Markov processes and the Kolmogorov equations

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1 Chapter 6 Markv prcesses ad the Klmgrv equatis 6. Stchastic Differetial Equatis Csider the stchastic differetial equati: dx(t) =a(t X(t)) dt + (t X(t)) db(t): (SDE) Here a(t x) ad (t x) are give fuctis, usually assumed t be ctiuus i (t x) ad Lipschitz ctiuus i x,i.e., there is a cstat L such that ja(t x) a(t y)j Ljx yj j(t x) (t y)j Ljx yj fr all t x y. Let (t 0 x) be give. A sluti t (SDE) with the iitial cditi (t 0 x) is a prcess fx(t)g tt0 satisfyig X(t 0 )=x X(t) =X(t 0 )+ t t 0 a(s X(s)) ds + t t 0 (s X(s)) db(s) t t 0 The sluti prcess fx(t)g tt0 will be adapted t the filtrati ff(t)g t0 geerated by the Brwia mti. If yu kw the path f the Brwia mti up t time t, the yu ca evaluate X(t). Example 6. (Drifted Brwia mti) Let a be a cstat ad =,s dx(t) =adt+ db(t): If (t 0 x) is give ad we start with the iitial cditi X(t 0 )=x 77

2 78 the X(t) =x + a(t t 0 )+(B(t) B(t 0 )) t t 0 : T cmpute the differetial w.r.t. t, treat t 0 ad B(t 0 ) as cstats: dx(t) =adt+ db(t): Example 6. (Gemetric Brwia mti) Let r ad be cstats. Csider dx(t) = rx(t) dt + X(t) db(t): Give the iitial cditi X(t 0 )=x the sluti is X(t) =x exp (B(t) B(t0 )) + (r )(t t 0 ) : Agai, t cmpute the differetial w.r.t. t, treat t 0 ad B(t 0 ) as cstats: dx(t) =(r )X(t) dt + X(t) db(t) + X(t) dt = rx(t) dt + X(t) db(t): 6. Markv Prperty Let 0 t 0 <t be give ad let h(y) be a fucti. Dete by IE t 0 x h(x(t )) the expectati f h(x(t )), give that X(t 0 )=x. Nw let IR be give, ad start with iitial cditi We have the Markv prperty X(0) = : IE h(x(t 0 )) F (t 0) = IE t 0 X(t 0 ) h(x(t )): I ther wrds, if yu bserve the path f the drivig Brwia mti frm time 0 t time t 0, ad based this ifrmati, yu wat t estimate h(x(t )), the ly relevat ifrmati is the value f X(t 0 ). Yu imagie startig the (SDE) at time t 0 at value X(t 0 ), ad cmpute the expected value f h(x(t )).

3 CHAPTER 6. Markv prcesses ad the Klmgrv equatis Trasiti desity Dete by p(t 0 t x y) the desity (i the y variable) f X(t ), cditied X(t 0 )=x. I ther wrds, IE t 0 x h(x(t )) = h(y)p(t 0 t x y) dy: IR The Markv prperty says that fr 0 t 0 t ad fr every, IE h(x(t 0 )) F(t 0) = h(y)p(t 0 t X(t 0 ) y) dy: IR Example 6.3 (Drifted Brwia mti) Csider the SDE dx(t) =adt+ db(t): Cditied X(t 0 ) =x, the radm variable X(t ) is rmal with mea x + a(t t 0 ) ad variace (t t 0 ), i.e., p(t 0 t x y) = p (t t 0 ) exp (y (x + a(t t 0 ))) (t t 0 ) Nte that p depeds t 0 ad t ly thrugh their differece t t 0. This is always the case whe a(t x) ad (t x) d t deped t. : Example 6.4 (Gemetric Brwia mti) Recall that the sluti t the SDE dx(t) =rx(t) dt + X(t) db(t) with iitial cditi X(t 0 )=x, is Gemetric Brwia mti: X(t )=xexp (B(t ) B(t 0 )) + (r )(t t 0 ) : The radm variable B(t ) B(t 0 ) has desity IP fb(t ) B(t 0 ) dbg = p exp (t t 0 ) b (t t 0 ) db ad we are makig the chage f variable y = x exp b +(r )(t t 0 ) r equivaletly, The derivative is dy db b = h lg y i x (r )(t t 0 ) : = y r equivaletly, db = dy y :

4 80 Therefre, p(t 0 t x y) dy = IP fx(t ) dyg h = p exp y (t t 0 ) (t t 0 ) lg y i x (r )(t t 0 ) dy: Usig the trasiti desity ad a fair amut f calculus, e ca cmpute the expected payff frm a Eurpea call: where Therefre, IE t x (X(T ) K) + = (y K) + p(t T x y) dy 0 IE 0 e r(tt) (X(T ) K) + F(t) = e r(tt) xn KN N () = p p T t p T t e x dx = p h x i lg K + r(t t) + (T t) h lg x K + r(t t) (T t) e x dx: = e r(tt) IE t X(t) (X(T ) K) + = X(t)N p lg X(t) T t K + r(t t) + (T t) e r(tt) KN p lg X(t) T t K + r(t t) (T t) i 6.4 The Klmgrv Backward Equati Csider dx(t) =a(t X(t)) dt + (t X(t)) db(t) ad let p(t 0 t x y) be the trasiti desity. The the Klmgrv Backward Equati p(t 0 t x y) =a(t p(t 0 t x y) + (t p(t 0 t x y): (KBE) The variables t 0 ad x i (KBE) are called the backward variables. I the case that a ad are fuctis f x ale, p(t 0 t x y) depeds t 0 ad t ly thrugh their differece = t t 0. We the write p( x y) rather tha p(t 0 t x y), ad p( x p( x y) + p( x (KBE )

5 CHAPTER 6. Markv prcesses ad the Klmgrv equatis 8 Example 6.5 (Drifted Brwia mti) dx(t) =adt+ db(t) p( x y) = p exp (y (x + a @ p = p (y x a ) = (y x a p exp (y x a ) = a(y x a ) (y x a ) + p = p x = y x @x p = p y x a xx = p + y x a p = (y x a ) p + p: Therefre, ap x + p xx = a(y x a ) = p : (y x a ) + p This is the Klmgrv backward equati. Example 6.6 (Gemetric Brwia mti) dx(t) = rx(t) dt + X(t) db(t): p( x y) = y p exp h lg y i x (r ) : It is true but very tedius t verify that p satisfies the KBE p = rxp x + x p xx : 6.5 Cecti betwee stchastic calculus ad KBE Csider dx(t) = a(x(t)) dt + (X(t)) db(t): (5.) Let h(y) be a fucti, ad defie v(t x) =IE t x h(x(t ))

6 8 where 0 t T. The v(t x) = h(y) p(t t x y) dy v t (t x) = h(y) p (T t x y) dy v x (t x) = h(y) p x (T t x y) dy v xx (t x) = h(y) p xx (T t x y) dy: Therefre, the Klmgrv backward equati implies v t (t x) +a(x)v x (t x) + (x)v xx (t x) = h i h(y) p (T t x y) +a(x)p x (T t x y) + (x)p xx (T t x y) dy =0 Let (0 ) be a iitial cditi fr the SDE (5.). We simplify tati by writig IE rather tha IE 0. Therem 5.50 Startig at X(0) =, the prcess v(t X(t)) satisfies the martigale prperty: IE v(t X(t)) F(s) = v(s X(s)) 0 s t T: Prf: Accrdig t the Markv prperty, IE h(x(t )) F(t) = IE t X(t) h(x(t )) = v(t X(t)) s IE [v(t X(t))jF(s)] = IE IE h(x(t )) F(t) F(s) = IE h(x(t )) F(s) = IE s X(s) h(x(t )) (Markv prperty) = v(s X(s)): Itô s frmula implies dv(t X(t)) = v t dt + v x dx + v xxdx dx = v t dt + av x dt + v x db + v xx dt:

7 CHAPTER 6. Markv prcesses ad the Klmgrv equatis 83 I itegral frm, we have v(t X(t)) = v(0 X(0)) t h i + v t (u X(u)) + a(x(u))v x (u X(u)) + (X(u))v xx (u X(u)) du 0 t + (X(u))v x (u X(u)) db(u): 0 i hv t + av x + v xx We kw that v(t X(t)) is a martigale, s the itegral R t 0 fr all t. This implies that the itegrad is zer hece v t + av x + v xx =0: du must be zer Thus by tw differet argumets, e based the Klmgrv backward equati, ad the ther based Itô s frmula, we have cme t the same cclusi. Therem 5.5 (Feyma-Kac) Defie v(t x) =IE t x h(x(t )) 0 t T where The dx(t) = a(x(t)) dt + (X(t)) db(t): v t (t x) +a(x)v x (t x) + (x)v xx (t x) =0 (FK) ad v(t x)=h(x): The Black-Schles equati is a special case f this therem, as we shw i the ext secti. Remark 6. (Derivati f KBE) We pluked dw the Klmgrv backward equati withut ay justificati. I fact, e ca use Itô s frmula t prve the Feyma-Kac Therem, ad use the Feyma-Kac Therem t derive the Klmgrv backward equati. 6.6 Black-Schles Csider the SDE With iitial cditi the sluti is ds(t) = rs(t) dt + S(t) db(t): S(t) =x S(u) =x exp (B(u) B(t)) + (r )(u t) u t:

8 84 Defie v(t x) =IE t x h(s(t )) = IEh x exp (B(T ) B(t)) + (r )(T t) where h is a fucti t be specified later. Recall the Idepedece Lemma: IfG is a -field, X is G-measurable, ad Y is idepedet f G, the IE h(x Y ) G = (X) where (x) =IEh(x Y ): With gemetric Brwia mti, fr 0 t T, we have We thus have S(t) =S(0) exp B(t) +(r )t S(T )=S(0) exp B(T )+(r )T = S(t) exp (B(T ) B(t)) + (r {z} )(T t) {z } F(t)-measurable idepedet f F (t) where S(T )=XY Nw X = S(t) Y =exp (B(T ) B(t)) + (r )(T t) : IEh(xY )=v(t x): The idepedece lemma implies IE h(s(t )) F (t) = IE [h(xy )jf(t)] = v(t X) = v(t S(t)):

9 CHAPTER 6. Markv prcesses ad the Klmgrv equatis 85 We have shw that v(t S(t)) = IE h(s(t )) F(t) 0 t T: Nte that the radm variable h(s(t )) whse cditial expectati is beig cmputed des t deped t. Because f this, the twer prperty implies that v(t S(t)) 0 t T, is a martigale: Fr 0 s t T, IE v(t S(t)) F(s) = IE IE h(s(t )) F (t) F(s) = IE h(s(t )) F(s) = v(s S(s)): This is a special case f Therem 5.5. Because v(t S(t)) is a martigale, the sum f the dt terms i dv(t S(t)) must be 0. By Itô s frmula, h i v t (t S(t)) dt + rs(t)v x (t S(t)) + S (t)v xx (t S(t)) dt dv(t S(t)) = This leads us t the equati + S(t)v x (t S(t)) db(t): v t (t x) +rxv x (t x) + x v xx (t x) =0 0 t<t x 0: This is a special case f Therem 5.5 (Feyma-Kac). Alg with the abve partial differetial equati, we have the termial cditi v(t x)=h(x) x 0: Furthermre, if S(t) = 0fr sme t [0 T], the als S(T ) = 0. This gives us the budary cditi v(t 0) = h(0) 0 t T: Fially, we shall evetually see that the value at time t f a ctiget claim payig h(s(t )) is at time t if S(t) =x. Therefre, u(t x) =e r(t t) IE t x h(s(t )) = e r(t t) v(t x) v(t x) =e r(t t) u(t x) v t (t x) =re r(t t) u(t x) +e r(t t) u t (t x) v x (t x) =e r(t t) u x (t x) v xx (t x) =e r(t t) u xx (t x):

10 86 Pluggig these frmulas it the partial differetial equati fr v ad cacellig the e r(t t) appearig i every term, we btai the Black-Schles partial differetial equati: ru(t x) +u t (t x) +rxu x (t x) + x u xx (t x) =0 0 t<t x 0: (BS) Cmpare this with the earlier derivati f the Black-Schles PDE i Secti 5.6. I terms f the trasiti desity ( p(t T x y) = y p (T t) exp y ) (T t) lg x (r )(T t) fr gemetric Brwia mti (See Example 6.4), we have the stchastic represetati u(t x) =e r(t t) IE t x h(s(t )) = e r(t t) h(y)p(t T x y) dy: 0 (SR) I the case f a call, h(y) =(y K) + ad u(t x) =xn p T t x lg K + r(t t) + (T t) e r(t t) KN p x lg T t K + r(t t) (T t) Eve if h(y) is sme ther fucti (e.g., h(y) =(K y) +, a put), u(t x) is still give by ad satisfies the Black-Schles PDE (BS) derived abve. 6.7 Black-Schles with price-depedet vlatility ds(t) =rs(t) dt + (S(t)) db(t) v(t x) =e r(t t) IE t x (S(T ) K) + : The Feyma-Kac Therem w implies that rv(t x)+v t (t x) +rxv x (t x) + (x)v xx (t x) =0 0 t<t x>0: v als satisfies the termial cditi v(t x)=(x K) + x 0

11 CHAPTER 6. Markv prcesses ad the Klmgrv equatis 87 ad the budary cditi v(t 0) = 0 0 t T: A example f such a prcess is the fllwig frm J.C. Cx, Ntes ptis pricig I: Cstat elasticity f variace diffusis, Wrkig Paper, Stafrd Uiversity, 975: ds(t) =rs(t) dt + S (t) db(t) where 0 <. The vlatility S (t) decreases with icreasig stck price. The crrespdig Black-Schles equati is rv + v t + rxv x + x v xx =0 0 t<t x>0 v(t 0) = 0 0 t T v(t x)=(x K) + x 0:

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