Mathematical Finance

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1 ETH Zürich, HS 2017 Prof. Josef Teichmann Matti Kiiski Mathematical Finance Solution sheet 14 Solution 14.1 Denote by Z pz t q tpr0,t s the density process process of Q with respect to P. (a) The second claim follows directly from the first claim together with the fact that yz T y dq dp P Dpyq since Z P Zp1q and the fact that the function V is decreasing. So it remains to show the first claim. Seeking a contradiction, suppose there exists Y T P Dpyq such that A : ty T ą yz T u has P ras ą 0. Set a QrAs ą 0 and define the Q-martingale M pm t q tpr0,t s by M t : E Q r1 A F t s. Then M is non-negative and M 0 a by the fact that F 0 is P -trivial. By the predictable representation property of S under Q, there exists H P LpSq such that M a ` H S. Thus, M P Vpaq. Now, on the one hand, by the definition of Dpyq, there exists a supermartingale Z r P Zpyq with Y T ď Z r T. Therefore, On the other hand, Thus, we arrive at the contradiction ErM T Y T s ď ErM T r ZT s ď ErM 0 r Z0 s ay. (1) ErZ T M T s E Q rm T s M 0 a. (2) 0 ě ErM T py T yz T qs Er1 tyt ąyz T upy T yz T qs ą 0. (3) (b) Note that 0 ď y 0 ă 8 and vpyq ă 8 on py 0, 8q. Moreover, recall that the function V is strictly decreasing, strictly convex and in C 1 on p0, 8q. First, define the function g : py 0, 8q Ñ r 8, 0s by gpsq ErZ T V 1 psz T qs. (4) This is well defined as Z T ą 0 P -a.s. and V 1 ă 0. Moreover, it is increasing as V 1 is increasing. Thus if gps 0 q ą 8 for some s 0 ą y 0, it follows by dominated convergence that it is continuous on ps 0, 8q. Next, for y 1, y 2 P py 0, 8q, y 1 ă y 2, the fundamental theorem of calculus gives V py 2 Z T q V py 1 Z T q ż y2 y 1 Z T V 1 psz T q ds. (5) Now, the left-hand side of (5) is integrable by assumption. Thus, the right-hand side is so, too, and since V 1 ă 0, the integrand on the right-hand side is strictly negative, and Fubini s theorem gives vpy 2 q vpy 1 q ż y2 y 1 gpsq ds. (6) In particular, the function g is finite a.e. on py 0, 8q, and thus continuous and finite on py 0, 8q. Now the claim follows from the fundamental theorem of calculus. (c) First, recall that X T P Cpxq if and only if sup ErX T hs ď x. (7) Y T PDp1q Updated: December 19, / 5

2 By part a), this is equivalent to Now, by part b) and the choice of pypxq, and so p X T P Cpxq. ErX T Z T s ď x. (8) Er p X T Z T s Er V 1 ppypxqz T qz T s v 1 ppypxqq x, (9) Next, fix X T P Cpxq. We may assume without loss of generality that ErUpX T qs ą 8. By the fact that p X T ą 0 P -a.s. and U is in C 1 and strictly concave on p0, 8q, UpX T q Up p X T q ď U 1 p p X T qpx T px T q, (10) where the equality is strict on tx T U 1 p V 1 q id and (8) and (9) gives p X T u. Taking expectations and using the fact that ErUpX T q Up p X T qs ď ErU 1 p p X T qpx T px T qs pypxqerz T px T px T qs ď 0. (11) If X T p X T P -a.s., then both inequalities are trivially equalities, and if P rx T p X T s ą 0, then the first inequality is strict. Solution 14.2 The situation on which the dual optimizer fails to be of the form py T : dq dp, (12) where dq dp is the Radon-Nikodym density for some EσMM Q is given by Exercise 7.4. Consider for instance the logarithmic utility U pxq logpxq for which the dual optimization problem is to minimize ErV py T qs E log py T q 1 E log py T q 1, or equivalently, maximize E log py T q over Y T P Dp1q. Take S : Z 1, where Z : X σ^τ, given by Exercise 7.4, fails to be uniformly integrable martingale and deploy the usual time change t{pt tq to obtain a finite time-horizon T. Clearly, Z T P Dp1q, but since ErZ T s ă 1 it does not define a probability measure, i.e., it fails be of the form (12). For any Y T P Dp1q, the process is a supermartingale starting from Y 0 S 0 1. Hence, by Jensen s inequality, we have «ˆ ff YT ErlogpY T qs E log ` ErlogpZ T qs ď log `ErY T S T s ` ErlogpZ T qs ď ErlogpZ T qs, Z T i.e., vp1q ErlogpZ T qs 1. Solution 14.3 (a) Since Erf{ ˆfs ď 1 holds for all f P C, by the Markov inequality we have for any f P C and any M ą 0 P r fˆf ě Ms ď 1 M Erfˆf s ď 1 M, which indeed implies that the set tf{ ˆf f P Cu is bounded in probability. Since lim MÑ8 P r ˆf ě Ms 0, from the fact that P rf ě Ms P r fˆf?? Ms ` P r ˆf ě Ms it follows that C is bounded in probability as well. ˆf ě Ms ď P r fˆf ě Updated: December 19, / 5

3 (b) The convexity and boundedness in probability of C 1 follow immediately from the definition of C 1 and the corresponding properties of C. So it remains to show that C 1 is closed in probability. Let pf n q be a sequence in C 1 convergent to an f P L 0` in probability. By passing to a subsequence we can and will assume that pf n q converges to f almost surely. By definition for each n there is an h n in C such that f n ď h n. Since C 1 is convex and bounded in probability, convph n n ě 1q Ă C 1 is also bounded in probability and we can apply the Komlos lemma for ph n q to find a sequence p r h n q such that r h n is a finite convex combination of h n, h n`1,... and r h n converges to an h in probability. Again, we will assume that this convergence also holds almost surely. By the closedness of C, we have h P C. Furthermore, if r h n a n 1 h n `... ` a n N n h Nn for convex weights a n 1,..., a n N n ě 0, a n 1 `... ` a n N n 1, then as h m ě f m for all m ě 1 we indeed have r h n ě r f n with r f n : a n 1 f n `... ` a n N n f Nn. On the other hand, it is clear that with f n Ñ f almost surely it holds that r f n Ñ f almost surely. Combining all arguments above we can conclude that h lim nñ8 r hn ě lim nñ8 r fn f, which implies that f P C 1. (c) Now we assume that C C 1 and 1 P C. Then it is clear that for each n ě 1, 1 P C n and therefore C n is nonempty. Moreover, using the same argument as in b) we can easily check that C n is closed in probability and convex. Also, since 1 P C, we have sup fpc n Erlog fs ě Erlog 1s 0. On the other hand, since each f P C n satisfies f ď n, we have sup fpc n Erlog fs ď log n. Now let pf m q be a sequence in C n such that Erlog f m s Ò sup fpc n Erlog fs as m tends to 8. Using Komlos lemma for pf m q we obtain a sequence p r f m q and an f n P L 0` such that r f m is a finite convex combination of f m, f m`1,... and r f m converges to f n in probability. Since x ÞÑ log x is a concave function, we have Erlog f m s ě Erlog f m s for each m (note that we assume Erlog f m s is increasing in m) for all m. Moreover, since C n is convex and in particular each r f m is in C n, it holds that r f m P C n. As a consequence of the closedness of C n, we have f n P C n and by the inverse Fatou s lemma (as log f ď log n for all f P C n ) we get Erlog f n s ě lim sup mñ8 Erlog f r m s ě lim sup Erlog f m s ě sup Erlog fs. mñ8 fpc n This gives Erlog f n s sup fpc n Erlog fs. Finally note that as Erlog f n s ě Erlog 1s 0 we must have f n P L 0``. (d) Now fix an n and we still denote by f n a maximizer for sup fpc n Erlog fs as above. For any ɛ P p0, 1 2 s and f P Cn, since p1 ɛqf n ` ɛf belongs to C n due ` to its convexity, the maximality of f n implies that Er ɛ pf f n qs ď 0 for ɛ pf f n q : log p1 ɛqf n`ɛf log f n ɛ. Furthermore, on tf ą f n u we indeed have ɛ pf f n q ą 0 while on tf ď f n u we can use the inequality log y log x ď y x x for all 0 ă x ă y to show that (set y f n, x p1 ɛqf n ` ɛf) ɛ pf f n f n f q ě f n ɛpf n fq ě f n f n f n 1 2 pf n fq 2f f f n ` f ě 2, where in the second inequality we use the assumption that ɛ ď 1 2. Now, by Fatou s lemma, we can derive that E lim inf ɛñ0 ɛ pf f n q E f f n f n ď lim inf E ɛ pf f n q ď 0, ɛñ0 ` where in the first equality above we used the fact that lim inf ɛñ0 log `p1 ɛqx`ɛy log x {ɛ py xq{x. Now we only need to note that E f f n f ď 0 is equivalent to E f n f ď 1 and as n h n P C n X L 0`` (see c)) the random variable 1{f n is well-defined. (e) Now for each n we obtain an f n P C n X L 0`` Ă C (keep in mind we still assume that C C 1 ) such that Erf{f n s ď 1 for all f P C n. Now, we apply the Komlos lemma for the sequence Updated: December 19, / 5

4 pf n q to get a sequence pf rn q and an ˆf P L 0` such that f rn is a finite convex combination of f n, f n`1,... and f rn converges to ˆf in probability. The convexity of C ensures that each f rn is in C and therefore the closedness in probability of C ensures that ˆf P C. Now, for any f P C k, for any n ě k, since C k Ă C n, we have Erf{f n s ď 1 and, furthermore, as x ÞÑ 1 x is convex, Jensen s inequality implies that Erf{ f rn s ď a n 1 Erf{f n s `... ` a n N n Erf{f Nn s ď a n 1 `... ` a n N n 1 for r f n : a n 1 f n `... ` a n N n f Nn with convex weights a n 1,...a n N n. This implies that for all f P Ť kě1 Ck, by Fatou s lemma we have Erf{ ˆfs ď lim inf nñ8 Erf{ r f n s ď 1. In particular, inserting 1 P Ť kě1 Ck we can get Er1{ ˆfs ď 1, which implies that ˆf P L 0`` X C. Further, Since for any f P C we have f ^ k P C k, the monotone convergence theorem provides that E f{ ˆf lim kñ8 E pf ^ kq{ ˆf ď 1. Finally, since ˆf P C and C is bounded in probability, we must have ˆf ă `8 almost surely. Hence 1{ ˆf is also in L 0``. (f) For a general C which is convex, closed and bounded in probability and C X L 0`` H, we first pick any g P C X L 0`` and define C g : f{g f P C (. Obviously this set C g contains the constant 1, and is convex, closed and bounded in probability as well. Now we apply our previous arguments for the solid set C g,1 : f f ď h for some h P C g( and get an element ĥ in C g,1 such that Erf{ĥs ď 1 for all f P Cg,1. Clearly this ĥ has to be an element in Cg, otherwise we would find a h P C g with h ě ĥ and P rh ą ĥs ą 0, which would yield that Erf{ĥs ą 1. Hence, ĥ has the form ĥ ˆf{g for some ˆf P C. Finally, since for any f P C, E f{ ˆf E p f ˆf g q{p g q E p f g q{ĥ ď 1, we see that ˆf P C satisfies all the requirements we want. Solution 14.4 (a) The decomposability simply means that the market modeled by X has the switching property. More precisely, the investors on this market are allowed to switch their portfolios at any time t P r0, T s from X to X 1 if the event A, which is observable up to time t, happens, otherwise they will keep their positions. (b) Let S be a semimartingale. Clearly then any process in 1 ` GpΘ 1 admpsqq is adapted, RCLL, nonnegative and starts at 1. As 0 is in X 1 Θ 1 adm psq, we must have 1 P X 1 which means that X 1 contains a strictly positive process. The convexity follows from the convexity of GpΘ 1 adm psqq. To show the decomposability, pick X 1 ` ϑ1 S and X 1 : 1 ` ϑ 1 S from X 1. For any t P r0, T s and A P F t, we define ϑ r X : 1 r0,ts ϑ ` 1 pt,t s p1 t A X ϑ 1 ` 1 t 1 A cϑq. It is easy to see that ϑ r is predictable and S-integrable. Furthermore, we have 1 ` pϑ r Xt_s Sq s 1 A cx 1 s ` 1 A Xt 1 X t^s for all s P r0, T s, which implies that ϑ r P Θ 1 adm and p1 A cx X s ` 1 1 t_s A X X t 1 t^s q spr0,t s is in X 1. (c) Since X is a wealth process set, it is clear that X T is convex and X T X L 0`` H. So, if X T is closed in probability and satisfies NUPBR, using the result from the previous exercise we can find an X p T P X T which corresponds to the final value of a wealth process X p such that ErX T { X p T s ď 1 holds for all X P X. Now for any X P X, for any t P r0, T s, the decomposability ensures that the process X Y s : p t_s X t^s, s P r0, T s px t Updated: December 19, / 5

5 belongs to X (choose A Ω). So from ErY T { p X T s ď 1 it follows that ErX t { p X t s ď 1. Hence, for t ď r in r0, T s, A P F t, for any strictly positive process X 1 P X, since Y s 1 A c p Xs ` 1 A X 1 t_s X 1 t px t^s, s P r0, T s is an element in X, and Y r 1 A c ` 1 A X 1 r X 1 t px t the inequality ErY r { p X r s ď 1 translates into E 1 A px 1 r{ p X r q{px 1 t{ p X t q ď P ras, which in turn implies that E pxr{ 1 X p r q{pxt{ 1 X p t q F t ď 1 and consequently X 1 { X p is a supermartingale for all strictly positive X 1 and we can extend this result to all X P X. Finally note that this X p is strictly positive so that 1 X x is R-valued, and since X p 1 t ă `8, is strictly xx positive. Updated: December 19, / 5