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Unformatted text preview: 1. Find dXt and check if X; is a martingale, where
(a) Xi : (em“2f;
(b) X; : 2tBteB‘ w f; sdBS; Solution. Many different ways to do the questions. Here is one way. (a) Let Y, = Bt — 73/2 and ﬂax} = e29“. So
ext = de/E) : f’{K)dlﬁ+%f”(K)d(Y}t = mamas, —t/2)+% X462Y‘dt = 2e2y3dBtIuewtdt
Hence, X: is NOT a martingale.
(b) deBr : eBtttBt + %eB"dt, and
dXt : 2[(BteB‘)dt + td(BteB*)] a MB,
: 2BteB‘dt + 2t[B,deBt + eB*dBt + d(B, (53),] — totst
: 2BteB‘dt + 2t[Bt{eB‘dBt + éeB‘dt) + eBtdBt + eBtdt] a MB,
: (213,63: + 2:3,.23: + teB‘)dt + [2tBteB‘ + 2t?» — t]dBt
Hence, Xt is NOT a martingale. 2. For a continuous semimartingale Z, denote 8(Zt) as the stochastic exponential of Z, i.e., E (Z) satisﬁes 2 (SIZE)de and 2 For any two continuous semirnartingaies X and Y, show that
(a) 8(Xt)5(Yt) = 5 (X22 + Y: + (Xayltl
(b) (6%))” : 8 (—Xt t (XM Proof. Approach I:
(a) Since d5£Xt) = £(X¢)dXt anci c3806) : 80/1)th we have d‘(8{Xt)€(Yt)) : 8(Xt)d5(Yt) + £(Yt)d5(Xt) + d(E(X),£(Y))t
: 5(Xt)5(Yt)dYt + 5(K)5(Xt)dXt + £(Xt)5(Yt)d(Xa 1%
: 8(Xt)5(Yt)d (Xt + K: + (X,Y)t).
That is, Z S (X: + Y} “i” (X, .
(b) StXt) 52(Xt) 2' 63(Xt) €(Xt) 5(Xt)
1
: “Xadpxt + 00,). That is, (8(X¢))’1 = .9 (—X, + (20,). d( 1 ) : _1 d€(Xt)+l.ﬂ£;2)82(Xt)d(X)t:_"§£+d<X>t Approach II: Solving the SDE to get: €(Zt) = eZ‘_%(Z)*. (a) LHS‘ : ext‘iyi‘iixlréiylt, and RHS : eX¢+Y*+(X’Y)t‘%<X+Yl*. The result follows from
(X + Ylt = (Xlt + (Y): + 2091/):
(b) LHS z e_X‘+(X)’, and HHS = eXr+<X>t%<X>t : e’X*+%<X}‘. 3. Let y; = f; Bats, and Wt : f; sign(Bs)st. Show that try, = 1/t'+ 2mm. Solution. Now {11”} : BtdBt, Y; I i t), (21’; +15) = Bf, «1214+ : Btsign(Bt). Also,
th = sign(Bt)dBt. Hence, V t “F I : BtdBt I 4. Solve for X; from the following SDE: ext = ca) — Xt)dt + odBt, Xe : wo Solution. One can solve this directly. dXt + chit = cbdt + 00113:
Alternatively, set Yt = X; — b, then d1”; = —thdt + 0033;, Whose solution is Y; : e”th + t
oe’ct/ ecsst. Hence,
0 t
Xi = Y} + b = b+ e'°t(xo — b) + (re—“if ecsst.
0 5. Let Xt = 3.: + sint + %t2, where 13,; is a PBrownian motion. Let Q be an equivalent measure to
CM? P such X: is a Q—Brownian motion. Give At = dP Ft Solution. Note that dXt : dB; + t+ cost (it : 0! Bi + t s + coss ds . Hence,
0 t 1 it
At : exp{ej {3—1—0033}st — 5] [3+ Goes]2 (13}. I
o 0
6. Let X15 and Yt be continuous semimartingales. Deﬁne the Stratonovich Integral
‘3 i 1 1
f XSOdI/s 1:] Xsdn+§(X,Y)t 4:} XgOiniz o e (a) Show that Stratonovich integrals obey the change of variables formula from ordinary calculus: rrBi):f(Bo)+/D f’(Bs}ost <:> dnaizf'rBaodBi. (you should assume that f is three times continuously differentiable). [Remark This makes Stratonovich integrals more convenient than Ito integrals for some pur—
poses, especially if one is trying to convert a result from ordinary calculus into stochastic
calculus. This sounds appealing, but there are drawbacks which mean that Ito integrals are
almost everyones preferred choice, in practice. For example, the integrands have to be semi
martingales, which is much more restrictive than simply being predictable. And unlike Ito
integrals, Stratonovich integrals typically arent martingalesl (b) Let Hg be a continuous semimartingale. For 0 = to 3 t1 g 5 tn : t, recall that
n t
In :: ZHmlUB}: — BM) Hf Hsng, in prob.
i=1 0 Show that
n H . H . t
J” :: Z “ﬁt—ELLE“ — Btiil) “4/1 H5 0 st, in prob.
i=1 2 0
Solution.
(a) We need to show that
1
df(Btl : fl(Bt)dBt + $150“le Bltv The LHS is alf(Bt) : f’(Bt}dBt + é—f”(Bt)dt. Now, df’{Bt) : f”(Bt)dBt + gfwsadt, hence,
a:(f'{B),B)t : f”(Bt)alt. Thus, RHS : f’(Bt)an + 5'13th : LHS. . l T’ t 1 t Jn = In, + E 2(Hti * Hti_1)(Bti — Bt171)4>\/ H.5st ‘5' Bh :/ HS C‘st lIl prob.
i=1 0 0 ...
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This note was uploaded on 02/01/2012 for the course MATH 5010 taught by Professor D during the Spring '11 term at HKU.
 Spring '11
 D

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