Theoretical Tutorial Session 2

Transcription

1 1 / 36 Theoretical Tutorial Session 2 Xiaoming Song Department of Mathematics Drexel University July 27, 216

2 Outline 2 / 36 Itô s formula Martingale representation theorem Stochastic differential equations

3 Itô s formula and martingale representation theorem 1. Using Itô s formula to show that M t = B 3 t 3 B sds is a martingale. Proof: Let f (x) = x 3 C 2 (R). Then f (x) = 3x 2, and f (x) = 6x. Applying Itô s formula to f (B t ) we have f (B t ) = f (B ) + f (B s )db s + 1 f (B s )ds, 2 that is, Then Bt 3 = 3 Bs 2 db s + 3 B s ds. (1) M t = Bt 3 3 B s ds = 3 Bs 2 db s. 3 / 36

4 Since ( ) ( ) E (Bs 2 ) 2 ds = E Bs 4 ds = 3s 2 ds < for all t, by the basic property of indefinite Itô integral, we can show that is a martingale. M t = Bt 3 3 B s ds = 3 Bs 2 db s 4 / 36

5 5 / Use Itô s formula to show that tb t = B s ds + Proof: Let f (t, x) = tx. Then f C 1,2 (R + R) and (t, x) = x, t (t, x) = t, x 2 f (t, x) =. x 2 sdb s. (2)

6 6 / 36 Applying Itô s formula f (t, B t ) = f (, B ) + we obtain tb t = t t (s, B s)ds + 2 f x 2 (s, B s)ds, B s ds + sdb s. x (s, B s)db s Note that the above equation give us an integration by parts formula.

7 7 / Check if the process X t = B 3 t 3tB t is a martingale. Solution: Using (1) and (2), we can write X t = B 3 t 3tB t = 3 = B 2 s db s + 3 (3B 2 s 3s)dB s. ( ) B s ds 3 B s ds + sdb s We can also show that ( ) E (3Bs 2 3s) 2 ds <, t. Therefore, the process X t = B 3 t 3tB t is a martingale.

8 8 / Find the stochastic integral representation on the time interval [, T ] of the square integrable random variable B 3 T. Solution: Using (1) and (2) with t = T, we have B 3 T = 3 = 3 = 3 = Bs 2 db s + 3 ( B 2 s db s + 3 B 2 s db s + 3 ( B s ds TB T T (3B 2 s + 3T 3s)dB s. 1dB s sdb s ) sdb s ) The above is the integral representation for BT 3 since the process {2B s + 3T 3s, s [, T ]} in in L 2 (P) and E(BT 3 ) =.

9 9 / Verify that the following processes are martingales: (a) X t = t 2 B t 2 sbsds (b) X t = e t/2 cos B t (c) X t = e t/2 sin B t (d) X t = B 1 (t)b 2 (t), where B 1 and B 2 are two independent Brownian motion. Solution 5(a): Let f (t, x) = t 2 x. Then f C 1,2 (R + R) and (t, x) = 2tx, t x (t, x) = t2, 2 f (t, x) =. x 2

10 1 / 36 Applying Itô s formula we get f (t, B t ) = f (, B ) Hence, the process is a martingale. t 2 B t = t t (s, B s)ds + 2 f x 2 (s, B s)ds, 2sB s ds + s 2 db s. X t = t 2 B t 2 sb s ds = s 2 db s x (s, B s)db s

11 Solution 5(b): Let f (t, x) = e t/2 cos x. Then f C 1,2 (R + R) and Note that t (t, x) = 1 f (t, x), 2 x (t, x) = et/2 sin x, 2 f (t, x) = f (t, x). x 2 t (t, x) f 2 x 2 (t, x) =, and f (, B ) = 1. Then we apply Itô s formula and show that the process X t = e t/2 cos B t = 1 e s/2 sin B s db s is a martingale since E( es sin B 2 s ds) es ds < for all t. 11 / 36

12 Solution 5(c): Let f (t, x) = e t/2 sin x. Then f C 1,2 (R + R) and Note that t (t, x) = 1 f (t, x), 2 x (t, x) = et/2 cos x, 2 f (t, x) = f (t, x). x 2 t (t, x) f 2 x 2 (t, x) =, and f (, B ) =. Then we apply Itô s formula and show that the process X t = e t/2 sin B t = e s/2 cos B s db s (3) is a martingale since E( es cos B 2 s ds) es ds < for all t. 12 / 36

13 13 / 36 Solution 5(d): For this exercise, we need to apply Itô s formula in multidimensional case. Let f (x 1, x 2 ) = x 1 x 2. Then f C 2 (R 2 ) and (x 1, x 2 ) = x 2 x 1 (x 1, x 2 ) = x 1 x 2 2 f (x 1, x 2 ) = x f (x 1, x 2 ) = x f x 1 x 2 (x 1, x 2 ) = 1

14 Applying multidimensional Itô s formula, one can obtain f (B 1 (t)b 2 (t)) = f (B 1 ()B 2 ()) (B 1 (s), B 2 (s))db 2 (s) x 2 2 f (B 1 (s), B 2 (s))ds x1 2 2 f (B 1 (s), B 2 (s))ds x 1 (B 1 (s), B 2 (s))db 1 (s) x2 2 2 f + (B 1 (s), B 2 (s))db 1 (s)db 2 (s), x 1 x 2 then noticing that db 1 db 2 =, we can show that the process X t = B 1 (t)b 2 (t) = B 2 (s)db 1 (s) + is ( a martingale, since ) ( t ) E B 1(s) 2 t ds = E B 2(s) 2 ds B 1 (s)db 2 (s) = sds <, t. 14 / 36

15 15 / If f (t, x) = e ax a2 2 t and Y t = f (t, B t ) = e ab t a2 2 t where a is a constant, then prove that Y satisfies the following linear SDE: Y t = 1 + a Y s db s. (4) Proof: Note that f (t, x) C 1,2 (R + R) and Note also that a2 (t, x) = f (t, x), t 2 (t, x) = af (t, x), x 2 f x 2 (t, x) = a2 f (t, x). t (t, x) f (t, x) =. 2 x 2

16 16 / 36 Applying Itô s formula, we have f (t, B t ) = f (, B ) = 1 + t t (s, B s)ds + 2 f x 2 (s, B s)ds x (s, B s)db s, x (s, B s)db s that is Y t = 1 + a Y s db s.

17 17 / 36 Remark: i). Note that ( E ) ( Y s 2 ds = E = = e a2s E ) e 2aBs a2s ds e a2s ds <, ) (2a)2 2aBs s (e 2 ds for all t. Hence, the Itô integral Y sdb s is well-defined. ii). The solution of the stochastic differential equation dy t = ay t db t, Y = 1 is not Y t = e ab t, but Y t = e ab t a2 2 t.

18 18 / Find the stochastic integral representation on the time interval [, T ] of the following square integrable random variables: (a) F = B T (b) F = B 2 T (c) F = e B T (d) F = sin B T (e) F = B tdt (f) F = tb2 t dt Solution 7(a): Since E(B T ) =, the stochastic integral representation for B T is B T = 1dB t = E(B T ) + 1dB t.

19 19 / 36 Solution 7(b): Let f (x) = x 2. Then f C 2 (R) and f (x) = 2x and f (x) = 2. Using Itô s formula and E(BT 2 ) = T, we have BT 2 = B2 + 2B s db s = 2B s db s + T = E(BT 2 ) + 2B s db s. 2dt

20 2 / 36 Solution 7(c): We can calculate that E(e B T ) = e T 2 In fact, we can NOT apply Itô s formula directly to get the stochastic integral representation, since if we choose f (x) = e x and apply Itô s formula to f (B T ), then we get e B T = e B + = 1 + e B t db t e B t db t e B t dt e B t dt. We can not get rid of the integral with respect to dt. Question: How can we get its stochastic integral representation?

21 21 / 36 In order to obtain the stochastic integral representation for e B T, we will need the result (4) in Exercise 6 with a = 1 and t = T : e B T T 2 = 1 + e B t t 2 dbt. Multiplying e T 2 on both sides of the above equation, we obtain the following stochastic integral representation e B T = e T T T 2 + e 2 = E(e B T ) + e B t t 2 dbt e B t + T t 2 dbt.

22 Solution 7(d): Note that E(sin B T ) =. Since sin x and e x are closely related in e ix = cos x + i sin x, we can foresee the same problem if we apply Itô s formula directly to f (B T ) = sin B T. Instead, we make use of (3) and obtain sin B T = e T 2 = E(sin B T ) + e t 2 cos Bt db t e T t 2 cos B t db t. 22 / 36

23 23 / 36 Solution 7(e): We have Using (2) with t = T we get ( ) T E B t dt =. B t dt = TB T = T tdb t 1dB t = (T t)db t ( ) T = E B t dt + tdb t (T t)db t.

24 Solution 7(f): Note that ( ) T E tbt 2 dt = From Part 7(b), we know B 2 t = 2 te(b 2 t )dt = B s db s + t, t. t 2 dt = T 3 3. Using the above equation and then changing the order of the integrals we have ( ) tb 2 t dt = t 2 B s db s + t = t 2 dt + 2 tb s db s dt = T 3 ( T ) T B s tdt db s s ( ) T =E tbt 2 dt + (T 2 s 2 )B s db s. dt 24 / 36

25 25 / Consider an n-dimensional Brownian motion B(t) = (B 1 (t), B 2 (t),, B n (t)) and constants α i, i = 1,..., n. Solve the following SDE: dx t = rx t dt + X t n α i db i (t), X = x, i=1 where x R. Solution: The coefficients in this SDE satisfy the Lipschitz and linear growth conditions, so there exists a unique solution. If α 1 = α 2 = = α n =, then the above SDE becomes an ODE dx t = rx t dt, X = x. and its unique solution is X t = xe rt.

26 If n i=1 α2 i, then by using the standard method mentioned in my first tutorial session we can show that the process 1 n B t = n α i B i (t) i=1 α2 i i=1 is a Brownian motion. Let Y t = e rt. Then Y satisfies dy t = ry t dt. Applying multidimensional Itô s formula to f (x, y) = xy we have d(x t Y t ) = X t dy t + Y t dx t + dx t dy t = X t dy t + Y t dx t n =X t ( ry t )dt + Y t (rx t dt + X t α i db i (t)) n = X t Y t α i db i (t) i=1 i=1 = n αi 2 (X t Y t )d B t. (5) i=1 26 / 36

27 27 / 36 Thus, X t Y t satisfies the linear SDE (5). Note also that X Y = x. Then the solution to (5) is X t Y t = x exp n α i 2 B t t n i=1 α2 i 2. i=1 Therefore, X t = Y 1 t x exp n i=1 = x exp rt + n i=1 α 2 i α 2 i { n ( = x exp α i B i (t) + i=1 B t t n i=1 α2 i 2 B t t n i=1 α2 i 2 r n i=1 α2 i 2 ) } t.

28 9. Solve the following stochastic differential equations dx t = 1 X t dt + αx t db t, X = x >. For which values of the parameter α the solution explodes? Solution: Let Y t = e αb t α2 t 2. Then Y t satisfies the following SDE: dy t = αy t db t, Y = 1. Using Itô s formula, we have d(x t Y t ) = X t dy t + Y t dx t + dx t dy t = X t ( αy t db t ) + Y t ( 1 X t dt + αx t db t ) α 2 X t Y t dt = Y t X t dt α 2 X t Y t dt, which implies 28 / 36

29 2(X t Y t )d(x t Y t ) = 2Y 2 t dt 2α 2 (X t Y t ) 2 dt. Then for each fixed ω, (X t (ω)y t (ω)) 2 solves the following linear ODE: ẏ = 2Y 2 t (ω) 2α 2 y, y() = x 2, whose solution is given by Then y(t) = e 2α2 t X t = Y 1 t e α2 t ( ) x e 2α2s Ys 2 (ω) ds. x e 2α2s Y 2 s ds. Since the trajectories of the process Y are continuous on [, ) almost surely, the above integral is well defined for all t, and hence X t will not explode for any parameter α. 29 / 36

30 3 / Solve the following stochastic differential equations dx t = X γ t dt + αx t db t, X = x >. For which values of the parameters γ, α the solution explodes? Solution: If α =, then the differential equation is an ODE d dt X = X γ, X = x >. This is a separable equation and we know that the solution explodes when γ > 1. If α and γ = 1, then this is a linear SDE and its solution is given by ( ) X t = e αb t + 1 α2 t 2, t.

31 If α and γ 1, then we will use very similar steps as in Problem 9 to obtain the solution. Let Y t = e αb t α2 t 2. Then Y t satisfies the following SDE: Using Itô s formula, we have dy t = αy t db t, Y = 1, d(x t Y t ) = X t dy t + Y t dx t + dx t dy t = X t ( αy t db t ) + Y t ( X γ t dt + αx t db t ) α 2 X t Y t dt =Y t X γ t dt α 2 X t Y t dt, which implies for each fixed ω Ω, y(t) = X t (ω)y t (ω) satisfies the following nonlinear ODE: ẏ = Y t (ω) 1 γ y γ α 2 y, y() = x, 31 / 36

32 32 / 36 or equivalently, ẏ + α 2 y = Y t (ω) 1 γ y γ, y() = x. Multiplying the above equation by e α2t and denoting z = e α2t y, we obtain ( ) 1 γ ż = Y t (ω)e α2 t z γ, z() = x. We can separate the variables to solve this ODE as follows: ż ( ) z γ = 1 γ Y t (ω)e α2 t = e α(1 γ)b t + α2 (1 γ)t 2, z() = x. (6)

33 33 / 36 Note that z(t) = X t (ω)e αb t (ω)+ 1 2 α2 t and e αb t (ω)+ 1 2 α2t is continuous in t. Then, X t (ω) explodes as t T (ω) if and only if z(t) explodes when t T (ω). Suppose that X t (ω) explodes as t T (ω) for some T (ω) <. Then integrating (6) on both sides, we should get x dz T z γ = e α(1 γ)b t + α2 (1 γ)t 2 dt <, (7) and hence, explosion might occur only if γ > 1.

34 For γ > 1, then dz x z γ = x 1 γ γ 1. Thus, if X t (ω) explodes as t T (ω) for some T (ω) <, the following equation holds e α(1 γ)b t + α2 (1 γ)t 2 dt = x 1 γ γ 1. Note also that for each t we have ( ) E e α(1 γ)bs+ α2 (1 γ)s 2 ds = = = e α2 (1 γ) 2 s 2 + α2 (1 γ)s 2 ds e α2 (1 γ)(2 γ)s 2 ds t, if γ = 2, 2 α 2 (1 γ)(2 γ) ( ) e α2 (1 γ)(2 γ)t 2 1, if γ 2. (8) 34 / 36

35 So, for α and γ 2, we have Define τ = inf ( ) lim E e α(1 γ)bs+ α2 (1 γ)s 2 ds =. (9) t { t, e α(1 γ)bs+ α2 (1 γ)s 2 ds = x 1 γ }. γ 1 That is, z (or equivalently, X) explodes at τ. Then τ is a stopping time, and moreover, from (9), we get P(τ < ) >, that is, X explodes on {τ < }. 35 / 36

36 For α and 1 < γ < 2, we get from (8) ( ) lim E e α(1 γ)bs+ α2 (1 γ)s 2 ds = t If α and γ satisfy 2 α 2 (γ 1)(2 γ) > x 1 γ γ 1, 2 α 2 (γ 1)(2 γ). then we also get P(τ < ) >, that is, X explodes on {τ < } in this case. 36 / 36