MA8109 Stochastic Processes in Systems Theory Autumn 2013

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1 Norwegian University of Science and Technology Department of Mathematical Sciences MA819 Stochastic Processes in Systems Theory Autumn MA819 Exam 23, problem 3b This is a linear equation of the form dx t = p t X t dt + q t db t, which is generally solvable using an integrating factor: h t dx t = d h t X t h t X t dt = h t dx t + h t tdb t. t Choose h such that h t = h t /t, e.g. h t = t 1. Then Xt d = db t t so that X t t The solution may then be written 1 1 = B t B 1. X t = t 1 + B t B 1 = t B 1,1 t, t 1. Here, B1,1 t means a regular BM starting at x = 1 for t = 1. 2 i This is a system of equations, which may be written out as 1 dx 1 t = dt + db 1 t, dx 2 t = X 1 t db 2 t. Let us assume that the initial values are given by X 1 = x 1 and X 2 = x 2, and as usual, B 1 = B 2 =, where B 1 and B 2 are independent standard Brownian motions. We observe that the first equation is completely independent of X 2 and B 2, and may therefore be solved right away, 2 X 1 t = x 1 + t + B 1 t. This gives us the second equation as 3 dx 2 t = [x 1 + t + B 1 t] db 2, which may also be solved by a simple integration, 4 X 2 t = x 2 + = x 2 + x 1 B 2 t + x 1 + s + B 1 s db 2 s s + B 1 s db 2 s, October 31, 213 Page 1 of 9

2 and this is the solution stated in B.Ø. There are several alternate forms of the solution. First of all, 5 see, e.g. Exercise 3.1. Thus, sdb 2 s = tb 2 t 6 X 2 t = x 2 + x 1 + t B 2 t B 2 s ds + B 2 s ds It is also possible to apply the product formula Exercise 4.3, B 1 s db 2 s. 7 d [X 1 t B 2 t] = X 1 t db 2 t + B 2 t dx 1 t + dx 1 t db 2 t. The last term is, according to the rules, equal to, 8 dx 1 t db 2 t = dt + db 1 t db 2 t =. Thus, 9 dx 2 = d [X 1 t B 2 t] B 2 t dx 1 t, leading to 1 X 2 t = x 2 + X 1 t B 2 t B 2 s ds + Show that this solution is the same as the one above! ii and iii Both these equations are of the form 11 dx t = p t X t dt + q t db t B 2 s db 1 s. and may be solved by introducing an integration factor, as discussed in the lecture. We multiply the equation by a function h t and use Itô s Formula, 12 d [h t X t ] = dh dt t X tdt + h t dx t, so that the equation becomes 13 d [h t X t ] dh dt t X tdt = h t p t X t dt + h t q t db t. The idea is then to choose h so that 14 dh dt and the resulting equation becomes = h t p t, 15 d h t X t = h t q t db t, with the solution 16 X t = X h + h s q s. h t October 31, 213 Page 2 of 9

3 For Exercise ii, Eqn. 14 becomes 17 dh dt = h and then 18 h t = e t, as stated in the book. The solution is then 19 X t = X h + e s e t = X e t + Similarly, for Exercise iii, the equation for h t is 2 dh dt = h, and h t = e t. From Eqn. 16 we then obtain e t s. 21 X t = X + es e s e t = X e t + e t B t, assuming B =. 3 Øksendal 5:5.5 a We multiply X t by the integrating factor e µt, and then employ Itô s formula: de µt X t = de µt X t + e µt dx t + de µt dx t = µe µt X t dt + e µt µx t dt + σdb t µe µt dtµx t dt + σdb t = σe µt db t. Hence e µt X t = X + σe µs, and X t = e µt X + σe µt s. By Ito s formula with fs, x = e µt s x integration by parts, we can simplify further since e µt s = B t + µ b For the stochastic process above, we have e µt s B s ds. EX t = Ee µt X + E σe µt s = e µt EX, since the Itô integral is a martingale when the integrand belongs to V[, t] ok since e µt s 2 ds < for any t >. October 31, 213 Page 3 of 9

4 If we assume X is independent of the Brownian Motion B t, V arx t = V are µt X + σe µt s = e 2µt V arx + V ar σe µt s 2 = e 2µt V arx + E σe µt s = e 2µt V arx + E σ 2 e 2µt s ds = e 2µt V arx + σ2 2µ e2µt 1. 4 It is stated in the exercise that also this equation may be solved by an integrating factor multiplying both sides, 22 F t dy t = F t rdt + αy t db t. However, here the dependent variable Y t is in the db t -term, and the factor is already given in the problem. Let us, nevertheless, try to find the integration factor directly by assuming that it is of the form 23 F t = f t, B t. From Itô s Formula we have 24 df t = f t f 2 x 2 dt + f x db t, and the idea for the integrating factor is again to make use of the product formula, 25 d F t Y t = F t dy t + Y t df t + df t dy t. In the present case, [ f 26 df t dy t = t f 2 x 2 dt + f ] x db t rdt + αy t db t = f x αy tdt Inserting this and the equation itself into Eqn. 25 leads to f d F t Y t = f rdt + αy t db t + Y t t f 2 x 2 dt + f x db f t + αy t x dt f = fr + Y t t + α f x f 2 x 2 dt + fα + f 27 Y t db t x The equation would be easy to solve if we were able to find some function f such that 28 fα + f x =, f t + α f x f 2 x 2 =. October 31, 213 Page 4 of 9

5 Any such function will do, and the first equation is always satisfied for 29 f t, x = g t e αx. Putting this into the second leads to 3 which is satisfied for dg dt + α α2 g = dg dt 1 2 α2 g =, 31 g t = e α2 t/2. We may therefore use the integrating factor α 2 32 F t = exp 2 t αb t, which turns out to be the same as the one given in the book. Eqn. 27 now simplifies to 33 d F t Y t = rf t dt, with the solution 34 F t Y t F Y = or 35 Y t = exp α2 2 t + αb t Y + rf s ds, 5 Øksendal By setting Y t = X t m, the equation is turned into dx t = m X t dt + σdb t dy t = Y t dt + σdb t. Here the integrating factor is ht = e t since and thus or α 2 r exp 2 s αb s ds d e t Y t e t Y t dt = e t dy t = e t Y t dt + σe t db t, e t Y t Y = σ e s X t = Y t + m = m + X m e t + σ e s t Since the Itô integral has expectation and variance given by the Itô isometry, we have, for X independent of B t, EX t = m + EX m e t, Var X t = e 2t Var X + σ 2 e 2s t ds = e 2t Var X + σ2 2 We note that for t, EX t m and Var X t σ e 2t. October 31, 213 Page 5 of 9

6 6 Øksendal 5:5.9 We use Theorem in Øksendal to conclude that there is a unique strong solution of { dx t = ln1 + Xt 2 dt + χ 36 {Xt>}X t db t t, ω [, R Z := X = a ω R, where a R and χ is the characteristic function. We set bt, X t = ln1 + X 2 t and σt, X t = χ {Xt>}X t, and check below that the assumptions A1 A3 of the theorem are satisfied. A1 bt, X t and σ are Lipschitz continuous: By the Mean Value Theorem bt, x bt, y x y = ln1 + x 2 ln1 + y 2 x y MVT = d dx ln1 + x2 x=ξ = max 2x x R 1 + x 2 1, and σt, x σt, y = χ {x>} x χ {y>} y x y. Since b, σ are independent of t, measurability follows from continuity in x. A2 Since b, σ are independent of t, linaer growth is a direct consequence of the Lipschitz estimate A1: By adding and subtracting terms and using the Lipschitz result, bt, x bt, + bt, x bt, = + 1 x, and σt, x σt, + σt, x σt, = + 1 x A3 Z = a is in L 2 since E a 2 = a 2 <. Moreover, a is independent of F m = F {Bs:s } since a is a constant constants are independent of any stochastic variable. 7 Øksendal 5:5.1 A slightly simpler argument than suggested in the hint. Suppose that b, σ, Z are as in theorem Then the stochastic differential equation dx t = bt, X t dt + σt, X t db t has a unique strong solution X t for t [, T ] satisfying X = Z. We wish to find a bound on E X t 2. By the Ito s formula we have d X t 2 = 2 X t, dx t + dx t, dx t = 2 X t, bt, X t dt + 2 X t, σt, X t db t + tr σt, X t T σt, X t dt = 2 X t, bt, X t + σt, X t 2 dt + 2 X t, σt, X t db t, whence [ ] E X t 2 = E Z 2 + E 2 X s, bs, X s + σs, X s 2 ds. October 31, 213 Page 6 of 9

7 Now, by the assumptions, we have 2 X s, bt, X s + σt, X s 2 X s 2 + bs, X s 2 + σs, X s 2 X s 2 + bs, X s + σs, X s 2 X s 2 + C X s 2 X s 2 + 2C X s 2 = 2C C 2 X s 2, and so E X t 2 E Z 2 + 2C 2 t C 2 Grönwall s lemma now implies that E X t 2 E Z 2 + 2C 2 te 1+2C2 t. E X s 2 ds. 8 The equation has a form we have not really addressed, but by writing dy t = b Y t 1 t dt + db t, 37 X t = b Y t, we obtain the linear SDE 38 dx t = 1 t 1 X tdt db t, X = b a. This equation may, as before, be solved by an integrating factor when < t < 1. The solution is 39 X t = X h + h s 1, h t where h satisfies 4 dh dt = e.g. h t = 1 t 1. Then 41 or h t 1, X t = t 1 1 = b a 1 t 1 t s 1 42 Y t = b X t = a 1 t + bt + 1 t which is the solution stated in the book. 1 s 1 s, October 31, 213 Page 7 of 9

8 This solves the equation for us, although the exercise only requires that this solution is verified. Putting the solution into the left hand side of the equation gives LHS = dy t = a + b dt dt 1 s + 1 t 1 1 t db t Similarly, for right hand side, 45 = a + b dt + db t dt RHS = b Y b t 1 t dt + db t = and LHS = RHS. = b a dt + db t dt 1 s. a 1 t + bt + 1 t 1 s, 1 t The limit when t 1 is not at all obvious, and requires that 46 lim 1 t ω t 1 1 s = a.s. Let 47 g t, ω = 1 t It is easy to see, from the Itô Isometry and E Var g t, ω = E g t, ω 2 ω 1 s. 1 s ω 1 s = that dt + db t = g t 2 L 2 Ω = 1 t 2 ds 1 s 2 = 1 t 2 [ 1 1 s ] t = 1 t t 1 = t 1 t. Thus, g t L 2 Ω t 1, but this is not quite sufficient for the limit in Eqn. 46. However, it is now at least possible to find a sequence {t k } such that t k 1 and lim k g t k, ω = a.s. Which means that if it converges at all, it has to be to. This is probably as far as we are able to come by elementary means. The book gives us a hint of applying the Doob s Martingale Inequality stated without proof in Thm Following B.Ø., we let the sequence {t n } be defined t n = 1 2 n, I n = [t n, t n+1, and set 5 M t = 1 s. Note that M t, since it is defined in terms of an Itô integral, is an L 2 -martingale. Note also that all I n are disjoint and that the union makes up the whole of [, 1. October 31, 213 Page 8 of 9

9 Then 51 P ω ; sup 1 t M t ω > ε < P ω ; 1 t n sup M t ω > ε t I n t I n < P ω ; 1 t n sup M t ω > ε <t<t n+1 ε = P ω ; sup M t ω > <t<t n+1 1 t n 1 t n 2 ε 2 E Mt 2 n+1 = 1 t n 2 1 ε 2 1 t n+1 = 2 2n 1 ε 2 2n+1 = 2ε 2 2 n, which is the stated inequality. The trick is now to consider { } 52 A n = ω ; sup 1 t M t ω > 2 n/4 = t I n From the inequality above we obtain that 53 P A n < 2 2 n/4 2 2 n = 2 2 n/2, { ω ; sup t I n g t, ω > 2 n/4 }. and hence 54 P A n <. n=1 The rest is piece of cake using the easy part of the Borel-Cantelli Lemma: For almost all paths, g t, ω there exists an N ω such that ω / A n for all n > N ω. Thus, in that case, 55 sup t I n g t, ω 2 n/4 for all n > N ω. We finally observe that n=nω I n = [1 2 Nω, 1, so that, indeed, 56 lim t 1 g t, ω = for these cases, that is, almost surely. October 31, 213 Page 9 of 9