(B t, S (t) t P AND P,..., S (p) t ): securities P : actual probability P : risk neutral probability Realtionship: mutual absolute continuity P P For example: P : ds t = µ t S t dt + σ t S t dw t P : ds t = µ t S t dt + σ t S t dw t Money market bond numeraire: µ t = r t Q : σ t =?? A : σ 2 t = σ 2 t. Why? Autumn 25 1 Per A. Mykland
CONTINUOUS CASE: σ t AND σ t P : P : ds t = µ t S t dt + σ t S t dw t ds t = µ t S t dt + σ t S t dw t P : d log S t =... dt + σ t dw t >: (d log S t ) 2 2 = σt (dw t ) 2 = σt 2 dt d[log S, log S] t P : same argument : d[log S, log S] t = σt 2 dt Process same under P, P : σt 2 dt = d[log S, log S] t = σt 2 dt >: σ 2 t In particular: if σ is constant: σ = σ = σ 2 t If numeraire = Money Market Bond: ds t = r t S t dt + σ t S t dw t d S t = σ t S t dw t Autumn 25 2 Per A. Mykland
CONTINUOUS CASE: THE MARKET PRICE OF RISK ds t = µ t S t dt + σ t S t dw t ds t = µ t S t dt + σ t S t dw t µ t S t dt + σ t S t dw t = µ t S t dt + σ t S t dw t >: µ t dt + σ t dw t = µ t dt + σ t dw t >: µ t µ t dt + dw t = dwt σ }{{ t } λ t Change from P to P : Market price of risk dw t P BM = dw t P BM +λ t dt Autumn 25 3 Per A. Mykland
OTHER NUMERAIRE ] d log S t =... dt + σ t dw t P, P d log Λ t =... dt + γ t dv t S t = S t Λ t log S t = log S t log Λ t (d log S t ) 2 = (d log S t ) 2 + (d log Λ t ) 2 2(d log S t )(d log Λ t ) = σ 2 t (dw t ) 2 + γ 2 t (dv t ) 2 2σ t γ t (dw t )(dv t ) ρ t dt = (σ 2 t + γ 2 t 2σ t γ t ρ t )dt ρ t = 1 dt d[w, V ] t = correlation (dw, dv ) (can be random) σ 2 t = σ 2 t + γ 2 t 2σ t γ t ρ t P, P σ 2 t ex γ t = : MONEY MARKET BOND Autumn 25 4 Per A. Mykland
OTHER NUMERAIRE d log S t =... dt + σ t dw t d log Λ t =... dt + γ t dv t so d log S t =... dt + σ t + σ t dw t γ t dv t =... dt + σ t d W t ] previous page or d S t =... S t dt + σ t S t d W t = if P for numeraire Λ or σ t d W t = σ t dw t γ t dv t d W t = σ t σ t dw t γ t σ t dv t Yet another Brownian motion Autumn 25 5 Per A. Mykland
P DEPENDS ON NUMERAIRE Λ t, B t : NUMERAIRES, SUPPOSE P SAME d 1 M t dmg NEED d[m, M] t dt term M t = Λ t B t = P MG 1 M t = B t Λ t = P MG = 1 Mt 2 dm t + 1 M 3 d[m, M] t t dmg dmg?? = dmg UNIQUENESS OF DOOB-MEYER [M, M] t = M t constant if continuous Λ, B are the same Autumn 25 6 Per A. Mykland
THE RADON-NIKODYM THEOREM Theorem. Let P, Q be probabilities, Q P. Then r.v. dq dp : if E Q X <, then E Q X = E P ( X dq dp ). dq dp is unique P a.s. Proof and elaborations: Billingsley, Section 32. Uniqueness: Suppose both Y, Z satisfy conditions on dq dp. Set A = {ω : Y (ω) > Z(ω)}. Then unless P (A) =. Q(A) = E P Y I A > E P ZI A = Q(A) The finite case: F = σ(p): when ω A P dq dp (ω) = Q(A) P (A) Autumn 25 7 Per A. Mykland
CONTINUOUS DISTRIBUTIONS P (Z A) = f(z)dz Q(Z A) = Then: A dq g(z) (ω) = dp f(z). A g(z)dz. Proof: g(z) E P I(Z A) = f(z) = g(z) I(z A)f(z)dz f(z) I(z A)g(z)dz Generally: Z 1,..., Z p P (A) = Q(A) = A A = Q(A). f(z 1,..., z p )dz 1... dz p g(z 1,..., z p )dz 1... dz p dq dp (ω) = g(z 1,..., Z p ) f(z 1,..., Z p ). Autumn 25 8 Per A. Mykland
NORMAL PROCESSES WITH DRIFT P θ : X t+1 = X t + θ t + σɛ t+1 ɛ t+1 N(, 1), F t. 1 f θ (x t+1 F t ) = exp { (x t+1 X t θ t ) 2 } 2π σ 2σ 2 { } f θ (x 1,..., x t ) = (2πσ 2 ) t 2 exp 1 t 1 2σ 2 (x u+1 x u θ u ) 2 u= dp θ = f θ(x 1,..., X t ) dp f (X 1,..., X t ) { = exp 1 t 1 2σ 2 (X u+1 X u θ u ) 2 + 1 t 1 2σ 2 (X u+1 X u= u= { } 1 t 1 = exp θ σ 2 u (X u+1 X u ) 1 t 1 θ 2σ 2 u 2 u= u= { = exp Y t 1 } 2 [Y, Y ] t where: Y t = 1 σ 2 t θ udx u and θ u = θ t for t < u t + 1 Autumn 25 9 Per A. Mykland
GIRSANOV S THEOREM dx t = dw t +θ t dt P BM Q BM dq dp = exp(m T 1 2 [M, M] T ) where M t = t dp dq = exp( M T + 1 2 [M, M] T ) = exp( = exp( T T θ u dx u = P MG θ u [dw u + 1 2 θ udu] + 1 2 θ u dw u 1 2 T θ 2 udu) T = exp( M T 1 2 [ M, M] T ) where M t = Q MG as function of T θ 2 udu) t θ u dw u Autumn 25 1 Per A. Mykland
SMALL INCREMENTS, θ CONSTANT P θ : X t+ = X t + θ + σ(b t+ B t ) θ replaced by θ σ replaced by σ { dp θ θ = exp dp = exp σ 2 X N 1 2 { θ σ 2 X T 1 2 θ 2 σ 2 T θ 2 2 σ 2 } } N T = N If T fixed, N, : P θ : X t = θt + σb t BROWNIAN MOTION WITH DRIFT Form of dp θ dp : simplest case of Girsanov s Theorem. Autumn 25 11 Per A. Mykland
APPLICATION: EVALUATION OF THE EURO-RUSSIAN OPTION Recall: V = e rt E [payoff] V = e rt E f( max S t) t T ( ) = e rt E f exp( max log S t) t T ( ) = e rt E f exp( max log S t) t T ( ( )) = e rt E f exp log S + σ max (νt + t T B t ) σν = r 1 2 σ2 ( ) dp = e rt E Q f exp(log S + σ max X t) t T dq P : X t = νt + B t Q : X t is Brownian motion. dp dq = exp{νx T 1 2 ν2 T }. Problem reduced to one involving maxima of Brownian motion. Autumn 25 12 Per A. Mykland
M T = max t T X t V = e rt E Q f (S exp(σm T )) exp ( νx T 1 ) 2 ν2 T f X,M (a, b) = { } 2(2b a) exp (2b a)2 2πT 3 2T b a, (Lecture 7 p. 16 from Autumn) V = e rt f(s exp(σb)) ( exp νa 1 ) 2 ν2 T f X,M (a, b) dadb Autumn 25 13 Per A. Mykland
ADDITIVE AND MULTIPLICATIVE MARTINGALES M t IS A MARTINGALE Itô: Z t = exp(m t 1 2 [M, M] t) X t = f(x t ) dz t = f (X t )dx t + 1 2 f (X t )d[x, X] t = exp(x t ) [dx t + 1 2 d[x, X] t] dm t = Z t dm t = MG Since dx t + 1 2 d[x, X] t {}}{ = dm t 1 {}}{ 2 d[m, M] 1 t + 2 d[m, M] t = dm t In particular: EZ T = EZ = E1 = 1 Autumn 25 14 Per A. Mykland
THE DISTRIBUTION OF HITTING TIMES P : X t = νt + σw t and Q : X t = σwt dp { ν = exp dq σ 2 X t 1 ν 2 } 2 σ 2 t. t Martingale property: dp dq A special case = exp τ { ν σ 2 X τ 1 2 ν 2 } σ 2 τ. τ =inf {t : X t = b}. P (τ u) = E I {τ u} { ν = E Q I {τ u} exp σ 2 X τ u 1 ν 2 } (τ u) 2 σ2 { ν = E Q I{τ u} exp σ b 1 ν 2 } 2 2 σ τ 2 u ( ν = exp σ b 1 ν 2 ) 2 2 σ t f 2 Q,τ (t) dt Autumn 25 15 Per A. Mykland
or: P (τ u) = u f P,τ (t) = exp exp ( ν σ 2 b 1 2 ( ν σ 2 b 1 2 ν 2 ) σ 2 t f Q,τ (t) dt ν 2 ) σ 2 t f Q,τ (t). From distribution of maximum: { } b f Q,τ (t) = 2πσ2 t exp b2 3 2σ 2, t t EXAMPLE: DOWN AND IN OPTIONS { (ST K) + of min S t K t T η = otherwise X t = log S t log S b = log K log S price = E e rt η = E E [e rt η F τ ]I (τ T ) = E e rτ E [ e r(t τ) η F τ ] I (τ T ) = E e rτ BS(T τ)i (τ T ) = T e rt BS(T t)f P,τ (t) dt Autumn 25 16 Per A. Mykland