B (B k B k 1 ) + k=1 The terms in this expression are of the form: n Hk S (S k S k 1 ) k=1 n f(t k 1 )(g(t k ) g(t k 1 )) k=1 What happens with such expressions if the number of trading periods is large and the trading intervals are small? In mathematical terms: Consider a Riemannian sequence of subdivisions. What happens with lim n n f(t n,k 1 )(g(t nk ) g(t n,k 1 ))? k=1 Module 1 Slide 96 - Integration c Helmut Strasser, May 30, 2009

97 Notion of an integral: Let [a, b] be an interval and consider a Riemannian sequence of subdivisions. Mathematics provides very effective calculation methods for dealing with terms called integrals. lim n n f(t n,k 1 )(g(t nk ) g(t n,k 1 )) =: k=1 b a f(t) dg(t) Theoretical mathematics provides answers to the following questions: Under what circumstances do such limits (called integrals) exist, and are independent of the particular sequence of subdivisions (being well-defined)? What are the properties of such integrals, what are the rules for calculation? How can we evaluate such integrals? Module 1 Slide 97 - Notion of an Integral c Helmut Strasser, May 30, 2009

98 Interpretation of integrals: There are many different intuitive interpretations of integrals: Geometry: Area under a function graph. Physics: Accumulated distance of a moving particle. Inventory: Accumulated cost of varying inventory. Finance: Accumulated capital of a cash flow. Module 1 Slide 98 - Interpretation c Helmut Strasser, May 30, 2009

99 Mathematics provides answers which go far beyond our present needs. Scope of our integral: For which kind of functions would we like to deal with integrals lim n n f(t n,k 1 )(g(t nk ) g(t n,k 1 )) =: k=1 Concerning the function g(t): b a f(t) dg(t)? At present we are only interested in continuous functions (without jumps). However, we are interested in using both smooth functions g(t) (ordinary integrals) and stochastic paths (stochastic integrals). Concerning the function f(t): We are interested to cover both continuous functions and functions with jumps. Module 1 Slide 99 - Scope of the integral c Helmut Strasser, May 30, 2009

100 Evaluation There are two approaches for evaluating integrals. Closed form evaluation: Sometimes it is possible to evaluate an integral in terms of expressions which are simpler than an integral. t t t t Examples: 0 s ds = t2 2 0 W s dw s = W 2 Closed form evaluation makes use of the properties and calculation rules for integrals. Numerical evaluation: Integrals can be evaluated numerically based on a subdivision: b a f(t) dg(t) 2 n f(t n,k 1 )(g(t nk ) g(t n,k 1 )) k=1 This can always be carried out by a computer and is subject to a discretization error. Example: b a e t2 dt can only be evaluated numerically. Module 1 Slide Evaluation c Helmut Strasser, May 30, 2009

101 The ordinary integral: lim n n f(t n,k 1 )(g(t nk ) g(t n,k 1 )) =: k=1 b a f(t) dg(t) Theorem: The integral exists and is well-defined if g(t) is a function of finite variation (FV). For historical reasons the ordinary integral has many different names: Leibniz-Newton integral: g(t) = t, f(t) continuous with isolated jumps (since about 1700) Riemann-Stieltjes integral: g(t) FV, f(t) continuous with isolated jumps (since about 1850). Lebesgue integral: g(t) = t, f(t) rather general (since about 1900). Lebesgue-Stieltjes integral: g(t) FV, f(t) rather general (since about 1930). The mathematical existence theory (so-called: abstract measure theory) becomes the more sophisticated the more general the functions g(t) and f(t) are chosen. Module 1 Slide The ordinary integral c Helmut Strasser, May 30, 2009

102 The stochastic integral: Let (H t ) and (X t ) be stochastic processes: n H tn,k 1 (X tnk X tn,k 1 ) =: lim n k=1 b a H t dx t Such integrals are ordinary integrals if the paths of the process (X t ) are of finite variation (FV-processes). The integral goes beyond ordinary integration (is a stochastic integral) only if the paths of (X t ) are not of finite variation (e.g. the Wiener process). Wiener integral: X t = W t Wiener process, H t = f(t) general non-stochastic function (since about 1930). Ito integral: X t = W t, H t a non-anticipating process (since about 1950). Stochastic integral: X t = M t a martingale, H t a non-anticipating process (since about 1960). Module 1 Slide The stochastic integral c Helmut Strasser, May 30, 2009

103 Requirements of the stochastic integral: lim n n H tn,k 1 (X tnk X tn,k 1 ) =: k=1 b a H t dx t The existence of the stochastic integral requires two fundamental conditions. Condition on the integrator (X t ): The integrator has to be a semimartingale. Condition on the integrand (H t ): The integrand has to be non-anticipating (adapted). Details later. Module 1 Slide Requirements c Helmut Strasser, May 30, 2009

104 Stochastic calculus The ordinary integral Module 1 Slide Title c Helmut Strasser, May 30, 2009

105 Rules for integrals: ((g(t) of finite variation and continuous) lim n n f(t n,k 1 )(g(t nk ) g(t n,k 1 )) =: k=1 b a f(t) dg(t) All rules for integration can be understood intuitively by considering the discrete counterpart (the corresponding sums for subdivisions). The most simple case: General rules: b a dg = b a 1 dg = g(b) g(a) Linearity: Integrals are linear operators both in f(t) and in g(t). Truncation rule: Let a < u < v < b. b a 1 (u,v] (t) f(t) dg(t) = v Problem: Show that integrals fulfil the concatenation rule! u f(t) dg(t) Module 1 Slide General rules c Helmut Strasser, May 30, 2009

106 Smooth integrators Theorem: If g(t) is differentiable with (piecewise) continuous derivative then for every (suitable) function f(t) b Differential notation: dg(t) = g (t) dt a f(t) dg(t) = b a f(t)g (t) dt Problem: Explain the statement of the theorem intuitively by considering the discrete counterpart. Problem: Deduce b a g (t) dt = g(b) g(a). Module 1 Slide Smooth integrators c Helmut Strasser, May 30, 2009

107 The concept of an integral fun=function(t) exp(-t^2) integrate(fun,-2,3) p1=pintegral("exp(-t^2)",tlim=c(-2,3),ngrid=20) p1[length(p1)] p2=pintegral("exp(-t^2)",tlim=c(-2,3),ngrid=200) p2[length(p2)] pplot(pbind(p1,preduce(p2,p1))) n=seq(from=50,to=1000,by=50) i=numeric(0) for (k in n) i=c(i,pintegral("exp(-t^2)",tlim=c(-2,3),ngrid=k,last=true)) plot(n,i) Module 1 Slide R: Concept of integral c Helmut Strasser, May 30, 2009

108 Riemannian sums and integrals fun=function(t) exp(-t^2)*2*t integrate(fun,-2,3) q=as.path("t^2",tlim=c(-2,3),ngrid=50) p1=pintegral("exp(-t^2)",driver=q,adjust="left");p1[length(p1)] p2=pintegral("exp(-t^2)",driver=q,adjust="right");p2[length(p2)] p3=pintegral("exp(-t^2)",driver=q,adjust="");p3[length(p3)] p4=pintegral("exp(-t^2)*2*t",tlim=c(-2,3),ngrid=50,adjust="left") p4[length(p4)] p5=pintegral("exp(-t^2)*2*t",tlim=c(-2,3),ngrid=50,adjust="right") p5[length(p5)] p6=pintegral("exp(-t^2)*2*t",tlim=c(-2,3),ngrid=50,adjust="") p6[length(p6)] pplot(pbind(p1,p2,p3,p4,p5,p6)) Module 1 Slide R: Riemannian sums and integrals c Helmut Strasser, May 30, 2009

109 Riemannian sums and integrals q=as.path("t^2",tlim=c(-2,3),ngrid=500) p1=pintegral("exp(-t^2)",driver=q,adjust="left");p1[length(p1)] p2=pintegral("exp(-t^2)",driver=q,adjust="right");p2[length(p2)] p3=pintegral("exp(-t^2)",driver=q,adjust="");p3[length(p3)] p4=pintegral("exp(-t^2)*2*t",tlim=c(-2,3),ngrid=500,adjust="left") p4[length(p4)] p5=pintegral("exp(-t^2)*2*t",tlim=c(-2,3),ngrid=500,adjust="right") p5[length(p5)] p6=pintegral("exp(-t^2)*2*t",tlim=c(-2,3),ngrid=500,adjust="") p6[length(p6)] pplot(pbind(p1,p2,p3,p4,p5,p6)) Module 1 Slide R: Riemannian sums and integrals c Helmut Strasser, May 30, 2009

110 Application: Cash flow Assume that a capital K(t) is accumulated by a cash flow with rate K (t) = c(t). How does the capital evolve in time? Discrete picture: 0 = t 0 < t 1 <... < t n = T n K T = K 0 + c(t k 1 )(t k t k 1 ) k=1 Continuous picture: K T = K 0 + T 0 c(t) dt Module 1 Slide Cash flow c Helmut Strasser, May 30, 2009

111 Application: Interest rates Assume that a capital K(t) is accumulated by an interest rate r(t). How does the capital evolve in time? Discrete picture Continuous picture K(t k ) K(t k 1 ) = K(t k 1 )r(t k 1 )(t k t k 1 ) K(t) = K(t)r(t) n T K T = K 0 + K(t k 1 )r(t k 1 )(t k t k 1 ) K T = K 0 + K(t)r(t) dt K T = K 0 n k=1 k=1 ( ) 1 + r(t k 1 )(t k t k 1 ) ( T K T = K 0 exp Problem: Deduce the exponential formula from the differential equation! 0 0 ) r(t) dt Module 1 Slide Interest rates c Helmut Strasser, May 30, 2009

112 Linear differential (integral) equation: Let R(t) = r(t) which means R(T ) = T 0 r(t) dt Linear differential (integral) equation for the capital: K (t) = K(t)R (t) K T = K 0 + T 0 K(t) dr(t) dk(t) = K(t)dR(t) The closed form solution is the exponential function: ( T ) K T = K 0 exp r(t) dt 0 = K 0 e R(T ) Module 1 Slide Linear equation c Helmut Strasser, May 30, 2009

113 Application: Cash flow with interest Differential equation: dk(t) = K(t)dR(t) + c(t)dt K (t) = r(t)k(t) + c(t) Integral equation: K T = K 0 + T 0 K(t) dr(t) + T 0 c(t) dt Closed form solution: (e R(t) K(t)) = e R(t) c(t) K T = K 0 e R(T ) + T 0 e R(T ) R(t) c(t) dt Module 1 Slide Cash flow with interest c Helmut Strasser, May 30, 2009

114 Integration by parts (product rule) Let both f(t) and g(t) be smooth functions. Product rule: (f(t)g(t)) = f (t)g(t) + f(t)g (t) The corresponding rule for integrals is Integration by parts: f(b)g(b) = f(a)g(a) + = f(a)g(a) + b a b a f (t)g(t) dt + g(t) df(t) + b a b a f(t)g (t) dt f(t) dg(t) Differential form: d(fg) = g df + f dg Module 1 Slide Integration by parts c Helmut Strasser, May 30, 2009

115 Self-financing portfolios: An investor holds H B t units of a bond with price B t = e rt, and H S t units of a stock with price S t. Portfolio value: V t = H B t B t + H S t S t How does the portfolio value change with time? Assuming smoothness of the involved functions we get from integration by parts: dv t = (H B t db t + H S t ds t ) + (B t dh B t + S t dh S t ) Interpretation: The value changes if either the assets prices change or the structure of the portfolio is changed. Self-financing portfolio: The value is influenced by the asset prices only. The second part of the value expression is zero (rebalancing condition). Module 1 Slide Application c Helmut Strasser, May 30, 2009

116 Transformation rule (chain rule) Let φ(t) and g(t) be smooth functions. Chain rule: φ(g(t)) = φ (g(t))g (t) The corresponding rule for integals is the Transformation rule: φ(g(b)) = φ(g(a)) + = φ(g(a)) + b a b a φ (g(t))g (t) dt φ (g(t)) dg(t) Differential form: dφ(g) = φ (g) dg Module 1 Slide Transformation rule c Helmut Strasser, May 30, 2009

117 Exponential function: Let φ(x) = exp(x) und denote f(t) := exp(g(t)). The transformation rule implies: exp(g(t)) = exp(g(0)) + t In differential notation: df(t) = f(t) dg(t) 0 exp(g(s)) dg(s) f(t) = f(0) + t 0 f(s) dg(s) The exponential f(t) = exp(g(t)) is the solution of the linear differential equation. Module 1 Slide Application c Helmut Strasser, May 30, 2009

118 Linear differential equation p=as.path("3*t-t^2/2",tlim=c(0,5),ngrid=50) q=papply("exp(x)",p) r=1+pintegral("exp(x)",driver=p) pplot(pbind(q,r)) p=as.path("3*t-t^2/2",tlim=c(0,5),ngrid=500) q=papply("exp(x)",p) r=1+pintegral("exp(x)",driver=p) pplot(pbind(q,r)) Module 1 Slide R: Linear differential equation c Helmut Strasser, May 30, 2009

119 Stochastic calculus The stochastic integral Module 1 Slide Title c Helmut Strasser, May 30, 2009

120 Question: For which processes (X t ) and (H t ) can we define integrals n H tn,k 1 (X tnk X tn,k 1 ) =: lim n with reasonable properties? k=1 b a H t dx t Stochastic integration: Integrals with adapted integrands. Condition on the integrand (H t ): The integrand has to be left-continuous and non-anticipating (adapted), i.e. for every time t the random variable H t may depend only on the past F t (X). Examples: H t = f(t), H(t) = φ(t, X t ), H t = φ(t, X t, X t 1 ), etc. Module 1 Slide Adapted processes c Helmut Strasser, May 30, 2009