Stochastic Calculus February 11, / 33

Transcription

1 Martingale Transform M n martingale with respect to F n, n =, 1, 2,... σ n F n (σ M) n = n 1 i= σ i(m i+1 M i ) is a Martingale E[(σ M) n F n 1 ] n 1 = E[ σ i (M i+1 M i ) F n 1 ] i= n 2 = σ i (M i+1 M i ) + σ n 1 E[M n M n 1 F n 1 ] i= = (σ M) n 1 Called Martingale Transform Fortune at time n using strategy σ Stochastic Integral σdb is continuous time version based on martingale B(t). () Stochastic Calculus February 11, 29 1 / 33

2 Note one can also define stochastic integrals with respect to continuous martingales M(t) other than Brownian motion. The only thing that changes is that we have the quadratic variation M, M t = lim (M(t i+1 ) M(t i )) 2. () Stochastic Calculus February 11, 29 2 / 33

3 Itô s Lemma Let f (x) be twice continuously differentiable. Then df (B) = f (B)dB f (B)dt Proof First of all we can assume without loss of generality that f, f and f are all uniformly bounded, for if we can establish the lemma in the uniformly bounded case, we can approximate f by f n so that all the corresponding derivatives are bounded and converge to those of f uniformly on compact sets. () Stochastic Calculus February 11, 29 3 / 33

4 Let s = t < t 1 < t 2 < < t n = t. We have f (B(t)) f (B(s)) = = n 1 [f (B(t j+1 )) f (B(t j ))] j= n 1 f (B(t j ))(B(t j+1 ) B(t j )) j= n 1 1 ξ j [t j, t j+1 ] + 2 f (B(ξ j ))(B(t j+1 ) B(t j )) 2, j= Let the width of the partition go to zero. By definition of the stochastic integral n 1 f (B(t j ))(B(t j+1 ) B(t j )) f db. j= s () Stochastic Calculus February 11, 29 4 / 33

5 Finally we want to show that n 1 f (B(ξ j ))(B(t j+1 ) B(t j )) 2 j= s f (B(u))du It s a lot like the computation of the quadratic variation. Suppose we had f (B(t j ) inside instead of f (B(ξ j )) n 1 [ )] 2 E f (B(t j )) (B(t j+1 ) B(t j )) 2 (t j+1 t j j= = n 1 i,j= E [ f (B(t i ))X i f (B(t j ))X j ] X j =(B(t j+1 ) B(t j )) 2 (t j+1 t j ) Suppose i < j E [ f (B(t i ))X i f (B(t j ))X j ] = E [ E [ f (B(t i ))X i f (B(t j ))X j F(t j ) ]] = E [ f (B(t i ))X i f (B(t j ))E [ X j F(t j ) ]] = () Stochastic Calculus February 11, 29 5 / 33

6 Hence so n 1 [ )] 2 E f (B(t j )) (B(t j+1 ) B(t j )) 2 (t j+1 t j j= n 1 = E j= [ [ ] ] 2 (f (B(t i )) 2 (B(t j+1 ) B(t j )) 2 (t j+1 t j ) n 1 f (B(t j ))(B(t j+1 ) B(t j )) 2 L 2 f (B(u))du s j= () Stochastic Calculus February 11, 29 6 / 33

7 We still have to show n 1 j= f (B(ξ j )) f (B(t j ) (B(t j+1 ) B(t j )) 2 Taking expectation we get n 1 E[ f (B(ξ j )) f (B(t j ) (B(t j+1 ) B(t j )) 2 ] j= n 1 E[(f (B(ξ j )) f (B(t j )) 2 ] E[(B(t j+1 ) B(t j )) 4 ] j= n 1 = C E[(f (B(ξ j )) f (B(t j )) 2 ](t j+1 t j ) j= by continuity So we have proved that f (B(t) f (B(s)) = which is Itô s formula. s f (B(u))dB(u) s f (B(u))du () Stochastic Calculus February 11, 29 7 / 33

8 1 In differential notation Itô s formula reads df (B) = f (B)dB f (B)dt. The Taylor series is df (B) = n=1 1 n! f (n) (B)(dB) n. In normal calculus we would have (db) n = if n 2, but because of the finite quadratic variation of Brownian paths we have (db) 2 = dt, while still (db) n = if n 3. 2 If the function f depends on t as well as B(t), the formula is df (t, B(t)) = f f (t, B(t))dt + t x (t, B(t))dB(t) The proof is about the same as the special case above. 2 f (t, B(t))dt. x 2 () Stochastic Calculus February 11, 29 8 / 33

9 Exponential Martingales M t = e λb t 1 2 λ2 t Ito s formula df (t, B(t)) = ( f t f f )(t, B(t))dt + (t, B(t))dB(t) 2 x 2 x f (t, x) = e λx 1 2 λ2 t f t f 2 = f x 2 x = λf M t = λe λbs 1 2 λ2s db s () Stochastic Calculus February 11, 29 9 / 33

10 Geometric Brownian motion σ2 (µ S t = S e 2 )t+σb t is called Geometric (or Exponential) Brownian Motion S t so it is (Samuelson) a better model of stock prices than B t (Bachelier) µ = drift σ = volatility Ito s formula f (t, x) = e df (t, B(t)) = ( f t f f )(t, B(t))dt + (t, B(t))dB(t) 2 x 2 x σ2 (µ )t+σx f 2 t f 2 f = µf x 2 x = σf ds t = µs t dt + σs t db t Sometimes people write ds t S t = µdt + σdb t but note that ds t S t d log S t () Stochastic Calculus February 11, 29 1 / 33

11 Local time f continuous function on R + L t (x) = δ x (f (s))ds = lim ɛ (2ɛ) 1 { s t : f (s) x ɛ} 1 A (f (s))ds = A L t (x)dx f C 1 L t (x) = s i [,t]:f (s i )=x f (s i ) 1 discontinuous in t Itô s lemma applied to B t x gives Tanaka s formula for Brownian Local Time L t (x) = B t x B x In particular, L t (x) continuous in t a.s. sgn(b s x)db s () Stochastic Calculus February 11, / 33

12 But x not bounded, so no fair!!! Proof. Itô f ɛ (x) = (2ɛ) 1 1 [ ɛ,ɛ] (2ɛ) 1 { s t : B s ɛ} = f ɛ (B t ) f ɛ (B ) f ɛ(b s )db s ɛ L t (x) = B t x B x sgn(b s x)db s () Stochastic Calculus February 11, / 33

13 Feynman-Kac formula V a nice function (say bounded). u C 1,2 solves u t = 1 2 u 2 x 2 + Vu, u(, x) = u (x) u(x) exp{ x 2 /2t}dx <. Then [er t ] u(t, x) = E x V (B(s))ds u (B(t)) Proof. For s t let Z (s) = u(t s, B(s))e R s V (B(u))du. By Îto s lemma Z (t) Z () = { u + s } u er 2 x 2 + Vu s V (B(u))du ds = u x (t s, B(s))e R s V (B(u))du ds = martingale so E x [Z (t)] = E x [Z ()] () Stochastic Calculus February 11, / 33

14 1 If V = V (t, x) [er t ] u(t, x) = E x V (t s,b(s))ds u (B(t)) 2 Historical remark. Feynman s thesis was that solution u of it Schrödinger equation u t = i[ 1 2 u 2 + Vu] should have x 2 representation u(x, t) = e i R t V (fs)ds i R t 2 f 2ds u (f t ) where is supposed to be average over functions starting at x. Kac pointed out that it is rigorous if i 1 () Stochastic Calculus February 11, / 33

15 Arcsin law ξ(t) = 1 t 1 [, )(B(s))ds = the fraction of time that Brownian motion is positive up to time t a < ; P(ξ(t) a) = 2 π arcsin a a 1; 1 a > 1. Simple explanation why distribution of ξ(t) indep of t ξ(t) = 1 1 [, ) (B(ts))ds = 1 1 [, ) ( 1 t B(ts))ds = 1 1 [, ) ( B(s))ds ξ(t) d = ξ(1) () Stochastic Calculus February 11, / 33

16 Proof By Feynman-Kac if we can find a nice solution of u t = 1 2 u 2 x 2 u1 [, ) u(, x) = 1 Then and u(t, x) = E x [e R t 1 [, )(B(s))ds ] u(t, ) = 1 e at dp(ξ a) α >, φ α (x) = α u(t, x)e αt dt 1 2 φ α + (α + 1 [, ) )φ α = α { α φ α (x) = α+1 + Aex 2(α+1) + Be x 2(α+1), x, 1 + Ce x 2α + De x 2α, x. u 1 φ α 1 A = D = () Stochastic Calculus February 11, / 33

17 Proof. φ α ( ) = φ α ( + ), φ α( ) = φ α( + ) B = α 1/2 (1+α)( α+ α+1), C = 1 φ α () = (1+α) 1/2 ( α+ α+1) 1/2 α α + 1 = By Fubini s theorem this reads E 1 [ α α+ξ ] E[e tξ αe αt ]dt = α α+1 or γa dp(ξ a) = γ Looking up a table of transforms we find dp(ξ a) = 2 π 1 a(1 a) da a 1 which is the density of the arcsin distribution () Stochastic Calculus February 11, / 33

18 Stochastic differential equations σ(x, t), b(x, t) mble Definition A stochastic process X t is a solution of a stochastic differential equation dx t = b(x t, t)dt + σ(x t, t)db t, X = x on [, T ] if X t is progressively measurable with respect to F t, T b(x t, t) dt <, T σ(x t, t) 2 dt < a.s. and T X t = x + b(x s, s)ds + σ(x s, s)db s t T The main point is that σ(ω, t) = σ(x t, t), b(ω, t) = b(x t, t) () Stochastic Calculus February 11, / 33

19 If B(t) is a d-dimensional Brownian motion and f (t, x) is a function on [, ) R d which has one continuous derivative in t and two continuous derivatives in x, then Ito s formula reads df (t, B(t)) = f t (t, B(t))dt + f ((t, B(t)) db(t) + 1 f (t, B(t))dt. 2 () Stochastic Calculus February 11, / 33

20 Bessel process (d = 2) Let B t = (Bt 1, B2 t ) be 2d Brownian motion starting at, r t = B t = (Bt 1)2 + (Bt 2)2. By Ito s lemma, dr t = B1 B db1 + B2 B db This is not a stochastic differential equation. 1 B dt. () Stochastic Calculus February 11, 29 2 / 33

21 Y (t) = Let f (t, y) be a smooth function B 1 B db1 + B 2 B db2 Use Itô s lemma. Intuitively df (t, Y t ) = t fdt + y fdy y f (dy ) 2 (dy ) 2 = ( B1 B db1 + B2 B db2 ) 2 ( ) B 1 2 = (db 1 ) B1 B 2 B B 2 db1 db 2 + = dt f (t, Y t ) = f (, Y ) + + y f B1 B db1 + ( ) B 2 2 (db 2 ) 2 B ( t f y f )(s, Y s )ds y f B2 B db2 () Stochastic Calculus February 11, / 33

22 Itô s lemma dx t = σ(t, X t )db t + b(t, X t )dt f (t, X t ) = f (, X ) + Lf (t, x) = d i,j=1 d i,j=1 { } s f (s, X s ) + Lf (s, X s ) ds σ ij (s, X s ) x i f (s, X s )db j s a ij (t, x) 2 f x i x j (t, x) + d i=1 b i (t, x) f x i (t, x) d a ij = σ ik σ jk k=1 a = σσ T () Stochastic Calculus February 11, / 33

23 dr t = B1 B db1 + B2 B db B dt = dy t r t 1 dt f (t, Y t ) = f (, Y ) + + y f B1 B db1 + ( t f y f )(s, Y s )ds y f B2 B db2 In particular e λy t λ 2 t/2 is a martingale So Y t is a Brownian motion. Therefore dr t = dy t r t 1 dt is a stochastic differential equation for the new Brownian motion Y t () Stochastic Calculus February 11, / 33

24 Itô s lemma f (t, X t ) f (, X ) = Proof { } s + L f (s, X s )ds + f (s, X s ) σdb s = i = i f (t i+1, X ti+1 ) f (t i, X ti ) f t (t i, X ti )(t i+1 t i ) + f (t i, X ti ) (X ti+1 X ti ) d j,k=1 2 f x j x k (t i, X ti )(X j t i+1 X j t i )(X k t i+1 X k t i ) + higher order terms () Stochastic Calculus February 11, / 33

25 Proof continued f t t (t f i, X ti )(t i+1 t i ) t (s, X s)ds i i f (t i, X ti ) (X ti+1 X ti ) i = i + i i+1 f (t i, X ti ) ( σ(s, X s )db s ) t i i+1 f (t i, X ti ) ( b(s, X s )ds) t i 2 f t (t i, X ti )(X j t x j x i+1 X j t i )(Xt k i+1 Xt k i ) k f σdb f b ds 2 f x j x k (s, X s )a jk (s, X s )ds () Stochastic Calculus February 11, / 33

26 Proof continued To show the last convergence, ie i g(t i, X ti )(X j t i+1 X j t i )(Xt k i+1 Xt k i ) g(s, X s )a jk (s, X s )ds i+1 Z (t i, t i+1 ) = ( t i i+1 t i l i+1 σ jl (s, X s )dbs)( l t i σ jl σ kl (s, X s )ds l E[ Z (t i, t i+1 ) 2 ] = O((t i+1 t i ) 2 ) σ km (s, X s )dbs m ) m () Stochastic Calculus February 11, / 33

27 Proof continued g(t i, X ti )E[ i i+1 t i a ij (s, X s )ds] g(s, X s )a ij (s, X s )ds E[( i g(t i, X ti )Z (t i, t i+1 )) 2 ] = i,j E[g(t i, X ti )Z (t i, t i+1 )g(t j, X tj )Z (t j, t j+1 )] i < j E[E[g(t i, X ti )Z (t i, t i+1 )g(t j, X tj )Z (t j, t j+1 ) F tj ]] = i = j E[E[g 2 (t i, X ti )Z 2 (t i, t i+1 ) F ti ]] = E[g 2 (t i, X ti )E[Z 2 (t i, t i+1 ) F ti ]] = O((t i+1 t i ) 2 ) () Stochastic Calculus February 11, / 33

28 { } f (t, X t ) f (, X ) s + L f (s, X s )ds = f (s, X s ) σdb s L = 1 2 d 2 a ij (t, x) + x i x j i,j=1 M t = f (t, X t ) { } s + L d i=1 b i (t, x) x i = generator f (s, X s )ds is a martingale { } = E[f (t, X t ) f (s, X s ) u + L f (u, X u )du F s ] s = f (t, y)p(s, x, t, y)dy f (s, x) { u + L} f (u, y)p(s, x, u, y)dydu, s X s = x () Stochastic Calculus February 11, / 33

29 For any f, = f (t, y)p(s, x, t, y)dy f (s, x) { u + L} f (u, y)p(s, x, u, y)dydu s Fokker-Planck (Forward) Equation t p(s, x, t, y) = 1 2 d i,j=1 d i=1 = L yp(s, x, t, y) 2 y i y j (a i,j (t, y)p(s, x, t, y)) y i (b i (t, y)p(s, x, t, y)) lim p(s, x, t, y) = δ(y x). t s () Stochastic Calculus February 11, / 33

30 Kolmogorov (Backward) Equation s p(s, x, t, y) = d i,j=1 d i=1 = L x p(s, x, t, y) a ij (s, x) 2 p(s, x, t, y) x i x j p(s, x, t, y) b i (s, x) x i lim p(s, x, t, y) = δ(y x). s t () Stochastic Calculus February 11, 29 3 / 33

31 Proof. f (x) smooth s u = L su s < t u(t, x) = f (x) Ito s formula: u(s, X(s)) martingale up to time t u(s, x) = E s,x [u(s, X(s))] = E s,x [u(t, X(t))] = f (z)p(s, x, t, z)dz Let f n (z) smooth functions tending to δ(y z). We get in the limit that p satisfy the backward equations. () Stochastic Calculus February 11, / 33

32 Example. Brownian motion d = 1 2 L = 1 2 x 2 Forward p(s, x, t, y t = 1 2 p(s, x, t, y) 2 y 2, t > s p(s, x, s, y) = δ(y x) Backward p(s, x, t, y s = 1 2 p(s, x, t, y) 2 x 2, s < t, p(t, x, t, y) = δ(y x) () Stochastic Calculus February 11, / 33

33 Example. Ornstein-Uhlenbeck Process L = σ2 2 2 x 2 αx x Forward p(s, x, t, y t = 1 2 p(s, x, t, y) 2 y 2 + (αyp(s, x, t, y), t > s, y p(s, x, s, y) = δ(y x) Backward p(s, x, t, y s = 1 2 p(s, x, t, y) 2 x 2 αx p(s, x, t, y), s < t, x p(t, x, t, y) = δ(y x) () Stochastic Calculus February 11, / 33