4
Itˆ o integral, Itˆ o’s formula and SDE’s
2, if 0 ≤ t ≤ 1, if 1 < t ≤ 3, 1. (a) Let X(t) = 3, −5, if 3 < t ≤ 4. or in oneR formula X(t) = 2I[0,1] (t) + 3I(1,3] (t) − 5I(3,4] (t). Give the Itˆo integral 04 X(t)dB(t) as a sum of random variables, give its distribution, specify the mean and the variance. Solution. Z 4
X(t)dB(t) =
0
Z 1
2dB(t) +
Z 3
0
3dB(t) +
Z 4
1
−5dB(t)
3
= 2(B(1) − B(0)) + 3(B(3) − B(1)) − 5(B(4) − B(3)). This can be seen as the sum of independent Normal random variables: Z 4
X(t)dB(t) = 2N (0, 1) + 3N (0, 2) − 5N (0, 1)
0
= N (0, 22 ) + N (0, 2(32 )) + N (0, (−5)2 ) = N (0, 47) Thus, the Itˆo integral and variance 47.
R4 0
X(t)dB(t) is Normally distributed with mean 0
2. Give the values of α for which the stochastic integral 01 s−α dB(s) is defined. In the case when the integral is defined, give its mean and variance. R
Solution. Stochastic integral 01 s−α dB(s) is defined if and only if 01 s−2α ds < ∞ (see Proposition on p. 34), i.e. 2α
Z 1
E
s−α dB(s)
R
= 0,
0
Z 1
V ar
s
−α
Z 1
dB(s)
= E
0
s
−α
2 !
dB(s)
=
0
=
Z 1 0
E s−2α ds
0
"
s−2α ds =
Z 1
s1−2α 1 − 2α
#1
= 0
1 . 1 − 2α
3. Using Itˆo’s formula, find the following stochastic differentials dXt and give their integral representations. (a) Xt = eBt
Solution. Take f (x) = ex . Then f 0 (x) = f 00 (x) = ex . So we have 1 dXt = d(eBt ) = f 0 (Bt )dBt + f 00 (Bt )dt 2 1 = eBt dBt + eBt dt. 2 Integral representation: Bt
e
Z t
1 Z t Bs e dBs + e ds = e + 2 0 0 Z t Z t 1 = 1+ eBs dBs + eBs ds. 2 0 0 B0
Bs
(b) Xt = sin(Bt ) Solution. Take f (x) = sin(x). Then f 0 (x) = cos(x) and f 00 (x) = − sin(x). So we have 1 dXt = d(sin(Bt )) = cos(Bt )dBt + (− sin(Bt ))dt 2 1 = cos(Bt )dBt − sin(Bt )dt. 2 Integral representation: sin(Bt ) = sin(B0 ) + =
Z t 0
Z t 0
cos(Bs )dBs −
1Z t sin(Bs )ds 2 0
1Z t cos(Bs )dBs − sin(Bs )ds. 2 0
(c) Xt = sin(Bt2 ) Solution. Take f (x) = sin(x2 ). Then f 0 (x) = 2x cos(x2 ) and f 00 (x) = 2 cos(x2 ) − 4x2 sin(x2 ). So we have 1 2 cos(Bt2 ) − 4Bt2 sin(Bt2 ) dt 2 2 = 2Bt cos(Bt )dBt + cos(Bt2 ) − 2Bt2 sin(Bt2 ) dt.
d(sin(Bt2 )) = 2Bt cos(Bt2 )dBt +
Integral representation: sin(Bt2 ) = sin(B02 ) + = 2
Z t 0
Z t 0
2Bs cos(Bs2 )dBs +
Bs cos(Bs2 )dBs +
Z t 0
Z t 0
cos(Bs2 ) − 2Bs2 sin(Bs2 ) ds
cos(Bs2 ) − 2Bs2 sin(Bs2 ) ds.
(d) Xt = t + Bt2 . Solution. d(t + Bt2 ) = dt + d(Bt2 ). For d(Bt2 ), take f (x) = x2 . Then f 0 (x) = 2x and f 00 (x) = 2. So we have 1 d(Bt2 ) = 2Bt dBt + (2)dt 2 = 2Bt dBt + dt. Hence, d(t + Bt2 ) = dt + d(Bt2 ) = dt + 2Bt dBt + dt = 2Bt dBt + 2dt. Integral representation: t+
Bt2
= 0+ = 2
Z t 0
4. Show that
Rt 0
B02
+2
Z t 0
Bs dBs + 2
Z t
ds
0
Bs dBs + 2t.
Bs dBs , 0 ≤ t ≤ T , is a martingale
(a) by using its closed form expression (b) by checking conditions for Itˆo integral to be a martingale Solution. (a) From Question 3(d) (or Example on p.37) we have Z t 0
1 Bs dBs = (Bt2 − t). 2
Let Mt = 12 (Bt2 − t). We check the integrability and the martingale property of Mt . Integrability: 1 E|Bt2 − t| 2 1 ≤ (E|Bt2 | + t) by triangular inequality 2 1 (t + t) = t < ∞. = 2
E|Mt | =
Martingale property: We first compute E (Bt2 |Bu , u ≤ s) for s < t,
E Bt2 |Bu , u ≤ s = E (Bs + (Bt − Bs ))2 |Bu , u ≤ s
= E Bs2 + 2Bs (Bt − Bs ) + (Bt − Bs )2 |Bu , u ≤ s
= Bs2 + 2Bs E (Bt − Bs |Bu , u ≤ s) + E (Bt − Bs )2 |Bu , u ≤ s
= Bs2 + 2Bs E (Bt − Bs ) + E (Bt − Bs )2
by independence
= Bs2 + 2Bs (0) + (t − s) as Bt − Bs ∼ N (0, t − s) = Bs2 + t − s. Then, for s < t, E (Mt |Mu , u ≤ s) = = = = Thus,
Rt 0
1 2 E (B − t)|Bu , u ≤ s 2 t 1 2 E Bt |Bu , u ≤ s − t 2 1 2 Bs + t − s − t 2 1 2 Bs − s = Ms 2
Bs dBs = 12 (Bt2 − t) is a martingale.
(b) Itˆo integral 0t Bs dBs , 0 ≤ t ≤ T is a martingale if Proposition at top of p. 38). Since we have R
Z T 0
the Itˆo integral
Rt 0
E(Bs2 )ds
=
Z T 0
s2 sds = 2 "
#T
= 0
RT 0
E(Bs2 )ds < ∞ (see
T2 < ∞, 2
Bs dBs is a martingale.
5. Show that eBt −t/2 is a martingale by using Ito’s formula for the function ex−t/2 and properties of Ito’s integral. Solution. Let Xt = Bt and f (x, t) = ex−t/2 . We use Ito’s formula for f (Xt , t) (see p. 41): df (Bt , t) = Since
∂ex−t/2 ∂x
= ex−t/2 ,
∂f ∂f 1 ∂ 2f (Bt , t)dBt + (Bt , t)dt + (Bt , t)dt ∂x ∂t 2 ∂x2 ∂ 2 ex−t/2 ∂x2
= ex−t/2 and
∂ex−t/2 ∂t
= − 21 ex−t/2 , we have
1 1 d(eBt −t/2 ) = eBt −t/2 dBt − eBt −t/2 dt + eBt −t/2 dt = eBt −t/2 dBt 2 2 Z t
⇒ eBt −t/2 = 1 +
0
eBs −s/2 dBs
Using exponential moment of Normal distribution (Theorem 3 p. 9), Z T 0
E(eBt −t/2 )2 dt =
Z T 0
E(e2Bt −t )dt =
Z T 0
e2t−t dt < ∞
thus the Ito integral 0t eBs −s/2 dBs is a martingale (see proposition at very top of p. 38). Because a martingale plus a constant is still a martingale (see question 8, section 3 of Exercise Book), eBt −t/2 is a martingale. R
6. Solve the SDE dXT = 0.2dt + 0.4Bt , X0 = 0 Give the distribtuion of Xt . Give the probability density function for X1 . Solution. The meaning of a stochastic differential is in its integral form, Z t 0
dXs =
Z t
0.2ds +
0
Z t 0
0.4dBs
⇒ Xt − X0 = 0.2(t − 0) + 0.4(Bt − B0 ) ⇒ Xt = 0.2t + 0.4Bt d
d
Hence Xt = N (0.2t, 0.16t), so X1 = N (0.2, 0.16) with probability density function (see p.6) fX1 (x) = √
(x−0.2)2 1 e− 0.32 . 2π0.4
7. Solve the SDE dXt = 0.2Xt dt + 0.4Xt dBt , X0 = 1 Give the distribution of Xt . Give the probability density function for X1 . Solution This is the Black-Scholes PDE. Ito’s formula with f (x) = ln(x) gives 1 1 dXt − (dXt )2 2 Xt 2Xt 1 1 = (0.2Xt dt + 0.4Xt dBt ) − 0.16Xt2 dt Xt 2Xt2 = 0.2dt + 0.4dBt − 0.08dt = 0.12dt + 0.4dBt
d(ln Xt ) =
With Yt = ln Xt we have in integral form Yt = Y0 + 0.12t + 0.4Bt , so taking exponential on both sides we have Xt = exp(Y0 + 0.12t + 0.4Bt ) = X0 exp(0.12t + 0.4Bt ) = exp(0.12t + 0.4Bt )
d
d
This has a lognormal distribution, Xt = LN (0.12t, 0.16t). Hence X1 = LN (0.12, 0.16) and its probability density function is given by (see p.23) (ln x−0.12)2 1 fX1 (x) = √ e− 0.32 2π0.4x
8. Solve SDE dXt = −Xt dt + dBt , X0 = 1 Solution This is the Ornstein-Uhlenbeck process. Use transformation Yt = Xt et and the product rule (Section 9.5, p. 42) dYt = et dXt + Xt d(et ) = et (−Xt dt + dBt ) + et Xt dt = et dBt Hence, Z t
Yt = Y0 + ⇒ Xt et = 1 +
Z 0
0 t
es dBs
es dBs
⇒ Xt = e−t + e−t
Z t 0
es dBs
9. The price of stock is given by St = 10eBt , 0 ≤ t ≤ 1, t denotes time in years. (a) Derive the SDE for St . Solution Use Ito’s formula with function f (x) = 10ex , then dSt = 10eBt dBt + 5eBt dt 1 = St dBt + St dt, S0 = 10 2 (b) Find the mean of St Solution St = 10eBt = eln 10+Bt , thus St is lognormally distributed and its mean is E[St ] = E[10eBt ] = 10E[eBt ] = 10et/2
(c) Find the probability that the stock at the end of the year will outperform its mean, i.e. find the probability P (S1 > E(S1 )). Solution P (S1 > E(S1 )) = = = =
P (S1 > 10e1/2 ) = P (10eB1 > 10e1/2 ) P (eB1 > e1/2 ) = P (B1 > 1/2) 1 − P (B1 ≤ 1/2) 1 − Φ(1/2) = Φ(−1/2)
where Φ denotes the CDF of a standard Normal (N (0, 1)).
