Stochastic differential equations solutions i, oksendal

Stochastic Differential Equations, Sixth Edition Solution of Exercise Problems Yan Zeng July 16, 2006 This is a solution manual for the SDE book by Øksendal, Stochastic Differential Equations, Sixth Edition It is complementary to the books own solution, and can be downloaded at www.math.fsu.edu/˜zeng If you have any comments or find any typos/errors, please email me at yz44@cornell.edu This version omits the problems from the chapters on applications, namely, Chapter 6, 10, 11 and 12 I hope I will find time at some point to work out these problems 2.8 b) Proof ∞ E[eiuBt ] = k=0 ik E[Btk ]uk = k! So E[Bt2k ] = t k k! (− ) (−1)k (2k)! = ∞ k=0 t (− )k u2k k! (2k)! k t k! · 2k d) Proof n E x [|Bt − Bs |4 ] (i) (i) E x [(Bt − Bs(i) )4 ] + = i=1 (j) E x [(Bt − Bs(i) )2 (Bt − Bs(j) )2 ] i=j 4! = n· · (t − s)2 + n(n − 1)(t − s)2 2! · = n(n + 2)(t − s)2 2.11 Proof Prove that the increments are independent and stationary, with Gaussian distribution Note for Gaussian random variables, uncorrelatedness=independence 2.15 d Proof Since Bt − Bs ⊥ Fs := σ(Bu : u ≤ s), U (Bt − Bs ) ⊥ Fs Note U (Bt − Bs ) = N (0, t − s) 3.2 Proof WLOG, we assume t = 1, then n B13 3 − B(j−1)/n ) (Bj/n = j=1 n [(Bj/n − B(j−1)/n )3 + 3B(j−1)/n Bj/n (Bj/n − B(j−1)/n )] = j=1 n n 3B(j−1)/n (Bj/n − B(j−1)/n ) (Bj/n − B(j−1)/n ) + = j=1 j=1 n 3B(j−1)/n (Bj/n − B(j−1)/n )2 + j=1 := I + II + III By Problem EP1-1 and the continuity of Brownian motion n (Bj/n − B(j−1)/n )2 ] max |Bj/n − B(j−1)/n | → a.s I≤[ 1≤j≤n j=1 To argue II → Bt2 dBt as n → ∞, it suffices to show E[ B(j−1)/n 1{(j−1)/n s, then E Mt E[eσBt−s ] |Fs = E eσ(Bt −Bs )− σ (t−s) |Fs = σ2 (t−s) = Ms e2 The second equality is due to the fact Bt − Bs is independent of Fs 4.4 Proof For part a), set g(t, x) = ex and use Theorem 4.12 For part b), it comes from the fundamental property of Itˆ o integral, i.e Itˆ o integral preserves martingale property for integrands in V Comments: The power of Itˆ o formula is that it gives martingales, which vanish under expectation 4.5 Proof t Btk = kBsk−1 dBs + k(k − 1) t Bsk−2 ds Therefore, k(k − 1) βk (t) = t βk−2 (s)ds This gives E[Bt4 ] and E[Bt6 ] For part b), prove by induction 4.6 (b) Proof Apply Theorem 4.12 with g(t, x) = ex and Xt = ct + constant coefficient n j=1 αj Bj Note n j=1 t t αj Bj is a BM, up to a 4.7 (a) Proof v ≡ In×n (b) Proof Use integration by parts formula (Exercise 4.3.), we have t Xt2 = X02 + t So Mt = X02 + t |vs |2 ds = X02 + Xs dX + |vs |2 ds Xs vs dBs + 0 Xs vs dBs Let C be a bound for |v|, then t t |Xs vs |2 ds ≤ C E E 0 t = C2 s |vu |2 du ds ≤ E t |Xs |2 ds = C 2 s vu dBu E ds C t2 So Mt is a martingale 4.12 Proof Let Yt = Y t t (n) u(s, ω)ds Then Y is a continuous {Ft }-martingale with finite variation On one hand, |Ytk+1 − Ytk |2 ≤ lim (total variation of Y on [0, t]) · max |Ytk+1 − Ytk | = = lim ∆tk →0 tk ≤t tk ∆tk →0 On the other hand, integration by parts formula yields t Yt2 = Ys dYs + Y t So Yt2 is a local martingale If (Tn )n is a localizing sequence of stopping times, by Fatou’s lemma, E[Yt2 ] ≤ lim E[Yt∧T ] = E[Y02 ] = n n So Y· ≡ Take derivative, we conclude u = 4.16 (a) Proof Use Jensen’s inequality for conditional expectations (b) T t Proof (i) Y = Bs dBs So Mt = T + Bs dBs T T T T t (ii) BT3 = 3Bs2 dBs + Bs ds = Bs2 dBs + 3(BT T − sdBs ) So Mt = Bs2 dBs + 3T Bt − t t sdBs = 3(Bs2 + (T − s) dBs (iii)Mt = E[exp(σBT )|Ft ] = E[exp(σBT − 12 σ T )|Ft ] exp( 12 σ T ) = Zt exp( 12 σ T ), where Zt = exp(σBt − 2 σ t) Since Z solves the SDE dZt = Zt σdBt , we have t Mt = (1 + t 1 Zs σdBs ) exp( σ T ) = exp( σ T ) + 2 σ exp(σBs + σ (T − s))dBs 5.1 (ii) Proof Set f (t, x) = x/(1 + t), then by Itˆ o’s formula, we have dXt = df (t, Bt ) = − dBt Xt dBt Bt dt + =− dt + (1 + t)2 1+t 1+t 1+t (iii) Proof By Itˆ o’s formula, dXt = cos Bt dBt − : Bs ∈ [− π2 , π2 ]} Then t∧τ Xt∧τ cos Bs dBs − = Xs ds 0 t∧τ − Xs2 dBs − So for t < τ , Xt = t − Xs2 dBs − t Xs ds Let τ = inf{s > t∧τ − sin2 Bs 1{s≤τ } dBs − = t Xs ds t = cos Bs 1{s≤τ } dBs − cos Bs dBs − t∧τ t = t sin Bt dt So Xt = 2 t∧τ Xs ds t∧τ Xs ds Xs ds (iv) Proof dXt1 = dt is obvious Set f (t, x) = et x, then dXt2 = df (t, Bt ) = et Bt dt + et dBt = Xt2 dt + et dBt 5.3 Proof Apply Itˆ o’s formula to e−rt Xt 5.5 (a) Proof d(e−µt Xt ) = −µe−µt Xt dt + e−µt dXt = σe−µt dBt So Xt = eµt X0 + (b) t σeµ(t−s) dBs Proof E[Xt ] = eµt E[X0 ] and t t e−µs dBs )2 + 2σe2µt X0 Xt2 = e2µt X02 + σ e2µt ( e−µs dBs 0 So t E[Xt2 ] = e−2µs ds e2µt E[X02 ] + σ e2µt t −µs e dBs since = = is a martingale vanishing at time e−2µt − −2µ 2µt e − e2µt E[X02 ] + σ 2µ e2µt E[X02 ] + σ e2µt So V ar[Xt ] = E[Xt2 ] − (E[Xt ])2 = e2µt V ar[X0 ] + σ e 2µt −1 2µ 5.6 Proof We find the integrating factor Ft by the follows Suppose Ft satisfies the SDE dFt = θt dt + γt dBt Then d(Ft Yt ) = Ft dYt + Yt dFt + dYt dFt = Ft (rdt + αYt dBt ) + Yt (θt dt + γt dBt ) + αγt Yt dt = (rFt + θt Yt + αγt Yt )dt + (αFt Yt + γt Yt )dBt (1) Solve the equation system θt + αγt = αFt + γt = 0, we get γt = −αFt and θt = α2 Ft So dFt = α2 Ft dt − αFt dBt To find Ft , set Zt = e−α t Ft , then 2 dZt = −α2 e−α t Ft dt + e−α t dFt = e−α t (−α)Ft dBt = −αZt dBt Hence Zt = Z0 exp(−αBt − α2 t/2) So 2 Ft = eα t F0 e−αBt − α t = F0 e−αBt + α t Choose F0 = and plug it back into equation (1), we have d(Ft Yt ) = rFt dt So t Yt = Ft−1 (F0 Y0 + r t eα(Bt −Bs )− α Fs ds) = Y0 eαBt − α t + r 0 5.7 (a) Proof d(et Xt ) = et (Xt dt + dXt ) = et (mdt + σdBt ) So t Xt = e−t X0 + m(1 − e−t ) + σe−t es dBs (b) (t−s) ds Proof E[Xt ] = e−t E[X0 ] + m(1 − e−t ) and t E[Xt2 ] = E[(e−t X0 + m(1 − e−t ))2 ] + σ e−2t E[ e2s ds] = e−2t E[X02 ] + 2m(1 − e−t )e−t E[X0 ] + m2 (1 − e−t )2 + σ (1 − e−2t ) Hence V ar[Xt ] = E[Xt2 ] − (E[Xt ])2 = e−2t V ar[X0 ] + 12 σ (1 − e−2t ) 5.9 Proof Let b(t, x) = log(1 + x2 ) and σ(t, x) = 1{x>0} x, then |b(t, x)| + |σ(t, x)| ≤ log(1 + x2 ) + |x| Note log(1 + x2 )/|x| is continuous on R − {0}, has limit as x → and x → ∞ So it’s bounded on R Therefore, there exists a constant C, such that |b(t, x)| + |σ(t, x)| ≤ C(1 + |x|) Also, |b(t, x) − b(t, y)| + |σ(t, x) − σ(t, y)| ≤ 2|ξ| |x − y| + |1{x>0} x − 1{y>0} y| + ξ2 for some ξ between x and y So |b(t, x) − b(t, y)| + |σ(t, x) − σ(t, y)| ≤ |x − y| + |x − y| Conditions in Theorem 5.2.1 are satisfied and we have existence and uniqueness of a strong solution 5.10 t t Proof Xt = Z + b(s, Xs )ds + σ(s, Xs )dBs Since Jensen’s inequality implies (a1 + · · · + an )p ≤ np−1 (ap1 + · · · + apn ) (p ≥ 1, a1 , · · · , an ≥ 0), we have t E[|Xt |2 ] ≤ E[|Z|2 ] + E σ(s, Xs )dBs 0 t ≤ E[|Z|2 ] + E[ t +E b(s, Xs )ds t |b(s, Xs )|2 ds] + E[ |σ(s, Xs )|2 ds] t ≤ 3(E[|Z|2 ] + C E[ t (1 + |Xs |)2 ds] + C E[ (1 + |Xs |)2 ds]) t = 2 (1 + |Xs |) ds]) 3(E[|Z| ] + 2C E[ t ≤ 3(E[|Z|2 ] + 4C E[ (1 + |Xs |2 )ds]) t ≤ 3E[|Z|2 ] + 12C T + 12C E[|Xs |2 ]ds t E[|Xs |2 ]ds, = K1 + K2 where K1 = 3E[|Z|2 ] + 12C T and K2 = 12C By Gronwall’s inequality, E[|Xt |2 ] ≤ K1 eK2 t 5.11 Proof First, we check by integration-by-parts formula, t −a + b − dYt = Set Xt = (1 − t) t dBs , 1−s dBs 1−s dt + (1 − t) dBt b − Yt = dt + dBt 1−t 1−t then Xt is centered Gaussian, with variance t E[Xt2 ] = (1 − t)2 ds = (1 − t) − (1 − t)2 (1 − s)2 So Xt converges in L to as t → Since Xt is continuous a.s for t ∈ [0, 1), we conclude is the unique a.s limit of Xt as t → 5.14 (i) Proof dZt = d(u(B1 (t), B2 (t)) + iv(B1 (t), B2 (t))) i = u · (dB1 (t), dB2 (t)) + ∆udt + i v · (dB1 (t), dB2 (t)) + ∆vdt 2 = ( u + i v) · (dB1 (t), dB2 (t)) ∂u ∂v ∂v ∂u = (B(t))dB1 (t) − (B(t))dB2 (t) + i( (B(t))dB1 (t) + (B(t))dB2 (t)) ∂x ∂x ∂x ∂x ∂v ∂v ∂u ∂u + i (B(t)))dB2 (t) = ( (B(t)) + i (B(t)))dB1 (t) + (i ∂x ∂x ∂x ∂x = F (B(t))dB(t) (ii) Proof By result of (i), we have deαB(t) = αeαB(t) dB(t) So Zt = eαB(t) + Z0 solves the complex SDE dZt = αZt dB(t) 5.15 t Proof The deterministic analog of this SDE is a Bernoulli equation dy dt = rKyt − ryt The correct substitu−2 −1 tion is to multiply −yt on both sides and set zt = yt Then we’ll have a linear equation dzt = −rKzt + r Similarly, we multiply −Xt−2 on both sides of the SDE and set Zt = Xt−1 Then − rKdt dBt dXt =− + rdt − β Xt2 Xt Xt and dZt = − dXt · dXt dXt + = −rKZt dt + rdt − βZt dBt + β Xt2 dt = rdt − rKZt dt + β Zt dt − βZt dBt Xt2 Xt3 Xt Define Yt = e(rK−β )t dYt = e(rK−β Zt , then )t (dZt + (rK − β )Zt dt) = e(rK−β )t (rdt − βZt dBt ) = re(rK−β )t dt − βYt dBt Now we imitate the solution of Exercise 5.6 Consider an integrating factor Nt , such that dNt = θt dt + γt dBt and d(Yt Nt ) = Nt dYt + Yt dNt + dNt · dYt = Nt re(rK−β )t dt − βNt Yt dBt + Yt θt dt + Yt γt dBt − βγt Yt dt Solve the equation θt = βγt γt = βNt , we get dNt = β Nt dt + βNt dBt So Nt = N0 eβBt + β d(Yt Nt ) = Nt re(rK−β Choose N0 = 1, we have Nt Yt = Y0 + Xt = Zt−1 = e(rK−β )t Yt−1 = t re(rK− β2 2 )t t t and dt = N0 re(rK− β )s+βBs e(rK−β Y0 + 2 )t )t+βBt dt ds with Y0 = Z0 = X0−1 So Nt re(rK− β 2 )s+βB s ds = e(rK− β x−1 + t )t+βBt re(rK− β )s+βB s ds 5.15 (Another solution) Proof We can also use the method in Exercise 5.16 Then f (t, x) = rKx − rx2 and c(t) ≡ β So Ft = e−βBt + β t and Yt satisfies dYt = Ft (rKFt−1 Yt − rFt−2 Yt2 )dt Divide −Yt2 on both sides, we have − dYt = Yt2 − rK + rFt−1 dt Yt So dYt−1 = −Yt−2 dYt = (−rKYt−1 + rFt−1 )dt, and d(erKt Yt−1 ) = erKt (rKYt−1 dt + dYt−1 ) = erKt rFt−1 dt Hence erKt Yt−1 = Y0−1 + r t rKs βBs − β s ds e e Xt = Ft−1 Yt = eβBt − β and erKt t Y0−1 +r t βBs +(rK− β )s ds e = e(rK− β x−1 + r )t+βBt t (rK− β )s+βBs ds e 5.16 (a) and (b) Proof Suppose Ft is a process satisfying the SDE dFt = θt dt + γt dBt , then d(Ft Xt ) = Ft (f (t, Xt )dt + c(t)Xt dBt ) + Xt θt dt + Xt γt dBt + c(t)γt Xt dt = (Ft f (t, Xt ) + c(t)γt Xt + Xt θt )dt + (c(t)Ft Xt + γt Xt )dBt Solve the equation c(t)γt + θt = c(t)Ft + γt = 0, we have γt = −c(t)Ft θt = c2 (t)F (t) So dFt = c2 (t)Ft dt − c(t)Ft dBt Hence Ft = F0 e integrating factor Ft and d(Ft Xt ) = Ft f (t, Xt )dt 10 t c2 (s)ds− t c(s)dBs Choose F0 = 1, we get desired 8.2 Proof By Kolmogorov’s backward equation (Theorem 8.1.1), it suffices to solve the SDE dXt = αXt dt + βXt dBt This is the geometric Brownian motion Xt = X0 e(α− ∞ β2 )t+βBt y2 (α− β2 )t+βy x f (xe u(t, x) = E [f (Xt )] = Then −∞ e− 2t )√ dy 2πt 8.3 Proof By (8.6.34) and Dynkin’s formula, we have E x [f (Xt )] = f (y)pt (x, y)dy Rn t = f (x) + E x [ Af (Xs )ds] t Ps Af (x)ds = f (x) + t = f (x) + ps (x, y)Ay f (y)dyds Rn Differentiate w.r.t t, we have f (y) Rn ∂pt (x, y) dy = ∂t pt (x, y)Ay f (y)dy = Rn Rn A∗y pt (x, y)f (y)dy, where the second equality comes from integration by parts Since f is arbitrary, we must have A∗y pt (x, y) ∂pt (x,y) ∂t = 8.4 Proof The expected total length of time that B· stays in F is ∞ T = E[ ∞ (Sufficiency) If m(F ) = 0, then F √ 1F (Bt )dt] = √1 e 2πt − x2t dx = for every t > 0, hence T = (Necessity) If T = 0, then for a.s t, Rn , we must have m(F ) = F F − x2 e 2t dxdt 2πt x √ e− 2t 2πt x2 dx = For such a t > 0, since e− 2t > everywhere in 8.5 Proof Apply the Feynman-Kac formula, we have u(t, x) = E x [e t ρds n f (Bt )] = eρt (2πt)− e− (x−y)2 2t f (y)dy Rn 8.6 Proof The major difficulty is to make legitimate using Feynman-Kac formula while (x − K)+ ∈ C02 For the conditions under which we can indeed apply Feynman-Kac formula to (x − K)+ ∈ C02 , c f the book of Karatzas & Shreve, page 366 19 8.7 Proof Let αt = inf{s > : βs > t}, then Xαt is a Brownian motion Since β· is continuous and limt→∞ βt = ∞ a.s., by the law of iterated logarithm for Brownian motion, we have lim sup √ t→∞ Xαβt 2βt log log βt = 1, a.s Assume αβt = t (this is true when, for example, beta· is strictly increasing), then we are done 8.8 Proof Since dNt = (u(t) − E[u(t)|Gt ])dt + dBt = dZt − E[u(t)|Gt ]dt, Nt = σ(Ns : s ≤ t) ⊂ Gt So E[u(t) − E[u(t)|Gt ]|Nt ] = By Corollary 8.4.5, N is a Brownian motion 8.9 Proof By Theorem 8.5.7, and αt = t2 , 1+ 23 t3 we have e αt αt es dBs = t αs e ˜s , where B ˜t is a Brownian motion Note eαt = αs dB + 23 t3 αt = t 8.10 Proof By Itˆ o’s formula, dXt = 2Bt dBt + dt By Theorem 8.4.3, and 4Bt2 = 4|Xt |, we are done 8.11 a) Proof Let Zt = exp{−Bt − t2 }, then it’s easy to see Z is a martingale Define QT by dQT = ZT dP , then QT is a probability measure on FT and QT ∼ P By Girsanov’s theorem (Theorem 8.6.6), (Yt )t≥0 is a Brownian motion under QT Since Z is a martingale, dQ|Ft = ZT dP |Ft = Zt dP = dQt for any t ≤ T This allows us to define a measure Q on F∞ by setting Q|FT = QT , for all T > b) ˆ is a Brownian motion, then Proof By the law of iterated logarithm, if B lim sup √ t→∞ Bt Bt = a.s and lim inf = −1, a.s t→∞ 2t log log t 2t log log t So under P , lim sup Yt = lim sup t→∞ t→∞ Bt t +√ 2t log log t 2t log log t 2t log log t = ∞, a.s Similarly, lim inf t→∞ Yt = ∞ a.s Hence P (limt→∞ Yt = ∞) = Under Q, Y is a Brownian motion The law of iterated logarithm implies limt→∞ Yt does’nt exist So Q(limt→∞ Yt = ∞) = This is not a contradiction, since Girsanov’s theorem only requires Q ∼ P on FT for any T > 0, but not necessarily on F∞ 8.12 Proof dYt = βdt + θdBt where β = u= −3 Put Mt = exp{− t 0 udBs − and θ = t −1 We solve the equation θu = β and get −2 u2 ds} = exp{3B1 (t) − B2 (t) − 5t} and dQ = MT dP on FT , ˜t with B ˜t = then by Theorem 8.6.6, dYt = θdB −3t + B(t) a Brownian motion w.r.t Q t 8.13 a) 20 Proof {Xtx ≥ M } ∈ Ft , so it suffices to show Q(Xtx ≥ M ) > for any probability measure Q which is equivalent to P on Ft By Girsanov’s theorem, we can find such a Q so that Xt is a Brownian motion w.r.t Q So Q(Xtx ≥ M ) > 0, which implies P (Xtx ≥ M ) > b) Proof Use the law of iterated logarithm and the proof is similar to that of Exercise 8.11.b) 8.15 a) Proof We define a probability measure Q by dQ|Ft = Mt dP |Ft , where t α(Bs )dBs − Mt = exp{ t α2 (Bs )ds} ∆ t ˆt = Then by Girsanov’s theorem, B Bt − α(Bs )ds is a Brownian motion So Bt satisfies the SDE dBt = ˆt By Theorem 8.1.4, the solution can be represented as α(Bt )dt + dB t x EQ [f (Bt )] = E x [exp( α(Bs )dBs − t α2 (Bs )ds)f (Bt )] Remark: To see the advantage of this approach, we note the given PDE is like Kolmogorovs backward equation So directly applying Theorem 8.1.1, we get the solution E x [f (Xt)] where X solves the SDE dXt = α(Xt)dt + dBt However, the formula E x [f (Xt)] is not sufficiently explicit if α is non-trivial and the expression of X is hard to obtain Resorting to Girsanovs theorem makes the formula more explicit b) Proof e t α(Bs )dBs − 21 t α2 (Bs )ds =e t γ(Bs )dBs − 21 t γ (Bs )ds So u(t, x) = e−γ(x) E x eγ(Bt ) f (Bt )e− = eγ(Bt )−γ(B0 )− t ( t ∆γ(Bs )ds− 21 γ (Bs )+∆γ(Bs ))ds t γ (Bs )ds c) Proof By Feynman-Kac formula and part b), v(t, x) = E x eγ(Bt ) f (Bt )e− t ( γ +∆γ)(Bs )ds = eγ(x) u(t, x) 8.16 a) Proof Let Lt = − T t n ∂h i i=1 ∂xi (Xs )dBs Then L is a square-integrable martingale Furthermore, L C01 (Rn ) | h(Xs )| ds is bounded, since h ∈ By Novikov’s condition, Mt = exp{Lt − martingale We define P¯ on FT by dP¯ = MT dP Then dXt = h(Xt )dt + dBt defines a BM under P¯ 21 T = L t } is a E x [f (Xt )] ¯ x [Mt−1 f (Xt )] = E ¯ x [e = E x = E [e t n ∂h i=1 ∂xi (Xs )dXsi − 21 t | h(Xs )|2 ds t n ∂h i=1 ∂xi (Bs )dBsi − 12 t | h(Bs )|2 ds f (Xt )] f (Bt )] Apply Itˆ o’s formula to Zt = h(Bt ), we get t n h(Bt ) − h(B0 ) = i=1 ∂h (Bs )dBsi + ∂xi So E x [f (Xt )] = E x [eh(Bt )−h(B0 ) e− t t n i=1 V (Bs )ds ∂2h (Bs )ds ∂x2i f (Bt )] b) Proof If Y is the process obtained by killing Bt at a certain rate V , then it has transition operator TtY (g, x) = E x [e− t V (Bs )ds g(Bt )] So the equality in part a) can be written as TtX (f, x) = e−h(x) TtY (f eh , x) 8.17 Proof dY (t) = dY1 (t) dY2 (t) = 1 2 β1 (t) dt + β2 (t) 2   dB1 (t)  dB2 (t) dB3 (t) So equation (8.6.17) has the form   u  1 u2 = u3 β1 (t) β2 (t) The general solution is u1 = −2u2 + β1 − 3(β1 − β2 ) = −2u2 − 2β1 + 3β2 and u3 = β1 − β2 Define Q by (8.6.19), then there are infinitely many equivalent martingale measure Q, as u2 varies 9.2 (i) Proof The book’s solution is detailed enough We only comment that for any bounded or positive g ∈ B(R+ × R), E s,x [g(Xt )] = E[g(s + t, Btx )], where the left hand side is expectation under the measure induced by Xts,x on R2 , while the right hand side is expectation under the original given probability measure P Remark: The adding-one-dimension trick in the solution is quite typical and useful Often in applications, the SDE of our interest may not be homogeneous and the coefficients are functions of both X and t However, to obtain (strong) Markov property, it is necessary that the SDE is homogeneous If we augment the original SDE with an additional equation dXt = dt or dXt = −dt, then the SDE system is an (n + 1)-dimension SDE driven by an m-dimensional BM The solution Yts,x = (Xt , Xt ) (X0 = s and X0 = x) can be identified with 22 a probability measure P s,x on Rn+1 , with P s,x = Y s,x (P ), where Y s,x (P ) means the distribution function of Y s,x With this perspective, we have E s,x [g(Xt )] = E[g(t + s, Btx )] Abstractly speaking, the (strong) Markov property of SDE solution can be formulated precisely as follows Suppose we have a filtered probability space (Ω, F, (Ft )t≥0 , P ), on which an m-dimensional continuous semimartingale Z is defined Then we can consider an n-dimensional SDE driven by Z, dXt = f (t, Xt )dZt If X x is a solution with X0 = x, the distribution X x (P ) of X x , denoted by P x , induces a probability measure on C(R+ , Rn ) The (strong) Markov property then means the coordinate process defined on C(R+ , Rn ) is a (strong) Markov process under the family of measures (P x )x∈Rn Usually, we need the SDE dXt = f (t, Xt )dZt is homogenous, i.e f (t, x) = f (x), and the driving process Z is itself a Markov process When Z is a BM, we emphasize that it is a standard BM (cf [8] Chapter IX, Definition 1.2) 9.5 a) Proof If 21 ∆u = −λu in D, then by integration by parts formula, we have −λ u, u = −λ D u2 (x)dx = 1 u(x) · u(x)dx ≤ So λ ≥ Because u is not identically zero, we must have D u(x)∆u(x)dx = − D λ > b) Proof We follow the hint Let u be a solution of (9.3.31) with λ = ρ Applying Dynkin’s formula to the process dYt = (dt, dBt ) and the function f (t, x) = eρt u(x), we get τ ∧n E (t,x) [f (Yτ ∧n )] = f (t, x) + E (t,x) Lf (Ys )ds Since Lf (t, x) = ρeρt u(x) + 21 eρt ∆u(x) = 0, we have E (t,x) [eρτ ∧n u(Bτ ∧n )] = eρt u(x) Let t = and n ↑ ∞, we are done Note ∀ξ ∈ bF∞ , E (t,x) [ξ] = E x [ξ] (cf (7.1.7)) c) Proof This is straightforward from b) 9.6 Proof Suppose f ∈ C02 (Rn ) and let g(t, x) = e−αt f (x) If τ satisfies the condition E x [τ ] < ∞, then by Dynkin’s formula applied to Y and y, we have τ E (t,x) [e−ατ f (Xτ )] = e−αt f (x) + E (t,x) ( That is, ∂ + A)g(s, Xs )ds ∂s τ E x [e−ατ f (Xτ )] = e−ατ f (x) + E x [ e−αs (−α + A)f (Xs )ds] Let t = 0, we get τ E x [e−ατ f (Xτ )] = f (x) + E x [ e−αs (A − α)f (Xs )ds] If α > 0, then for any stopping time τ , we have τ ∧n E x [e−ατ ∧n f (Xτ ∧n )] = f (x) + E x [ e−αs (A − α)f (Xs )ds] Let n ↑ ∞ and apply dominated convergence theorem, we are done 9.7 a) 23 Proof Without loss of generality, assume y = First, we consider the case x = Following the hint and note ln |x| is harmonic in R2 \{0}, we have E x [f (Bτ )] = f (x), since E x [τ ] = 21 E x [|Bτ |2 ] < ∞ If we define τρ = inf{t > : |Bt | ≤ ρ} and τR = inf{t > : |Bt | ≥ R}, then P x (τρ < τR ) ln ρ + P x (τρ > τR ) ln R = ln |x|, P x (τρ < τR ) + P x (τρ > τR ) = ln R−ln |x| ln R−ln ρ Hence ln R−ln |x| limR→∞ limρ→0 ln R−ln ρ = So P x (τρ < τR ) = P x (τ0 < ∞) = limR→∞ P x (τρ < τR ) = limR→∞ limρ→0 P x (τρ < τR ) = For the case x = 0, we have P (∃ t > 0, Bt = 0) = P (∃ > 0, τ0 ◦ θ < ∞) = P (∪ >0, ∈Q+ {τ0 ◦ θ P (τ0 ◦ θ < ∞) < ∞}) = lim = lim E [P B (τ0 < ∞)] →0 →0 z2 = lim = e− √ P z (τ0 < ∞)dz 2π →0 b) ˜t = Proof B −1 0 Bt and −1 0 ˜ is also a Brownian motion is orthogonal, so B c) Proof P (τD = 0) = lim →0 P (τD ≤ ) ≥ lim (1) P (∃ t ∈ (0, ], Bt = P (∃ t ∈ (0, ], = (2) Bt (1) →0 P (∃ t ∈ (0, ], Bt (2) ≥ 0, Bt (1) = 0) + P (∃ t ∈ (0, ], Bt = 0) + P (∃ t ∈ (0, ], (1) (2) ≥ 0, Bt (1) Bt = 0, (2) Bt (2) = 0) Part a) implies (2) ≤ 0, Bt = 0) = 0) (1) (2) And part b) implies P (∃ t ∈ (0, ], Bt ≥ 0, Bt = 0) = P (∃ t ∈ (0, ], Bt ≤ 0, Bt = 0) So (1) (2) P (∃ t ∈ (0, ], Bt ≥ 0, Bt = 0) = 21 Hence P (τD = 0) ≥ 12 By Blumenthal’s 0-1 law, P (τD = 0) = 1, i.e is a regular boundary point d) (2) Proof P (τD = 0) ≤ P (∃ t > 0, Bt = 0) ≤ P (∃ t > 0, Bt boundary point (3) = Bt = 0) = So is an irregular 9.9 a) Proof Assume g has a local maximum at x ∈ G Let U ⊂⊂ G be an open set that contains x, then g(x) = E x [g(XτU )] and g(x) ≥ g(XτU ) on {τU < ∞} When X is non-degenerate, P x (τU < ∞) = So we must have g(x) = g(XτU ) a.s This implies g is locally a constant Since G is connected, g is identically a constant 9.10 24 Proof Consider the diffusion process Y that satisfies dt dXt dYt = dt αXt dt + βXt dBt = dt + dBt αXt βXt = Let τ = inf{t > : Yt ∈ (0, T ) × (0, ∞)}, then by Theorem 9.3.3, τ K(Xs )e−ρs ds] = E (t,x) [e−ρτ φ(Xτ )] + E (t,x) [ f (t, x) T −t K(Xsx )e−ρ(s+t) ds], = E[e−ρ(T −t) φ(XTx −t )] + E[ where Xtx = xe(α− β2 )t+βBt Then it’s easy to calculate T −t f (t, x) = e−ρ(T −t) E[φ(XTx −t )] + e−ρ(s+t) E[K(Xsx )]ds 9.11 a) Proof First assume F is closed Let {φn }n≥1 be a sequence of bounded continuous functions defined on ∂D such that φn → 1F boundedly This is possible due to Tietze extension theorem Let hn (x) = E x [φn (Bτ )] ¯ and ∆hn (x) = in D So by Poisson formula, for z = reiθ ∈ D, Then by Theorem 9.2.14, hn ∈ C(D) hn (z) = 2π 2π Pr (t − θ)hn (eit )dt Let n → ∞, hn (z) → E x [1F (Bτ )] = P x (Bτ ∈ F ) by bounded convergence theorem, and RHS → 2π it 2π Pr (t − θ)1F (e )dt by dominated convergence theorem Hence P z (Bτ ∈ F ) = 2π 2π Pr (t − θ)1F (eit )dt Then by π − λ theorem and the fact Borel σ-field is generated by closed sets, we conclude P z (Bτ ∈ F ) = 2π 2π Pr (t − θ)1F (eit )dt for any Borel subset of ∂D b) Proof Let B be a BM starting at By example 8.5.9, φ(Bt ) is, after a change of time scale α(t) and under the original probability measure P, a BM in the plane ∀F ∈ B(R), P (B exits D from ψ(F )) = P (φ(B) exits upper half plane from F ) = P (φ(B)α(t) exits upper half plane from F ) = Probability of BM starting at i that exits from F = µ(F ) So by part a), µ(F ) = 2π 2π 1ψ(F ) (eit )dt = f (ξ)dµ(ξ) = R 2π 2π 2π 1F (φ(eit ))dt This implies 2π f (φ(eit ))dt = 25 2πi ∂D f (φ(z)) dz z c) Proof By change-of-variable formula, f (ξ)dµ(ξ) = R π f (ω) ∂H ∞ dω = |ω − i|2 π dx x2 + f (x) −∞ d) Proof Let g(z) = u + vz, then g is a conformal mapping that maps i to u + vi and keeps upper half plane invariant Use the harmonic measure on x-axis of a BM starting from i, and argue as above in part a)-c), we can get the harmonic measure on x-axis of a BM starting from u + iv 9.12 dXt , then the generator of Y is Aφ(y1 , y2 ) = Ly1 φ(y) + q(Xt )dt q(y1 ) ∂y∂ φ(y), for any φ ∈ C02 (Rn × R) Choose a sequence (Un )n≥1 of open sets so that Un ⊂⊂ D and Un ↑ D Define τn = inf{t > : Yt ∈ Un × (−n, n)} Then for a bounded solution h, Dynkin’s formula applied to h(y1 )e−y2 (more precisely, to a C02 -function which coincides with h(y1 )e−y2 on Un ×(−n, n)) yields Proof We consider the diffusion dYt = (1) τn ∧n (2) E y [h(Yτn ∧n )e−Yτn ∧n ] = h(y1 )e−y2 − E y (2) g(Ys(1) )e−Ys ds , since A(h(y1 )e−y2 ) = −g(y1 )e−y2 Let y2 = 0, we have τn ∧n (2) (1) h(y1 ) = E (y1 ,0) [h(Yτn ∧n )e−Yτn ∧n ] + E (y1 ,0) (2) g(Ys(1) )e−Ys ds (2) Note Yt = y2 + t q(Xs )ds ≥ y2 , let n → ∞, by dominated convergence theorem, we have h(y1 ) τD (2) = E (y1 ,0) [h(Yτ(1) )e−YτD ] + E (y1 ,0) D = E[e− τD q(Xs )ds φ(XτyD1 )] + E τD τD τD q(Xs )ds g(Xsy1 )e− s y1 q(Xu )du ds Hence h(x) = E x [e− (2) g(Ys(1) )e−Ys ds φ(XτD )] + E x g(Xs )e− s q(Xu )du ds Remark: An important application of this result is when g = 0, φ = and q is a constant, the Laplace transform of first exit time E x [e−qτD ] is the solution of Ah(x) − qh(x) = on D limx→y h(x) = y ∈ ∂D In the one-dimensional case, the ODE can be solved by separation of variables and gives explicit formula for E x [e−qτD ] For details, see Exercise 9.15 and Durrett [3], page 170 9.13 a) 26 Proof w(x) solves the ODE µw (x) + σ2 w (x) = −g(x), a < x < b; w(x) = φ(x), x = a or b 2µ x σ2 (x) = − 2g(x) on both sides, we get σ Multiply e 2µ σ2 w The first equation gives w (x) + 2µ 2µ (e σ2 x w (x)) = −e σ2 x 2µ 2µ x a So w (x) = C1 e− σ2 x − e− σ2 x 2g(x) σ2 2µ e σ2 ξ 2g(ξ) σ dξ Hence w(x) = C2 − x 2µ σ2 C e− σ x − 2µ y 2µ e− σ2 y 2µ e σ2 ξ a a 2g(ξ) dξdy σ2 By boundary condition, φ(a) = C2 − φ(b) = C2 − 2µ σ2 Let 2µ −σ σ2 2a 2µ C1 e 2µ 2µ b b − σ σ2 − a e− σ y 2µ C1 e y a 2µ e σ2 ξ 2g(ξ) σ dξdy = θ and solve the above equation, we have θ b y θ(ξ−y) g(ξ)dξdy µ a a e , −θa −θb e −e θ[φ(b) − φ(a)] + C1 = C2 = φ(a) + C1 −θa e θ b) Proof b a g(y)G(x, dy) = E x [ τD C1 g(Xt )dt] = w(x) in part a), when φ ≡ In this case, we have = = = θ2 µ(e−θa − e−θb ) θ2 µ(e−θa − e−θb ) θ2 µ(e−θa − e−θb ) b = a b y eθ(ξ−y) g(ξ)dξdy a a b b e−θy dydξ eθξ g(ξ) a ξ b eθξ g(ξ) a e−θξ − e−θb dξ θ θ g(ξ) (1 − eθ(ξ−b) )dξ, −θa µ(e − e−θb ) and b C2 = g(ξ) a e−θa (1 − eθ(ξ−b) )dξ − e−θb ) µ(e−θa 27 (2) So b g(y)G(x, dy) a C2 − C1 e−θx − θ = C1 (e−θa − e−θx ) − θ b = a g(ξ) a b a a θ 1{a : Xt ∈ K}) for all compacts K ⊂ D and all x ∈ D, then −Rα (α ≥ 0) is the inverse of characteristic operator A on Cc2 (D): (A − α)(Rα f ) = Rα (A − α)f = −f, ∀f ∈ Cc2 (D) Note when D = Rn , we get back to the resolvent equation in B Application of diffusions to obtaining formulas The following is a table of computation tricks used to obtain formulas: BM w/o drift general diffusion, esp BM with drift Distribution of first passage time reflection principle Girsanovs theorme Exit probability P (τa < τb ), P (τb < τa ) BM as a martingale Dynkins formula / boundary value problems Expectation of exit time Wt2 − t is a martingale Dynkins formula / boundary value problems Laplace transform of first passage time exponential martingale Girsanovs theorem Laplace transform of first exit time exponential martingale FK formula for boundary value problems 33 ... L Gong Introduction to stochastic differential equations Second edition Peking University Press, Beijing, 1995 [6] S W He, J G Wang and J A Yan Semimartingale theory and stochastic calculus Science... stochastic calculus Science Press, Beijing; CRC Press, Boca Raton, 1992 [7] B Øksendal Stochastic differential equations: An introduction with applications Sixth edition SpringerVerlag, Berlin, 2003... E x [u(BτD )] = u(y)µxD (dy) = ∂D u(y)σ(dy) ∂D c) Proof See, for example, Evans: Partial Differential Equations, page 26 7.8 a) Proof {τ1 ∧ τ2 ≤ t} = {τ1 ≤ t} ∪ {τ2 ≤ t} ∈ Nt And since {τi ≥