The concentration of a drug in blood. Exponential decay. Different realizations. Exponential decay with noise. dc(t) dt.

Transcription

1 The concentration of a drug in blood Exponential decay C12 concentration C12 concentration dc(t) dt = µc(t) C(t) = C()e µt time in minutes time in minutes 1 2 Exponential decay with noise Different realizations C12 concentration dc(t) = µc(t)dt + σc(t)dw (t) C(t) = C() exp ( (µ σ2 )t + σw (t) ) C12 concentration dc(t) = µc(t)dt + σc(t)dw (t) C(t) = C() exp ( (µ σ2 )t + σw (t) ) time in minutes time in minutes 3 4

2 Dynamic noise Motivating example: Population Dynamics Deterministic vs stochastic models A simple population growth model: Dynamic vs measurement noise Is the system deterministic or stochastic in nature? Irrelevant! N t : a t : dn t dt size of population at time t = a t N t ; N() = N (e.g. size of a tumor or concentration of a drug in blood) relative rate of growth (or decay) at time t 5 6 If a t = a is constant: N t = N e at Stochastic extension 2 N(t) 15 1 a> Maybe a t is not completely known, but subject to some random environmental effects: 5 a t a t + noise time a< E.g. noise = σw t, W t = white noise, σ constant, a t = a constant 7 8

3 If a t = a + σw t : dn t = an t dt + σn t dw t or N t = N exp{(a σ 2 /2)t + σw t } sigma =.1 If a t = a + σw t : dn t = an t dt + σn t dw t or N t = N exp{(a σ 2 /2)t + σw t } sigma =.3 N(t) N(t) time time 9 1 dn t dt = a t N t = (a + σw t )N t = an t + σn t W t Randomization of parameters Random variation in a parameter a: a a + σ noise We can write for a zero mean noise process. N t = N + an s ds + σn s dw s } {{ }?? A stochastic process: dx t dt = b(t, X t ) + σ(t, X t ) noise 11 12

4 Natural requirements: dx t dt = b(t, X t ) + σ(t, X t ) W t (1) Let = t < t 1 < < t n = t. Discretization of equation 1: W t1 and W t2 are independent for t 1 t 2 W t is a stationary process E[W t ] = for all t X k+1 X k = b(t k, X k ) t k + σ(t k, X k )W k t k This leads us to a white noise process where X j = X(t j ), W k = W (t k ), t k = t k+1 t k Replace W k t k by V k = V k+1 V k where {V t } t is a suitable stochastic process Our discretized version becomes: V t should have stationary, independent increments with mean. If we require V t to be continuous it turns out that only one solution exists: Brownian Motion B t k 1 k 1 X k = X + b(t j, X j ) t j + σ(t j, X j ) B j j= Is there a limit when t j? If so: j= Thus V t = B t. X t = X + b(s, X s )ds + σ(s, X s )db s } {{ }?? 15 16

5 Basic properties I Basic properties II B t is a Gaussian process: B t has independent increments: For all t 1 t k the random variable Z = (B t1,..., B tk ) has a multinormal distribution, and B t1, B t2 B t1,..., B tk B tk 1 E[B t ] = B ; Var[B t ] = t. are independent for all t 1 t k We also call it a standard Wiener process: W = {W t } t, Basic properties III a Gaussian process with independent increments for which W = w.p. 1, E[W t ] =, Var[W t W s ] = t s There exists a continuous version, so we simply assume that B t is such a continuous version. for all s t. It can be shown that any continuous time stochastic process with independent increments and finite second moments E[X 2 t ] for all t, is a Gaussian process if X t is Gaussian. 19 2

6 Construction of the Itô integral We will define Let us try the usual tricks from ordinary calculus: or more generally for S T. S f(s) dw s f(s) dw s define the integral for a simple class of functions extend by some approximation procedure to a larger class of functions Assume f is a step-function of the form: f(t) = a j I {j/2 n,(j+1)/2 n }(t) j Then it will be natural to define S f(t) dw t = j a j [W tj+1 W tj ] Example: We want to calculate Problems!!! W t dw t Choose two different, but reasonable approximations: where k/2 n if S k/2 n T t k = S if k/2 n < S T if k/2 n > T f 1 (t) = j W tj I {j/2 n,(j+1)/2 n }(t) (Left end point) f 2 (t) = j W tj+1 I {j/2 n,(j+1)/2 n }(t) (Right end point) 23 24

7 But [ ] T E f 2 (t)dw t = j E[W tj+1 (W tj+1 W tj )] Then [ ] T E f 1 (t)dw t = j E[W tj (W tj+1 W tj )] = = E[W tj+1 (W tj+1 W tj )] j E[W tj (W tj+1 W tj )] j = j E[(W tj+1 W tj ) 2 ] since W t has independent increments. = j (t j+1 t j ) = T It is natural to approximate a given function f(t) by a step-function of the form: The variations of the paths of W (t) are too big to define the integral in the ordinary sense. The problem is that a Wiener process W (t) is nowhere differentiable. Worse still: the sample paths have unbounded variation on any bounded time interval. f(t) f(t j ) I {tj,t j+1}(t) j where the points t j belong to the interval [t j, t j+1 ]. Define S f(t)dw (t) = lim n j f(t j ) [W tj+1 W tj ] 27 28

8 We just saw - unlike ordinary integrals - that it makes a difference what t j we choose!!! Two useful and common choices: The Itô integral: t j = t j, the left end point. The Stratonovich integral: t j = (t j + t j+1 )/2, the mid point. Heuristics Mathematics Itô vs Stratonovich integrals 29 3 Stochastic Differential Equations The process X solves dx t = b(x t, t)dt + σ(x t, t)dw t, X = U if dx t = b(x t, t)dt + σ(x t, t)dw t, W t = W (1) t.. W (m) t X t = b : R d R + R d σ : R d R + R d m X = U X (1) ṭ. X (d) t X (i) t = U (i) + for all i = 1,, d and t >. Necessary condition: for all t > ; σ 2 = tr σσ T. b i (X s, s)ds + m j=1 { b(x s, s) + σ(x s, s) 2 }ds < T denotes transposition of vectors and matrices. σ i,j (X s, s)dw (j) s 31 32

9 The Ito integral (Ω, F, {F t }, P ) Z stochastic process adapted to the filtration {F t }: Z t is F t -measurable for all t > W Wiener process with respect to {F t } W t is F t -measurable for all t > For t > s, W t W s is independent of {F s } and N(, t s)-distributed (W = ) Random adapted step function = t < t 1 < < t n = T partition of [, T ] Z t = A i for t i t < t i+1, i =,..., n 1 where A i is an F ti -measurable random variable, i =,..., n 1 E(A 2 i ) <, i =,..., n 1 Z t dw t = n Z ti 1 (W ti W ti 1 ) i=1 How can an integral Z t dw t be defined? Z continuous adapted process It is not difficult to prove that ( ) T E Z t dw t = ( E ) 2 Z t dw t = n E(Zt 2 i 1 )(t i t i 1 ) i=1 Z s dw s is an {F t }-martingale E(Z 2 t )dt < = t (n) < t (n) 1 < < t (n) n = T a sequence of partitions of [, T ] satisfying that ( ) max t (n) j t (n) j 1 as n 1 j n Define an adapted step function by Z (n) t = Z t (n) i for t (n) i t < t (n) i+1, i =,..., n 1 Z (n) t dw t = n i=1 Z t (n) i 1 ( ) W (n) t W (n) i t i

10 The mean square limit of Z(n) t dw t as n exists and is almost surely unique. This limit defines the Ito integral Z (n) t dw t L 2 Z t dw t ( E ) 2 Z t dw t ( ) T E Z t dw t = =lim n n i=1 = ( ) ( E Z 2 t (n) t (n) i i 1 E(Z 2 t )dt Z s dw s is an {F t }-martingale t Z s dw s is continuous ) t (n) i Some names In order to define the Ito integral it is sufficient that Z t dw t ( ) T P Zt 2 dt < = 1 In general the Ito integral is a local martingale only We call a stochastic process X t for: An Itô integral if X t = X + An Itô process or a stochastic integral if X t = X + bds + σdw s }{{}}{{} drift diffusion σdw s or dx t = σdw t or dx t = bdt + σdw t 39 4

11 X d-dimensional. Itô s formula: dx t = b(x t, t)dt + σ(x t, t)dw t, g : R d R + R X = U g(x, t) continuously differentiable w.r.t. t and twice continuously differentiable w.r.t. x Y t = g(x t, t) dy t = d i=1 g (X t, t)dx (i) t + g x i t (X t, t)dt m l=1 i,j=1 d 2 g σ i,l (X t, t)σ j,l (X t, t) (X t, t)dt x i x j Let d = 1 in The Itô formula dx t = b(t, X t )dt + σ(t, X t )dw t Let g(x, t) be twice continuously differentiable on R R +. Then Y t = g(x t, t) is again an Itô process, and { g dy t = t (X t, t) + 1 } 2 σ2 2 g x 2 (X t, t) dt + g x (X t, t)dx t Calculate Example: Choose X t = W t and g(x, t) = 1 2 x2. Then W s dw s Y t = g(w t, t) = 1 2 W 2 t Apply Itô s formula: { g dy t = t (X t, t) σ2 2 g { = + 1 } dt + W t dw t 2 } x 2 (X t, t) dt + g x (X t, t)dx t Hence or Finally ( ) 1 dy t = d 2 W t W 2 t = 1 2 t + = 1 2 dt + W tdw t W s dw s. W s dw s = 1 2 W 2 t 1 2 t

12 Example: Let us return to the population growth model dn t = an t dt + σn t dw t Choose X t = N t and g(x, t) = log x. Apply Itô s formula: { g dy t = d(log N t ) = t (X t, t) + 1 } 2 σ2 (X t, t) 2 g x 2 (X t, t) dt + g x (X t, t)dx t { = σ2 Nt 2 ( 1 } Nt 2 ) dt + 1 dn t N t Hence Finally log N t = log N + (a 12 ) σ2 t + σw t N t = N exp {(a 12 ) } σ2 t + σw t = 1 2 σ2 dt + 1 (an t dt + σn t dw t ) N t = (a 12 ) σ2 dt + σdw t (Corresponding Stratonovich solution is N t = N exp {at + σw t }) Exercise: Show that e Ws dw s = e Wt e W 1 2 e Ws ds Hint: apply Itô s formula with X t = W t and g(x, t) = e x 47