MA 811 Stokastiske metoder i systemteori AUTUMN TRM 3 Suggested solution with some extra comments The exam had a list of useful formulae attached. This list has been added here as well. 1 Problem In this problem we are considering a standard Brownian motion B t in R 1 starting at : (a) State the basic properties of the Brownian motion. De ne, for a xed a >, the process Verify that also X t is a standard Brownian motion. X t ab ta : (1) A standard Brownian motion starting at is a Gaussian stochastic process de ned for t [; 1) ful lling 1. B t ;. Cov (B t B s ) min (t; s) : From () it follows that a B.M. has orthogonal increments. There exists a version of B.M. with continuous paths. It is obvious that X t is a Gaussian process (This actually requires that all nite collections (X t1 ; ; X tn ) are multivariate Gaussian, but this follows since B t has such a property). Moreover, 1. clearly true. Finally, Cov (X t ; X s ) a Cov B sa B ta s a min a ; t a () min (s; t) : (b) Let t < t 1 < < t N+1 T be a partition of the interval [; T ] and ' the elementary function NX ' (t;!) e j (!) [tj+1 t j ) (t) : (3) j What does it mean that ' is in the class V[; T ], and what is then the value of the Itô integral Show that for all f V[; T ]. ' (t;!) db t (!)? (4) f (t;!) db t (!) : (5) 1
The class V[; T ] consists of B F-measurable functions f (t;!) L ( [; T ]) such that f (t;!) is F t -measurable for all t [; T ]. Here this will be the case if e j < 1 and e j is F tj -measurable for all j ; ; N. Also, ' (t;!) db t (!) Since e j and B tj+1 B tj are independent, Thus, NX e j (!) B tj+1 (!) B tj (!) : (6) j e j Btj+1 B tj (ej ) B tj+1 B tj : (7) ' (t;!) db t (!) : (8) In general, the Itô integral is a limit of integrals of simple functions. This is about all we requite for the exam, but the full argument is as follows: We nd a sequence f' n g such that R 'n db R fdb! n!1 : Then Z ' n db Z Z fdb Z fdb Z (' n (' n Z (' n f) db f) db (9) f) db!! 1! : n!1 (c) Compute the variance of the integral tb t (!) db t (!) : (1) This follows immediately from Itô s Isometry since we know that the expectation is : Problem Var tb t (!) db t (!) tb t (!) db t (!) We consider two Itô processes X t and Y t on R 1. (tb t ) dt (a) Let X t and Y t be two Itô processes X t and Y t on R 1. Prove that t tdt 1 4 : (11) d (X t Y t ) X t dy t + Y t dx t + dx t dy t : (1)
For this formula we apply the D Itô formula for the function g (x; y) xy:then (b) Let d (X t Y t ) @g @x dx t + @g @y dy t + @ g @x@y dx t dy t Show that X t can be written as an Itô integral. We compute dx t using Ito s formula: Y t dx t + X t dy t + 1 dx t dy t : (13) X t e t sin (B t ) : (14) dx t 1 X tdt + e t cos (B t ) db t + 1 et ( sin B t ) dt Hence, e t cos (B t ) db t : (15) X t Z t since it is clear that e t cos (B t ) V [; T ]. e s cos (B s ) db s ; (16) (c) The conclusion in (.b) implies that X t is a Martingale with respect to the ltration of the Brownian motion, F t. Prove this directly by applying the de nition of a Martingale to the expression for X t in qn. 14. The rst is to observe that (X t ) (jx t j) e t jsin (B t )j e t < 1: (17) Since X t is a determininistic, continuous function of B t, X t is clearly F t -measureable. We nally need to show that (X t+t jf t ) X t for t >. Let us write B t+t B t + B. Then (X t+t jf t ) e (t+t) sin (B t + B) jf t e (t+t) (sin (B t ) cos (B) + cos (B t ) sin (B) jf t ) (i) e (t+t) [sin (B t ) (cos (B) jf t ) + cos (B t ) (sin (B) jf t )] (ii) e (t+t) [sin (B t ) (cos (B)) + cos (B t ) (sin (B))] (18) (iii) h i e (t+t) sin (B t ) e t + e t sin (B t ) X t : Here (i) and (ii) are formulae for the conditional expectation. Moreover, for (iii), (cos (B)) is listed and (sin (B)) is obviously. 3
3 Problem (a) Show that a linear stochastic di erential equation dx t p (t) X t dt + q (t) db t (19) may be solved by an integrating factor h (t) such that d [h (t) X t ] h (t) q (t) db t : () It follows from Itô s formula that d (h (t) X t ) h (t) X t dt + h (t) dx t : (1) We then multiply qn. 19 by h (t): h (t) dx t d (h (t) X t ) h (t) X t dt h (t) p (t) X t dt + h (t) q (t) db t : () The stated form follows if we nd h such that h (t) h (t) p (t) : (3) Then the solution to qn. 19 is: or h (t) X t h (t ) X t Z t t h (t) q (t) db t ; (4) X t h (t ) X t + R t t h (t) q (t) db t : (5) h (t) (b) Apply the method in (a) to solve the equation dx t 1 t X tdt + tdb t ; t 1; X 1 1: (6) The function h has to satisfy the equation that is, h (t) t 1. In this case the start is at t 1 such that 4 Problem X t X 1 + R t 1 1 s sdb s 1t h + 1 h ; (7) t t (1 + B t B 1 ) t ~ B 1;1 t ; t 1: (8) Consider the following geometric Brownian motion in R 1 : X t x exp ( t + B t ) ; t. (a) Show that X t is an Itô di usion with generator A x d dx + x d dx : (9) 4
The di usion representation of X t is dx t X t dt + X t db t + 1 X tdt The formula for the generator follows from when dx t dt + db t. (b) We start X t at x 1 and de ne, for x < 1, X t dt + X tdb t : (3) A d dx + 1 d dx ; (31) (!) minft ; X t (!) x g: (3) Prove that log x : (33) We are going to apply Dynkin s Formula and need to know that < 1. Here the part t will e ectively "kill" the Brownian motion part Bt. Of course, X t will also cross the level x a.s. This argument is su cient for the exam, but the full proof could be based on the law of the iterated logarithm, or the following simple estimate: P ( u) P (X u x ) P (B u log x + u) Bu P u 1 log x + u u 1 O u 1 e u ; (34) when u! 1; using the inequality in the list. This is su cient for to be nite. The next step is to nd an f such that Af is equal to a constant. Here f (x) log x will do since x d log x + x d log x dx dx 1: (35) By Dynkin s Formula we then have Z 1 (f (X )) log (x ) log 1 + 1 ( 1) ds ; (36) or log (x ) : (37) (c) Let a; b be two positive numbers, a < b. We start X t at x [a; b] and let Determine p for all x [a; b]. p Pr (X t hits level b before it hits level a) 5
We need to nd a function f (x) such that Af, and try x : x dx dx + x d x dx x + 1 ( 1) ; (38) for. We then apply Dynkin s formula with the function f (x) x,. The expected escape time from the interval [a; b] is clearly nite since the situation will be similar to the case in (b) for the lower level a. Thus (X ) p b + (1 p) a x + ; (39) and hence, a p x b a : (4) 6
A list of useful formulae Note: The list does not state requirements for the formulae to be valid. The probability density of a 1D Gaussian variable with mean and variance :! (x) p 1 (x ) exp (41) A de nite integral: 1 x cos (x) exp dx p e : (4) An inequality: Let X be N (; 1) and x >. Then P (X x) Two formulae for Conditional xpectations: The Itô isometry: r 1 1 x e x : (43) (i) If Y is H-measurable, then (XY jh) Y (XjH) : (ii) If X is independent of H, then (XjH) (X) : Itô s D formula: f (t;!) db t (!) dg (t; X t ; Y t ) @g @g dt + @t @x dx t + @g @y dy t + 1 jf (t;!)j dt kfk L ([;T ]) (44) @ g @x (dx t) + @ g @x@y dx tdy t + 1 @ g @y (dy t) : (45) and "the rules". The generator: A (f) (x) Dynkin s Formula: nx i1 i (x) @f @x i (x) + 1 nx i;j1 x (f (X )) f (x) + x Z (x) (x) @ f (x) : (46) i;j @x i @x j Af (X s ) ds : (47) 7