Mathematical Finance Option Pricing under the Risk-Neutral Measure Cory Barnes Department of Mathematics University of Washington June 11, 2013
Outline 1 Probability Background 2 Black Scholes for European Call/Put Options 3 Risk-Neutral Measure 4 American Options and Duality
We work in (Ω, F, P), with a 1-d Brownian Motion W t, on time interval [0, T ]. W 0 = 0 a.s. For partition 0 = t 0 < t 1,..., < t k = T, increments W tj W tj 1 independent. For s < t, increment W t W s N(0, t s). For ω Ω, the mapping t W t (ω) is continuous a.s.
Definition A filtration of F is a an increasing sequence of sub σ-algebras of F, i.e. (F t ) where F s F t for s < t. We take (F t ) to be the filtration generated by W t. All the information available at time t is the data that W s attained for 0 s t. Intuitively, we can make choices on the past and present, but we don t know the future.
Definition A filtration of F is a an increasing sequence of sub σ-algebras of F, i.e. (F t ) where F s F t for s < t. We take (F t ) to be the filtration generated by W t. All the information available at time t is the data that W s attained for 0 s t. Intuitively, we can make choices on the past and present, but we don t know the future.
Definition A stochastic process is a family (X t ) of real valued random variables indexed by time (we take t [0, T ]). It is continuous if t X t (ω) is continuous almost surely. We will construct various stochastic processes from (W t ) Definition A stochastic process (X t ) is adapted to the filtration (F t ) if for every s [0, T ], the random variable X s is F s measurable.
Definition A stochastic process is a family (X t ) of real valued random variables indexed by time (we take t [0, T ]). It is continuous if t X t (ω) is continuous almost surely. We will construct various stochastic processes from (W t ) Definition A stochastic process (X t ) is adapted to the filtration (F t ) if for every s [0, T ], the random variable X s is F s measurable.
Definition An F t -adapted stochastic process (X t ) is a martingale with respect to F t if for all 0 s t, E[X t F s ] = X s. As an example, take (W t ) on [0, T ], w.r.t. F t. Verify E[W t F s ] = E[W t W s + W s F s ] = E[W t W s F s ] + E[W s F s ] = W s.
We wish to make sense of the following object: ds t = µ t S t dt + σ t S t dw t. (S t ) stock price as an adapted continuous stochastic process Deterministic growth µ t S t, proportional to current stock price Random Brownian noise, volatility σ t S t, proportional to current stock price Both (µ t ) and (σ t ) adapted continuous processes (constants in simple case)
We wish to make sense of the following object: ds t = µ t S t dt + σ t S t dw t. (S t ) stock price as an adapted continuous stochastic process Deterministic growth µ t S t, proportional to current stock price Random Brownian noise, volatility σ t S t, proportional to current stock price Both (µ t ) and (σ t ) adapted continuous processes (constants in simple case)
How does dw t behave? Look at quadratic variation. For partition s 0 < s 1 <... < s n of interval s n s 0 = t, n 1 ( Q n = Wsj+1 W ) 2 s j t. j=0 The convergence here happens in mean-square, since E(Q n ) t and Var(Q n ) 0. Thus we write the following formal multiplication rule: dw dw = dt.
Similarly, we can show that as n, n 1 ( Wsj+1 W ) ( ) s j sj+1 s j 0, j=0 n 1 ( ) 2 sj+1 s j 0. j=0 Thus we add the following to our multiplication rules: dw dt = 0, dt dt = 0.
We make sense of the stochastic differential equation dx t = a t dt + b t dw t, where a t and b t are adapted to the Brownian filtration, by t t X t = X 0 + a s ds + b s dw s. 0 0 But what does this last term mean? Turns out it will be a martingale.
[ T As long as E 0 b2 t dt] <, we can approximate the integral with elementary functions. Take n 1 φ t = e j 1 [tj,t j+1 )(t), j=0 where e j is F tj measurable and φ has finite squared expectation. Define t 0 n 1 φ s dw s = e j (W tj+1 W tj ). j=0
Our SDE is a little bit more complicated: Need some sort of chain rule... Theorem (Ito s Formula) ds t = µ t S t dt + σ t S t dw t. Let f (t, x) have continuous second partial derivatives. Then t f (t, W t ) = f (0, W 0 ) + + 1 2 t 0 0 2 f x 2 (s, W s) ds. f t t (s, W s) ds + 0 f x (s, W s) dw s +
The SDE ds t = µ t S t dt + σ t S t dw t has solution S t = S 0 exp(x(t)), X t = t 0 σ s dw s + t 0 (µ s 1 2 σ2 s) ds. To verify this we differentiate f (X t ) where f (x) = S 0 e x using Ito s formula.
ds t = df (X t ) = f (X t ) dx t + 1 2 f (X t ) dx t dx t = S t dx t + 1 2 S t dx t dx t = S t (σ t dw t + (µ t 1 2 σ2 t ) dt + 1 2 S t ) + ( σ t dw t + (µ t 1 ) 2 2 σ2 t ) dt = µ t S t dt + σ t S t dw t.
Black Scholes for European Call/Put Options
European style call option is the right to purchase S for K at T Strike Price K Time to maturity T Underlying stock price at t = 0, denoted S 0 Payoff (S T K ) + = max{s T K, 0} Put option is the right to sell, payoff (K S T ) +
Market assumptions/characterizations: Single underlying stock with geometric brownian motion Stock has expected growth µ t and volatility σ t No taxes, transaction costs or bid-ask spread Can borrow/lend at risk free rate r f No arbitrage
The goal is to construct f (t, S t ) that measures the value of the option at time t, based on the stock price path. We do this with a hedge. Purchase some amount,, of the underlying stock and sell one option. The value of this portfolio at time t is then Π t = S t f (t, S t ). Over an infinitesimal time window dt, the portfolio s value changes dπ t = ds t df (t, S t ). Use Ito s formula to track this change.
The goal is to construct f (t, S t ) that measures the value of the option at time t, based on the stock price path. We do this with a hedge. Purchase some amount,, of the underlying stock and sell one option. The value of this portfolio at time t is then Π t = S t f (t, S t ). Over an infinitesimal time window dt, the portfolio s value changes dπ t = ds t df (t, S t ). Use Ito s formula to track this change.
The goal is to construct f (t, S t ) that measures the value of the option at time t, based on the stock price path. We do this with a hedge. Purchase some amount,, of the underlying stock and sell one option. The value of this portfolio at time t is then Π t = S t f (t, S t ). Over an infinitesimal time window dt, the portfolio s value changes dπ t = ds t df (t, S t ). Use Ito s formula to track this change.
( f dπ t = ds t t ( f = ds t = ( f S t f dt + ds t + 1 2 f S t 2 S 2 t ds 2 t ) ) σ 2 St 2 dt f dt + ds t + 1 2 f S t 2 St 2 ( f t + 1 2 ) f σ 2 St 2 dt. 2 t ) ds t S 2 t The only Brownian randomness comes from ds t terms. Take = f S t. Continuously update number of stocks to form a Delta Hedge.
With = f S t stocks in the portfolio, Π t grows deterministically. Because there is no arbitrage, Π t must grow at the risk free rate: dπ t = rπ t dt. Substitute previous expressions and obtain the Black-Scholes PDE: r f = f t + r f S t + 1 2 f σ 2 St 2. S t 2 Notice the lack of µ in this PDE! S 2 t
With = f S t stocks in the portfolio, Π t grows deterministically. Because there is no arbitrage, Π t must grow at the risk free rate: dπ t = rπ t dt. Substitute previous expressions and obtain the Black-Scholes PDE: r f = f t + r f S t + 1 2 f σ 2 St 2. S t 2 Notice the lack of µ in this PDE! S 2 t
With = f S t stocks in the portfolio, Π t grows deterministically. Because there is no arbitrage, Π t must grow at the risk free rate: dπ t = rπ t dt. Substitute previous expressions and obtain the Black-Scholes PDE: r f = f t + r f S t + 1 2 f σ 2 St 2. S t 2 Notice the lack of µ in this PDE! S 2 t
With = f S t stocks in the portfolio, Π t grows deterministically. Because there is no arbitrage, Π t must grow at the risk free rate: dπ t = rπ t dt. Substitute previous expressions and obtain the Black-Scholes PDE: r f = f t + r f S t + 1 2 f σ 2 St 2. S t 2 Notice the lack of µ in this PDE! S 2 t
Boundary conditions (for call option): f (t, 0) = 0 f (T, S T ) = (S T K ) + f (t, S t ) as S t How can we solve this? rf = f t + r f S t + 1 S t 2 2 f σ 2 St 2. S 2 t
Change variables... S = Ke x, t = T 2τ σ 2, f (T, S) = K v(x, τ). Change variables again and set c = 2r/σ 2... ( v(x, τ) = exp 1 2 (c 1)x 1 ) 4 (c + 1)2 τ u(x, τ). Miraculously this reduces to u τ = 2 u x 2.
Change variables... S = Ke x, t = T 2τ σ 2, f (T, S) = K v(x, τ). Change variables again and set c = 2r/σ 2... ( v(x, τ) = exp 1 2 (c 1)x 1 ) 4 (c + 1)2 τ u(x, τ). Miraculously this reduces to u τ = 2 u x 2.
Change variables... S = Ke x, t = T 2τ σ 2, f (T, S) = K v(x, τ). Change variables again and set c = 2r/σ 2... ( v(x, τ) = exp 1 2 (c 1)x 1 ) 4 (c + 1)2 τ u(x, τ). Miraculously this reduces to u τ = 2 u x 2.
Can use the Fourier Transform to solve this! f (t, S t ) = SΦ(d + ) Ke r(t t) Φ(d ) f (t, S t ) = Ke r(t t) Φ( d ) SΦ( d + ) where Φ(x) is the normal CDF, Φ(x) = x 1 exp ( x 2 ) dx, 2π 2 (call), (put), d ± = log(s/k ) + (r ± σ2 /2)(T t) σ. T t
Risk-Neutral Measure
Suppose Z is a RV with E(Z ) = 1 and Z > 0 a.s. on (Ω, F, P). Can define P(A) = Z (ω)dp(ω), for A F. A We call Z the Radon-Nikodym derivative of P w.r.t P. Definition If Z is the R-N derivative of P w.r.t P, then Z t = E[Z F t ], 0 t T is the R-N derivative process w.r.t. F t.
Lemma The R-N derivative process is a martingale w.r.t F t. Proof. E[Z t F s ] = E[E[Z F t ] F s ] = E[Z F s ] = Z s. Lemma Let Y be an F t measurable RV. Then for 0 s t, Ẽ[Y F s ] = 1 Z s E[YZ t F s ].
Lemma The R-N derivative process is a martingale w.r.t F t. Proof. E[Z t F s ] = E[E[Z F t ] F s ] = E[Z F s ] = Z s. Lemma Let Y be an F t measurable RV. Then for 0 s t, Ẽ[Y F s ] = 1 Z s E[YZ t F s ].
Theorem (Girsanov) On (Ω, F, P) with W t and F t, let Θ t be an adapted process for 0 t T. Define t W t = W t + Θ s ds, i.e. d W t = dw t + Θ t dt. 0 Then there exists an explicit Z t that defines P for which E(Z T ) = 1 and under P the process W t is a BM.
Recall our stock process ds t = µ t S t dt + σ t S t dw t. The risk free rate can be given as a stochastic process r t, so we have discount factor ( t ) D t = exp r s ds. 0 Discount process has zero quadratic variation, so ( t t D t S t = S 0 exp σ s dw s + µ s r s 1 ) 0 0 2 σ2 s ds.
In differential form, the discounted stock process is d(d t S t ) = σ t D t S t (Θ t dt + dw t ), Θ t = µ t r t σ t. Use Girsanov with Θ t to change to P, then d(d t S t ) = σ t D t S t d W t. Discounted stock process is a martingale! From this we get discounted payoff D t V t also a martingale.
The solution to this SDE is ( t S t = S 0 exp σ s d W s + 0 t 0 r s 1 ) 2 σ2 s ds. Evaluate call option price from martingale statement: ] f (0, S 0 ) = [e Ẽ rt (S T K ) + F 0. Straightforward integral to recover Black-Scholes.
Theorem (Martingale Representation) On (Ω, F, P) with W t and F t for 0 t T, let M t be a martingale w.r.t. F t. Then there exists a unique F t adapted process Γ t such that t M t = M 0 + Γ s dw s, 0 t T. 0 We really want the existence of F t adapted Γ for martingale M under P, such that t M t = M 0 + Γs d W s. 0
General hedging problem for European style option: Let V T be F T measurable - derivative payoff Start with initial capital X 0 and form portfolio process X t Want X T = V T Call V 0 = X 0 the price of the option at time zero
Change to risk-neutral measure, get D t V t to be a martingale: t D t V t = V 0 + Γs d W s, 0 t T. 0 On the other hand, we want the portfolio process X t to be self-financing : dx t = t ds t + r t (X t t S t ) dt, where we hold t shares of the stock, and invest/borrow at r t to finance with X 0 initial capital.
Write this under P and get t D t X t = X 0 + s σ s D s S s d W s, 0 t T. 0 Equating this with our previous expression we should pick t D t V t = V 0 + Γs d W s, 0 t T, 0 t = Γ t σ t D t S t, 0 t T.
Theorem (Fundamental Theorem of Asset Pricing) A market has a risk-neutral measure if and only if it does not admit arbitrage. In an arbitrage free market, every derivative can be hedged if and only if the risk neutral measure is unique.
American Options and Duality
An American style option can be exercised at any time up to maturity T. Closed form solutions rarely available Approximate with discrete exercise times G = {0, 1, 2,..., T } Discounted payoff h t (S t ) if exercised at time t Discounted value V t (S t ), inherent value/price at time t Notice that V t (S t ) h t (S t ), V 0 = sup Ẽ(h τ (S τ )). τ
An American style option can be exercised at any time up to maturity T. Closed form solutions rarely available Approximate with discrete exercise times G = {0, 1, 2,..., T } Discounted payoff h t (S t ) if exercised at time t Discounted value V t (S t ), inherent value/price at time t Notice that V t (S t ) h t (S t ), V 0 = sup Ẽ(h τ (S τ )). τ
The value of an American option follows V T (S T ) = h T (S T ) V i (S i ) = max{h i (S i ), Ẽ[V i+1(s i+1 ) F i ]} Use some sort of dynamic programming algorithm to find V 0. How do you compute this continuation value? Lattice Least squares Whatever works...
The value of an American option follows V T (S T ) = h T (S T ) V i (S i ) = max{h i (S i ), Ẽ[V i+1(s i+1 ) F i ]} Use some sort of dynamic programming algorithm to find V 0. How do you compute this continuation value? Lattice Least squares Whatever works...
Consider martingale M t with M 0 = 0 and τ attaining values in G. Then by Optional Sampling, Ẽ(h τ (S τ )) = Ẽ(h τ (S τ ) M τ ) Ẽ(max k G (h k(s k ) M k )). Take inf over M and sup over τ to arrive at V 0 = sup τ Claim: This is actually an equality! Ẽ(h τ (S τ )) inf Ẽ(max (h k(s k ) M k )). M k G
Theorem There exists a martingale M t with M 0 = 0 for which V 0 = Ẽ(max k G (h k(s k ) M k )). Proof. For i = 1,..., T define N i = V i (S i ) Ẽ[V i(s i ) F i 1 ]. Then take M 0 = 0, M i = i N j, i = 1,..., T. j=1 Notice that Ẽ[N i F i 1 ] = 0. Thus M t is a martingale.
continued... With induction: V i (S i ) = max{h i (S i ), h i+1 (S i+1 ) N i+1,..., h T (S T ) N T N i+1 }. True at T since V T (S T ) = h T (S T ). Assuming true at i, Therefore V i 1 (S i 1 ) = max{h i 1 (S i 1 ), Ẽ[V i(s i ) F i 1 ] = max{h i 1 (S i 1 ), V i (S i ) N i } V 0 Ẽ[V 1(S 1 ) F 0 ] = V 1 (S 1 ) N 1 = max (h k(s k ) M k ). k=1,...,t
Duality gives a way to form lower/upper bounds: sup τ Ẽ(h τ (S τ )) = V 0 = inf Ẽ(max (h k(s k ) M k )). M k G An approximate stopping time gives a lower bound An approximate martingale gives an upper bound What constitutes a good approximation?