STK4510 - Eksamen 2009
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Exercise 1
(a)
Show that Brownian motion Bs is Ito integrable on any interval [0, t].
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Definition 3.1.4 (Oksendal): Ito integrability.
A function f is Ito integrable, denoted f ∈ V = V[S, T ] if the following three
properties are satisfied:
(i) (t, ω) 7→ f (t, ω) is B × F -measurable.
(ii) f (t, ω) is Ft -adapted.
Z T
(iii) E[
f (t, ω)2dt] < ∞.
S
We have to verify these three properties for f (s, ω) = Bs (ω). Properties (i)
and (ii) are obvious for Brownian motion. Brownian motion Bt is Ft -adapted
and B × F -measurable. For property (iii), finite variance, we calculate the
result:
Z t
Z t
Z t
t2
2
2
E[ Bs ds] =
E[Bs ]ds =
sds =
< ∞.
2
0
0
0
The three properties are verified for BM, so Bs is Ito integrable for any
interval [0, t].
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(b)
Rt
Apply the Ito Formula to calculate 0 Bs dBs . Find the expectation and variance of the integral.
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We apply Ito’s formula to the function g(t, x) = x2 where x = Bs . Finding
the partial derivatives:
∂g
= 0,
∂t
∂g
= 2x,
∂x
∂2g
= 2.
∂x2
∂g
∂g
1 ∂2g
(t, Bt )dt +
(t, Bt )dBt +
(t, Bt )(dBt )2
d g(t, Bt ) =
∂t
∂x
2 ∂x2
1
= 0 + 2Bt dBt + (2)dt
2
= 2Bt dBt + dt
1
Writing out in integral form, and exchanging g(t, Bt ) = Bt2 :
Z t
Z t
2
2
Bt = B0 + 2
Bs dBs +
ds
0
BM starts in 0, so B02 = 0 and
Bt2
Rt
0
0
ds = t.
=2
Z
t
Bs dBs + t
0
Shifting terms and dividing by 2:
Z t
1
1
Bs dBs = Bt2 − t.
2
2
0
Finding the expectation:
Z t
E[ Bs dBs ] = 0
0
since the expectation to stochastic integrals always is 0. Alternatively,
1
1
1
1
1
1
E[ Bt2 − t] = E[Bt2 ] − t = t − t = 0.
2
2
2
2
2
2
For the variance, denoting the integral with I:
"Z
Var[I] = E[I 2 ] − E[I]2 = E
| {z }
=0
Ito isometry:
=
Z
t
E[Bs2 ]ds
=
0
Z
0
2
t
t
Bs dBs
0
t2
sds = .
2
2 #
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(c)
Use directly the definition of Ito integration to show that
Z t
Z t
sdBs = tBt −
Bs ds.
0
0
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Definition 3.1.6 (Oksendal): The Ito Integral.
For f ∈ V[0, t] the Ito integral of f from 0 to t is defined by:
Z t
Z t
f (s, ω)dBs = lim
φn (s, ω)dBs(ω)
n→∞
0
0
where {φn } is a sequence of elementary functions such that
Z t 2 E
f (s, ω) − φn (s, ω) dt → 0 as n → ∞.
0
We choose a partition of [0, t] with 0 = s0 < s1 < . . . < sN = t. We define
the elementary process:
φn (s) =
N
−1
X
j=0
sj 1(sj ≤ s < sj+1).
We must show that φn → s, which means the following expectation must
converge to 0:
Z t 2 E
φn (s) − s ds
0
Inserting the definition of the elementary process:
"Z
#
−1
2
tN
X
E
sj 1(sj ≤ s < sj+1 ) − s ds
0
j=0
— Unknown steps —
N
−1 Z sj+1
X
j=0
— Unknown steps —
sj
(sj − s)2 ds
N −1
1X
(sj+1 − sj )3
3 j=0
3
Telescoping sum.
=
n−1
X
1
max (sj+1 − sj )
(sj+1 − sj )2
≤
3 0≤j≤N −1
j=0
1
max(sj+1 − sj )t2 −→ 0 as n → ∞ or ∆sj → 0
3 j
Equality verified:
Z t
sdBs =
0
Z
n−1
X
t
φn (s)dBs =
0
j=0
(exp 2?)
(exp 2?)
sj (Bsj+1 − Bsj ).
Define Bj = Bsj .
tBt =
N
−1
X
j=0
=
N
−1
X
j=0
=
N
−1
X
j=0
=
N
−1
X
j=0
=
N
−1
X
j=0
sj+1 Bj+1 − sj Bj
sj+1 Bj+1 −sj Bj+1 + sj Bj+1 − sj Bj
sj+1 Bj+1 − sj Bj+1 +
Bj+1 (sj+1 − sj ) +
Bj+1 (sj+1 − sj ) +
N
−1
X
j=0
N
−1
X
j=0
Z
sj Bj+1 − sj Bj
sj (Bj+1 − Bj )
t
sdBs
0
Recall that the Ito integral needs the sample point must the the leftmost point
in the interval. For deterministic integrals, (Riemann/Stieltjes) the sample
point can be arbitrary. Remains to verify the last integral: the following
expectation must converge to 0.

!2 
Z t
N
−1
X
E
Bj+1(sj+1 − sj ) −
Bs ds 
0
j=0
Note that Bj+1 (sj+1 − sj ) =
calculus.

N
−1 Z
X

E
j=0
R sj+1
Bj+1ds by the fundamental theorem of
!2 
Z t
sj+1
Bj+1 ds −
Bs ds 
sj
0
sj
4
We can partition the second integral into N − 1 smaller integrals, and collect
them together.

!2 
N
−1 Z sj+1
X
E
Bj+1 − Bs ds 
j=0
sj
Split the indices:
N
−1
X
"Z
E
i,j=0
si+1
(Bi+1 − Bs )ds
si
Z
sj+1
sj
(Bj+1 − Bs )ds
#
Consider two possibilities: i = j and i 6= j, by symmetry we can collect the
i > j and j > as two times one of the cases. We assume j > i.

!2 
Z sj+1
N
−1 Z si+1 Z sj+1
N
−1
X
X
2
E [(Bi+1 − Bu )(Bj+1 − Bv )] dudv+
E
Bj+1 − Bj ds 
i,j=0
si
sj
j=0
sj
In the first integral, we have, for j > i and u ∈ (i, i + 1) and v ∈ (j, j + 1):
h
i
E Bi+1 Bj+1 − Bi+1 Bv − Bj+1 Bu + Bu Bv = (i + 1) − (i + 1) − u + u = 0
so the entire first integral is equal to 0, since the integrands are 0. (We used
that E[Bs Bu ] = min(s, u). The second integral:

!2 
Z
N
−1
s
j+1
X
E
Bj+1 − Bj ds  =
sj
j=0
N
−1 Z sj+1
X
j=0
sj
Z
sj+1
sj
h
i
E (Bj+1 − Bu ) (Bj+1 − Bv ) dudv
Multiplying and calculating the integrand:
h
i
2
E Bj+1
− Bj+1 Bv − Bj+1 Bu + Bv Bu = sj+1 − v − u + min(u, v) ⇒
N
−1 Z sj+1
X
j=0
sj
Z
sj+1
sj
(sj+1 − v − u + min(u, v)) dudv
*
Assuming now that s = u = v, we get back to the single integral:
N
−1 Z sj+1
N
−1 Z sj+1
X
X
(sj+1 − 2s + s) ds =
(sj+1 − s) ds
j=0
sj
j=0
Combining the summed integrals:
Z
0
sj
t
(t − s)ds
5
???????
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Exercise 2
Consider the process
dSt = σ(t)dBt ,
where σ(t) is a deterministic function being continuous and positive. This
dynamics could be considered as the Bachelier model for stock prices, with
time-dependent volatility.
(a) Define martingale processes, and use this definition to show that St is a
martingale. Formulate the martingale representation theorem, and show that
St satisfies this.
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Definition 3.2.2 (Oksendal): Definition of Martingales.
A stochastic process {Mt } is called a martingale wrt the filtration Ft if
(i) Mt is Ft -measurable for all t.
(ii) E[|Mt |] < ∞ for all t.
(iii) s ≥ t, E[Ms |Ft ] = Mt . (Martingale property).
Writing out St in the integral form.
St = S0 +
Z
t
σ(s)dBs .
0
(i) This does not take any values from after t, so it is Ft -adapted.
Before verifying (ii), we need Cauchy-Schwarz’s inequality: E[|XY |] ≤
1
1
E[X 2 ] 2 E[Y 2 ] 2 .
Z t
Z t
E[|St |] = E S0 +
σ(s)dBs ≤ E[|S0 |] + E σ(s)dBs 0
0
Applying Cauchy-Schwarz’s inequality on the second term with X = 1 and
Y as the integral.
"Z
2 # 21
Z t
21
t
2
≤ E[|S0 |] + E
σ(s)dBs
= |S0 | + E
σ(s) ds < ∞.
0
0
Since σ(s) is a continuous function it is bounded on the interval [0, t], so the
integral is also bounded. Now it remains to show the martingale property,
(iii). Assume t > s.
Z t
E[St |Fs ] = E S0 +
σ(u)dBu Fs
0
6
S0 + E
Z
s
0
Z t
σ(u)dBu Fs + E
σ(u)dBu Fs
s
In the first expectation we have measurability, so we can factor the integral
out of the conditional expectation. In the second term, we have a continuous,
deterministic function with an elementary function representation. We define
the partition: s = t0 < t1 < . . . < tN = t and write the integral as a sum:
"N −1
# N −1
Z t
i
h
X
X
E
σ(u)dBu Fs = E
ej E (Btj+1 − Btj ) Fs
ej (Bj+1 − Bj ) Fs =
s
j=0
j=0
Since Brownian motion has independent increments, it is independent from
the filtration and the conditional expectation becomes a normal expectation,
and the normal expectation of ∆Bj is 0, so this terms vanishes.
= S0 +
Z
s
σ(u)dBu + 0 = S0 +
0
Z
t
σ(u)dBu = Ss .
0
Property (iii) verified: St is a martingale.
Definition 4.3.4 (Oksendal): Martingale Representation Theorem.
Suppose Mt is a martingale and Mt ∈ L2 (P ) (or E[Mt2 ] < ∞). Then there
exists a unique, Ito integrable, stochastic process φ, such that
Z t
Mt = E[M0 ] +
φ(s, ω)dBs
0
To apply the MRT to St we need to show that variance is finite. We have
already shown it is a martingale.
"
2 #
Z t
E[St2 ] = E S0 +
σ(u)dBu
=
0
Z t
hZ t
i2 2
E S0 + 2S0
σ(u)dBu +
σ(u)dBu
0
0
Since the expectation is linear:
Z t
Z t
2 2
E S0 + 2S0 E
σ(u)dBu + E
σ(u)dBu
0
=
S02
+0+
0
Z
t
σ(u)2 du < ∞.
0
7
Hence, St satisfies the conditions of the MRT, so we can write:
Z t
St = E[S0 ] +
φ(s, ω)dBs
0
or,
S0 +
Z
t
σ(s)dBs = S0 +
0
Z
t
φ(s, ω)dBs
0
so φ(s, ω) = σ(s).
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(b)
Use Girsanov’s Theorem to find a probability Q such that exp(−rt)St is a
Q-martingale. Here, r > 0 is the risk-free interest rate. You do not need to
verify your change of probability.
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Using Girsanov’s Theorem, we get a new Brownian motion under Q:
dWt = θt dt + dBt ⇐⇒ dBt = dWt − θt dt
for some unknown θt . We get:
dSt = σ(t) dWt − θt dt = −θt σ(t)dt + σ(t)dWt
Lecture notes: It is a martingale measure if the discounted expected value is
a martingale. We check with Ito: f (x, t) = xe−rt .
d e−rt St = −re−rt St dt + e−rt dSt
We use dSt under Q.
= −re−rt St dt + e−rt − θt σ(t)dt + σ(t)dWt
−rt
=e
− θt σ(t) − rSt dt + σ(t)e−rt dWt
This is a martingale if the drift term is 0.
−θt σ(t) − rSt = 0 =⇒ θt = −
Using this for θt , we have the Q dynamics for St :
dSt = −
rSt
dt + σ(t)dWt
σ(t)
8
rSt
σ(t)
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(c)
Find St explicitly under Q, and show that St is normally distributed. Find
mean and variance.
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Solving the SDE.
rSt
dt + σ(t)dWt
σ(t)
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(d)
Find the price at time zero, P0 of a digital option on St which pays 1 at time
T if ST > K, and zero otherwise. Here, K > 0 is a constant.
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dSt = −
⌣
¨
9
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Exercise 3
(a) State the Black & Scholes formula for call options.
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Theorem 4.3 (Benth): B&S Pricing Formula.
For a call option (ST − K)+ the price Pt at time t is given by:
Pt = St Φ(u1 ) − Ke−r(T −t) Φ(u2 )
where
ln(St /K) + (r − 21 σ 2 )(T − t)
√
u2 =
σ T −t
ln(St /K) + (r + 21 σ 2 )(T − t)
√
,
u1 =
σ T −t
where St is the stock price at time t, K is the strike price, T is the strike
time, r is the interest rate, σ is the market volatility and Φ(·) is the normal
cumulative distribution function.
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(b) What happens with the option price when the volatility tends to infinity
or 0?
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As we can see from the formula above, the volatility only affects the
parameters u1 and u2 . To simplify the notation we define a := ln(St /K) and
τ := (T − t). First, when σ → ∞:
a + rτ + 21 σ 2 τ
a/σ + rτ /σ + 12 στ
√
√
= lim
= ∞.
σ→∞
σ→∞
σ τ
τ
lim u1 = lim
σ→∞
This means Φ(u1 ) → 1 as σ → ∞.
a/σ + rτ /σ − 12 στ
√
lim u2 = lim
= −∞.
σ→∞
σ→∞
τ
Which means Φ(u2 ) → 0 as σ → ∞. From the formula of the option price
we get:
Pt = St (1) − Ke−r(T −t) (0) = St .
The price of the option is always the same as the stock price. Now when
σ → 0:
rτ
1 √
a
lim u1 = lim √ + lim √ + lim σ τ = ∞
σ→0 σ τ
σ→0 2
σ→0
σ→0 σ τ
| {z } | {z } | {z }
→∞
→∞
10
→0
Similarly,
a
rτ
1 √
lim u1 = lim √ + lim √ − lim σ τ = ∞
σ→0
σ→0 σ τ
σ→0 σ τ
σ→0 2
| {z } | {z } | {z }
→∞
→∞
→0
so when σ → 0 both u1 → ∞ and u2 → ∞, so Φ(u1 ) = Φ(u2 ) = 1, and the
option price becomes:
Pt = St (1) − Ke−r(T −t) (1) = St − Ke−r(T −t) .
Possibility of negative prices? No. Since the volatility is 0, we can perfectly
predict the evolution of the stock price, so if the stock price would become
less than the strike price with certainty, then the option would be worthless.
11