Mat 324 Advanced Fnancal Matematcs Sprng 28 Fnal Exam Solutons May 2, 28 Ts s an open book take-ome exam. You may work wt textbooks and notes but do not consult any oter person. Sow all of your work and put your name on all papers. Te exam s due back by 5 PM on Tursday May 8. You may place t n my box n te faculty malroom or under my o ce door.. Let W (t) be a rownan moton and F(t) te assocated ltraton. Sow n two d erent ways tat e W (t) 2 2 t s an F(t) martngale were s a non-negatve constant.. (a) Frst, sow t usng Itô s lemma (Itô-Doebln formula) (b) Second, sow t wtout usng Itô s lemma n any way, usng only te de ntons of W (t) and of a martngale Soluton: de W (t) 2 2 t = e W (t) 2 2 t 2 2 dt + dw (t) + 2 2 dt = e W (t) 2 2t dw (t) wc s a martngale because coe cent of dt s zero. E e W (t) 2 2t jf (s) = E e W (s) 2 2s e (W (t) W (s)) = e W (s) 2 2s e 2 2 (t s) E = e W (s) 2 2 s Secondly, 2 2 (t s) jf (s) (W (t) e W (s)) jf (s) because W (t) W (s) s normal wt varance t s and mean of te lognormal s exp( 2 varance). ut ts proves te martngal property. 2. Let W (t) be a rownan moton and F(t) te assocated ltraton. Wtout assumng any knowledge of te moments of te standard normal dstrbuton, use Itô s lemma and your knowledge of stocastc calculus to sow tat E W 6 (t) = 5t 3. (Hnt: rst use stocastc calculus to gure out wat E W 2 (t) and E W 4 (t) are equal to.)
Soluton: dw 2 (t) = 2W (t)dw (t) + dt W 2 (t) = 2 E W 2 (t) = W (s)dw (s) + ds = t ds dw 4 (t) = 4W 3 (t)dw (t) + 6W 2 (t)dt W 4 (t) = 4 E W 4 (t) = 6 W 3 (s)dw (s) + 6 sds = 3t 2 dw 6 (t) = 6W 5 (t)dw (t) + 5W 4 (t)dt W 6 (t) = 6 E W 6 (t) = 5 W 5 (s)dw (s) + 5 3s 2 ds = 5t 3 W 2 (s)ds W 4 (s)ds 3. Derve te general form for all solutons S(t) to te equaton Soluton: ds(t) = (t)s(t)dw (t) + (t)s(t)dt d ln S(t) = ln S(t) ln S() = S(t) ds(t) 2 S 2 (t) ds(t)ds(t) (t)dw (t) + (t)dt 2 2 (t)dt (s)dw (s) + (s) S(t) = S()e (s)dw (s)+ 2 2 (s) ((s) 2 2 (s))ds ds 4. You know tat te prce V (t) at tme t for a dervatve securty tat as payo V (T ) at tme T > t s gven by R V (t) = E ~ T e t R(u)du V (T ) j F(t) were R(t) s te rsk-free sort-term nterest rate. Usng formulas were necessary, explan te d erence between E and E ~ and wat te connecton s between E ~ and te process S(t) for te underlyng stock prce. Soluton: E ~ s de ned under a measure P ~ R equvalent to te orgnal P but for t wc e R(u)du V (t) s a martngale. Te connecton wt S(t) s tat were 2
ds(t) = (t)s(t)dw (t)+(t)s(t)dt,.e. S(t) = S()e Z we de ne for any measurable set A, P(A) ~ d P = ~ (!)dp(!) were d ~ P dp (!) = e dp (s) R(s) dw(s) (s) 2 ( (s)dw (s)+ (s) R(s) (s) ) 2 ds and te relaton between rownan motons s d W ~ (t) R(t) (t) = dw (t) + dt (t) ((s) 2 2 (s))ds, 5. In te same stuaton as queston 4 above, explan wy te exstence of a portfolo process X(t) = V (t) wt dx(t) = (t)ds(t) + R(t) [X(t) (t)s(t)] dt depends upon te assumpton tat (t) > almost surely, were (t) s te volatlty process n te geometrc rownan moton process S(t) for te underlyng stock prce. e sure to explan wat any new symbols tat you ntroduce are, were tey come from, and wat just es your usng tem. Soluton: Te martngale representaton teorem gves us a process ~ (t) wt V (t) A = ~ (t)d W ~ (t) X(t) A = R(t)e ut f X(t) exsts as gven ten dtx(t) + e dx(t) + = R(t)e = e = e +e +R(t) [X(t) X(t)dt f(t) [(t)s(t)dw (t) + (t)s(t)dt] (t)s(t)] dtg (t)s(t) f(t)dw (t) + ((t) (t)s(t)(t)d ~ W (t) R(t)) dtg 3
(t) = So f X(t) exsts as gven ten ~ (t) must old and for tat e S(t)(t) (t) 6= must old almost surely. 6. Let te random varable A be te value at tme T of an asset and assume tat A s almost-surely postve, were A s F(T ) measurable n te ltraton determned by a rownan moton W (t). Assume tat tere s a rsk free rate process R(t) and a unque rsk-neutral measure. Sow tat tere exst random processes V (t); (t) and (t) so tat dv (t) = (t)v (t)dt + (t)v (t)dw (t) (.e. V (t) s a generalzed geometrc rownan moton) and V (T ) = A. Ts means tat all postve assets measurable n te ltraton generated by a rownan moton can be represented by a generalzed geometrc rownan moton based on te orgnal rownan moton. Soluton: De ne V (t) = e t A so V (T ) = A and V (t) 6= almost surely. Snce a rsk-neutral P ~ exts, e V (t) s P-martngale ~ and te representaton teorem gves a process ~ (t) wt V (t) A = ~ (t)d W ~ (t) so R(t)e dtv (t) + e V (t) = ~ (t)d ~ W (t) dv (t) = R(t)V (t)dt + e ~(t)d ~ W (t) ~(t) De ne (t) = e V (t) and (t) = R(t)+(t)(t) were d W ~ (t) = (t)dt+ dw (t) and ten dv (t) = (t)v (t)dt + (t)v (t)dw (t) 7. Do exercse 6. from te textbook. 4
Soluton: Z(t) = e = dz(u) = Z(u) (u)dw (u) + (b(u) = (u)z(u)dw (u) + b(u)z(u)du Y (t) = x dy (u) = a(u) (u)(u) du + (u) dw (u) Z(u) Z(u) d (Y (u)z(u)) = dy (u)z(u) + Y (u)dz(u) + dy (u)dz(u) 2 2 (u))du + 2 2 (u)du = (a(u) (u)(u)) du + (u)dw (u) + (u) (Y (u)z(u)) dw (u) +b(u) (Y (u)z(u)) du + (u)(u)du = (a(u) + b(u) (Y (u)z(u))) du + ((u) + (u) (Y (u)z(u))) dw (u) 5