brownian motion: heuristic motivationpages.stat.wisc.edu/~yzwang/ito.pdffigure 7: geometric brownian...
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1
Brownian motion: Heuristic motivation
The lognormal distribution goes back to Louis Bachelier’s (1990)
dissertation at the Sorbonne called The Theory of Speculation. Bache-
lier’s work anticipated Einstein’s (1905) theory of Brownian motion.
In 1827, Robert Brown, a Scottish botanist, observed the erratic un-
predictable motion of pollen grains (in water) under a microscope.
In 1905 Einstein understood that the movement was due to bom-
bardment by water molecules and he developed a mathematical the-
ory. Later, Norbert Wiener, an M.I.T. mathematician, developed a
more precise mathematical model of Brownian motion, now called
the Wiener process.
Random Walk
Suppose εj are i.i.d. standard normal random variables. Partition
interval [0, T ] into n subintervals of length ∆ = T/n. Let tk = k ∆,
2
B1
B2
-0.5 0.0 0.5 1.0
-0.4
-0.2
0.0
0.2
0.4
0.6
Brownian Motion
Figure 1: Simulated Brownian motion
B1
B2
-0.20 -0.15 -0.10 -0.05
0.0
0.05
0.10
0.15
0.20
0.25
Brownian Motion: the first 50 steps
Figure 2: The first 50 steps of Brownian motion
3
t
B(t
)
0.0 0.2 0.4 0.6 0.8 1.0
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
Two sample paths of Brownian motion
Figure 3: Simulated 2000 step Random Walk
t
S(t
)
0.0 0.2 0.4 0.6 0.8 1.0
12
34
Two sample paths of geometric Brownian motion
Figure 4: Simulated 2000 step geometric Random Walk
4
k = 1, · · · , n, and define normalized random walk
Bntk
=
√
T
n
k∑
`=1
ε`,
t
Ran
dom
Wal
k
0.0 0.2 0.4 0.6 0.8 1.0
-0.6
-0.4
-0.2
0.0
0.2
Figure 5: Simulated 21 step Random Walk
Plot (tk, Bntk
).
Properties of {Bntk
: tk = k/n ∈ [0, T ], k = 1, · · · , n}
1. Bn0 = 0,
2. Independent increment: Bntk−Bn
tjis independent of Bn
ti, ti ≤ tj ≤
tk,
3. Normality: Bntk
are jointly normal, Bntk∼ N(0, tk), Bn
tk− Bn
tj∼
5
N(0, tk − tj), and Cov(Bnti, Bn
tj) = min(ti, tj).
Proof. ti ≤ tj ≤ tk ↔ i ≤ j ≤ k.
ε1, · · · , εi, εi+1, · · · , εj, εj+1, · · · , εk
Bnti
=
√
T
n
i∑
`=1
ε`
relies on ε1, · · · , εi,
Bntk− Bn
tj=
√
T
n
k∑
`=j+1
ε`
depends on εj+1, · · · , εk. Thus, they are independent.
E[Bntk− Bn
tj] =
√
T
n
k∑
`=j+1
Eε` = 0
V ar[Bntk− Bn
tj] =
T
n
k∑
`=j+1
V ar(ε`) =T
n(k − j) = tk − tj.
Cov(Bnti, Bn
tj) = Cov(Bn
ti, Bn
tj− Bn
ti+ Bn
ti)
= Cov(Bnti, Bn
tj− Bn
ti) + Cov(Bn
ti, Bn
ti)
= V ar(Bnti) = ti
As n → ∞ [or ∆ → 0],
{Bntk
: tk = k/n ∈ [0, T ], k = 1, · · · , n} −→ {Bt : t ∈ [0, 1]}.
6
t
Ran
dom
Wal
k
0.0 0.2 0.4 0.6 0.8 1.0
-1.0
-0.5
0.0
0.5
Figure 6: Brownian motion: Simulated 300 step Random Walk
Brownian motion
The continuous time stochastic process Bt, t ∈ [0, T ] is called
standard Brownian motion (or Wiener process).
Properties of {Bt : t ∈ [0, T ]}
1. B0 = 0,
2. Independent increment: Bt−Bs is independent of Br, r ≤ s ≤ t,
3. Normality: Bt − Bs ∼ N(0, t − s).
(a). For any 0 < s1 < s2 < · · · < sm, Bsjare jointly normal, (b).
Cov(Bs, Bt) = min(s, t).
7
Geometric Brownian motion
The most common model by far in finance is one where the security
price is based on a geometric Brownian motion.
If one invests $1,000 in a stock selling for $1 and it goes up to $2,
one has the same profit, namely $1,000, as if one invests $1,000 in a
stock selling for $100 and it goes up to $200.
Our opportunities for gain are the same for both. It is the pro-
portional increase one wants. that is, it is ∆St/St matters, not ∆St.
Therefore, one set ∆St/St to be the quantity related to a Brownian.
Different stocks have different volatilities. In addition, one expects a
mean rate of return µ on ones investment that is positive. In fact,
one expects the mean rate of return to be higher than the risk-free
interest rate because one expects something in return for undertaking
risk.
For modeling stock price, the following model is frequently used
St+∆ − St
St= µ ∆ + σ
√∆ εt,
8
t
Sto
ck P
rice
0.0 0.2 0.4 0.6 0.8 1.0
1.0
1.2
1.4
1.6
1.8
Figure 7: Geometric Brownian motion
As ∆ → 0, it is convenient to write
dSt
St= µ dt + σ dBt.
Such a continuous process is an idealization of the discrete version
and is called a geometric Brownian motion.
St = S0 exp{(µ − σ2/2) t + σ Bt}.
Given S0,
log ST ∼ N(log S0 + (µ − σ2/2) T, σ2 T ).
E(ST ) = S0 exp{µ T},
9
V ar(ST ) = S20 exp{2 µ T}
[
exp{
σ2 T}
− 1]
.
Proof. For X ∼ N(ν, τ 2), its moment generating function
E(
ea X)
= exp(a ν + a2 τ 2/2).
Let X = log ST or ST = eX . Then
E(ST ) = E(eX), V ar(ST ) = E(S2T ) − [E(ST )]2.
10
Brownian motion
A process Wt is called a Brownian motion (or Wiener process), if
(1). W0 = 0;
(2). for r < s < t, Wt − Ws and Wr are independet, that is, inde-
pendent increments.
(2). for s < t, Wt−Ws follows a normal distribution with mean zero
and variance σ2 (t − s), where σ is a positive constant.
For Brownian motion Wt, let Ft = σ{Ws, s ≤ t}, t ∈ [0, 1].
Martingale: Wt is a martingale:
E[Wt|Fs] = Ws.
Proof.
E[Wt|Fs] = Ws + E[Wt − Ws|Fs] = Ws + E[Wt − Ws] = Ws.
Remark. E[W 2t ] = t, and W 2
t is a submartingale:
E[W 2t |Fs] ≥ W 2
s
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Quadratic Variation:
[W, W ]t = limk→∞
k∑
i=1
(
Wsi− Wsi−1
)2, in probability,
where the limit is taken for any sequence of partitions 0 = s0 < s1 <
s2 < · · · < sk = t with supj(sj − sj−1) → 0 as k → ∞.
[W, W ]t = t
Proof. We prove t = 1. Let si = i/k. Then Wsi− Wsi−1
∼i.i.d
N(0, 1/n). Let Zi = k (Wsi−Wsi−1
)2. Then Zi are i.i.d. with mean
1 and variance 2. By LLN, we getk
∑
i=1
(Wsi− Wsi−1
)2 =
∑ki=1 Zi
k→ 1.
Thus [W, W ]1 = 1.
Remark For smooth function H(t), if si is a partition of [0, t],
thenk
∑
i=1
|H(si) − H(si−1)| →∫ t
0
|H ′(s)| ds
This immediately impliesk
∑
i=1
|H(si)−H(si−1)|2 ≤ maxi
|H(si)−H(si−1)|k
∑
i=1
|H(si)−H(si−1)| → 0
12
W 2t − t is a martingale:
E[W 2t − t|Fs] = W 2
s − s
Proof.
E[W 2t − t|Fs] = E[(Ws + Wt − Ws)
2|Fs] − t
= E[W 2s + (Wt − Ws)
2 + 2 Ws (Wt − Ws)|Fs] − t
= W 2s + E[(Wt − Ws)
2|Fs] + 2 Ws E[(Wt − Ws)|Fs] − t
= W 2s + E[(Wt − Ws)
2] − t
= W 2s + t − s − t = W 2
s − s
Remark. In general, given a continuous martingale Mt, its
quadratic variation [M, M ]t is a unique non-decreasing process such
that M 2t − [M, M ]t is a martingale.
13
Stochastic Integral
Heuristic motivation: If one purchases ∆0 shares (possibly a neg-
ative number) at time t0 = 0, then changes the investment to ∆1
shares at time t1, then changes the investment to ∆2 at time t2, etc.
At time t0 one has origin wealth Wt0. One buys ∆0 shares and cost
is ∆0 St0. At time t1 one sells ∆0 shares for the price of St1 per share,
and so one wealth is now Wt0 + ∆0 (St1 −St0). One now pays ∆1 St1
for ∆1 shares at time t1 and continues. So one’s wealth at time t = tn
will be
Wt0 + ∆0 (St1 − St0) + ∆1 (St2 − St1) + · · · + ∆tn−1(Stn − Stn−1
),
which is the same as
Wt0 +
∫ t
0
∆s dSs
where ∆s = ∆ti for ti ≤ s < ti+1. In other words, our wealth is
given by a stochastic integral with respect to the stock price. The
requirement that the integrand of a stochastic integral be adapted is
very nature: we can not base the number of share we own at time
s on information that will not be available until the future. The
14
continuous time model of finance is then that the security price is
given by geometric Brownian motion, that there is no transaction
costs, but one can trade as many as one wants and vary amount held
in a continuous time fashion. This is clearly not the way the market
actually works, for example, stock prices are discrete, but this model
has proved to be a very good one.
∫ t
0
Hs dSt
represents the net profit or loss if St is the stock price and Ht is the
number of shares held at time t. This is the direct analogy of
∑
∆i (Sj+1 − Sj).
15
Definition:
Predictable process: a stochastic process H(t) = H(t, ω) is said
to be predictable, if for each t, H(t) is Ft measurable.
Case A: Simple Process:
H(t) = ξ 1(a,b](t)
where ξ is Fa measurable.
Stochastic Integral Definition:
Xt =
∫ t
0
H(s) dWs = ξ (Wt∧b−Wt∧a) =
ξ (Wb − Wa), t > b
ξ (Wt − Wa), a ≤ t ≤ b
0, t < a∫ t
0 H(s) dWs is martingale: for t < T ,
E[XT |Ft] = Xt
Proof. Case 1: for t ≥ b: XT = Xt = ξ (Wb − Wa). Thus,
E[XT |Ft] = ξ (Wb − Wa) = Xt
Case 2: for t ≤ a: Xt = 0, XT = ξ (Wb − Wa) if T > b or
ξ (WT − Wa) if a < T < b or 0 if T ≤ a.
E[Wb − Wa|Ft] = 0, E[WT − Wa|Ft] = 0
16
Thus,
E[XT |Ft] = 0 = Xt
Case 3: for a < t < b: Xt = ξ (Wt − Wa), XT = ξ (Wb − Wa) if
T > b or ξ (WT − Wa) if a < t < T < b.
E[Wb − Wa|Ft] = Wt − Wa, E[WT − Wa|Ft] = Wt − Wa
Thus,
E[XT |Ft] = ξ (Wt − Wa) = Xt
17
Quadratic Variation:
[X, X ]t =
∫ t
0
H2(s) ds = ξ2 (t∧b−t∧a) =
ξ2 (b − a) t > b
ξ2 (t − a) a ≤ t ≤ b
0 t < a
Proof. Case 1: for t ≥ b: XT = Xt = ξ (Wb − Wa), and [X, X ]T =
[X, X ]t = ξ2 (b − a). Thus,
E[X2T − [X, X ]T |Ft] = ξ2 [(Wb − Wa)
2 − (b − a)] = X2t − [X, X ]t
Case 2: for t ≤ a: Xt = 0 and [X, X ]t = 0. XT = ξ (Wb −Wa) if
T > b or ξ (WT − Wa) if a < T < b or 0 if T ≤ a, and [X, X ]T =
ξ2 (b − a) if T > b or ξ2 (T − a) if a < T < b or 0 if T ≤ a.
E[X2T − [X, X ]T |Ft] = ξ2 E[(Wb − Wa)
2 − (b − a)|Ft
= E[(Wb − Wa)2 − (b − a)] = 0
E[X2T − [X, X ]T |Ft] = ξ2 E[(WT − Wa)
2 − (T − a)|Ft
= E[(WT − Wa)2 − (T − a)] = 0
Case 3: for a < t < b: Xt = ξ (Wt − Wa), XT = ξ (Wb − Wa)
if T > b or ξ (WT − Wa) if a < t < T < b. [X, X ]t = ξ2 (t − a),
18
[X, X ]T = ξ2 (b − a) if T > b or ξ2 (T − a) if a < t < T < b.
E[X2T − [X, X ]T |Ft] = ξ2 E[(Wb − Wa)
2 − (b − a)|Ft]
= ξ2{
E[(Wb − Wt)2] + (Wt − Wa)
2
+ 2 (Wt − Wa) E[Wb − Wt|Ft] − (b − a)}
= ξ2{
b − t + (Wt − Wa)2 − (b − a)
}
= ξ2{
(Wt − Wa)2 − (t − a)
}
= X2t − [X, X ]t
E[X2T − [X, X ]T |Ft] = ξ2 E[(WT − Wa)
2 − (T − a)|Ft]
= ξ2{
E[(WT − Wt)2] + (Wt − Wa)
2
+ 2 (Wt − Wa) E[WT − Wt|Ft] − (T − a)}
= ξ2{
T − t + (Wt − Wa)2 − (T − a)
}
= ξ2{
(Wt − Wa)2 − (t − a)
}
= X2t − [X, X ]t
Case B: Step Process:
H(t) = ξ0 1{0}(t) +
m∑
i=1
ξi 1(ai,bi](t)
where ξi is Faimeasurable.
19
Stochastic Integral Definition:
Xt =
∫ t
0
H(s) dWs =
m∑
i=1
ξi (Wt∧bi − Wt∧ai)
[X, X ]t =
∫ t
0
H2(s) ds =
m∑
i=1
ξ2i (t ∧ bi − t ∧ ai)
Martingale property: for t < T ,
E[XT |Ft] = Xt, E[X2T − [X, X ]T |Ft] = X2
t − [X, X ]t
Proof. From Case A, we have that for t < T ,
E[ξi (WT∧bi − WT∧ai)|Ft] = ξi (Wt∧bi − Wt∧ai
)
E[ξ2i (WT∧bi − WT∧ai
)2 − ξ2 (T ∧ bi − T ∧ ai)|Ft]
= ξ2i (Wt∧bi − Wt∧ai
)2 − ξ2 (t ∧ bi − t ∧ ai)
The results are proved by summing over both sides.
Case C: General predictable process: H(t) with
∫
H2(t) dt < ∞
There exist a sequence of step process Hn(t) such that
∫
E|Hn(t) − H(t)|2 dt → 0
20
Stochastic Integral Definition:
Xt =
∫ t
0
H(s) dWs = limn→∞
∫ t
0
Hn(s) dWs, in L2
First show the L2 limit exists.
E
[
{∫ t
0
Hn(s) dWs −∫ t
0
Hm(s) dWs
}2]
= E
[
{∫ t
0
[Hn(s) − Hm(s)] dWs
}2]
= E
[
{∫ t
0
[Hn(s) − Hm(s)] dWs
}2]
=
∫ t
0
E[Hn(s) − Hm(s)]2 ds → 0
21
Martingale property: for t < T ,
E
[∫ T
0
Hn(s) dWs
∣
∣
∣
∣
Ft
]
=
∫ t
0
Hn(s) dWs
E
[
(∫ T
0
Hn(s) dWs
)2
−∫ T
0
H2n(s) ds
∣
∣
∣
∣
∣
Ft
]
=
(∫ t
0
Hn(s) dWs
)2
−∫ t
0
H2n(s) ds
We obtain martingale property by letingt n → ∞
E[XT |Ft] = Xt, E[X2T − [X, X ]T |Ft] = X2
t − [X, X ]t
22
Ito Lemma
∫ t
0
Bs dBs = (B2t − t)/2
Proof. Take t = 1 and si = i/n. Then∫ t
0 Hs dBs is the limit of
n∑
i=1
Bsi−1(Bsi
− Bsi−1)
=1
2
n∑
i=1
[(Bsi+ Bsi−1
) − (Bsi− Bsi−1
)] (Bsi− Bsi−1
)
=1
2
n∑
i=1
[(B2si− B2
si−1) − (Bsi
− Bsi−1)2]
=1
2
n∑
i=1
[(B2si− B2
si−1)] − 1
2
n∑
i=1
(Bsi− Bsi−1
)2
=1
2(B2
1 − B20) −
1
2
n∑
i=1
(Bsi− Bsi−1
)2
→ 1
2(B2
1 − 1)
n∑
i=1
(Bsi− Bsi−1
)2 =1
n
n∑
i=1
[N(0, 1)]2 → 1
Thus,∫ 1
0
Bs dBs = (B21 − 1)/2
23
Ito Formula: Suppose h(x) is twice differentialble, and Xt = X0 +∫ t
0 µs ds +∫ t
0 Hs dBs, with < X, X >t=∫ t
0 Hs ds. Then
h(Xt) = h(X0) +
∫ t
0
h′(Xs) dXs +1
2
∫ t
0
h′′(Xs) d < X, X >s
Ito formula: Suppose a stochastic process X follows
dXt = a(t, Xt) dt + b(t, Xt) dBt,
and H(t, Xt) is a function of Xt and t. Then
dH(Xt, t) =∂H
∂XdX +
∂H
∂tdt +
1
2
∂2H
∂X2d < X, X >t
=∂H
∂XdX +
∂H
∂tdt +
1
2
∂2H
∂X2b2 dt
=
(
∂H
∂Xa +
∂H
∂t+
1
2
∂2H
∂X2b2
)
dt +∂H
∂Xb dB,
where we use the fact
d < X, X >t= b2(t, Xt) dt, or equivalently < X, X >t=
∫ t
0
b2(s, Xs) ds.
Application Example For geometric Brownian motion, take H(t, St) =
log St. Then
∂H
∂S=
1
S,
∂2H
∂S2= − 1
S2,
∂H
∂t= 0.
24
Thus,
dlog St =
(
1
Stµ St +
1
2
−1
S2t
σ2 S2t
)
dt+1
Stσ St dBt = (µ−σ2/2) dt+σ dBt.
Example H(t, Bt) = B2t − t,
d(B2t − t) = 2 Bt dBt.
Heuristic Justification Calculus. Consider a continuous and
differentiable function H(x) of a variable x. If ∆x is a small change
in x and ∆H is the resulting small change in H , it is well known
that
∆H ≈ dH
dx∆x.
In other words, ∆H is approximately equal to the rate of change of
H with respect to x multiplied by ∆x. The error involves terms of
order (∆x)2. If more precision is required, a Taylor series expansion
of ∆H can be used:
∆H =dH
dx∆x +
1
2
d2H
dx2(∆x)2 +
1
6
d3H
dx3(∆x)3 + · · ·
For a continuous and differentiable bivariate function H(x, y) of two
25
variables x and y, the result analogous is
∆H ≈ ∂H
∂x∆x +
∂H
∂y∆y
and the Taylor series expansion of ∆H is
∆H =∂H
∂x∆x+
∂H
∂y∆y+
1
2
∂2H
∂x2(∆x)2+
∂2H
∂x∂y∆x ∆y+
1
2
∂2H
∂y2(∆y)2+· · ·
In the limit as ∆x and ∆y goes to zero,
dH =∂H
∂xdx+
∂H
∂ydy, or H(x, y) =
∫ x
−∞
∫ y
−∞
(
∂H
∂udu +
∂H
∂vdv
)
.
Stochastic calculus. Now H is a function of a stochastic process
Xt and t, and we will consider a stochastic version of differential form
for H(Xt, t). With small increment ∆t in time, let
∆X = Xt+∆t − Xt, ∆H = H(Xt+∆t, t + ∆t) − H(Xt, t).
From Taylor series expansion, we can write
∆H =∂H
∂X∆X+
∂H
∂t∆t+
1
2
∂2H
∂X2(∆X)2+
∂2H
∂X∂t∆X ∆t+
1
2
∂2H
∂t2(∆t)2+· · ·
Xt equation can be discretized as
∆Xt = a(t, Xt)∆t + b(t, Xt)√
∆t εt,
26
or if arguments are dropped,
∆X = a ∆t + b√
∆t ε.
This equation reveals an important difference between the determin-
istic and stochastic situations. When the limiting arguments were
used from Taylor series expansion to differentiable form, term with
(∆x)2 were ignored, because they were second-order terms and are
negligible. For the stochastic case,
(∆X)2 = b2 ε2 ∆t + o(∆t),
which shows that the term involving (∆X)2 has a component that
is of order ∆t and can’t be ignored. The expected value of ε2 ∆t is
∆t, with variance of order (∆t)2. As a result of this, ε2 ∆t becomes
nonstochastic and equal to its expected value of ∆t as ∆t tends to
zero. It follows that (∆X)2 becomes nonstochastic and equals to
b2 dt as ∆t tends to zero. Taking limits as ∆X and ∆t go to zero,
27
we obtain
dH =∂H
∂XdX +
∂H
∂tdt +
1
2
∂2H
∂X2b2 dt
=
(
∂H
∂Xa +
∂H
∂t+
1
2
∂2H
∂X2b2
)
dt +∂H
∂Xb dB.
28
Girsnov theorem
Suppose that Z is a standard normal random variable. Let Y =
σ Z and X = µ + σ Z.
P (Y ≤ y) = Φ(y
σ
)
, P (X ≤ x) = Φ
(
x − µ
σ
)
Define a new probability Q such that
EQ[H ] = EP [M H ],
where M is a nonnegative random variable with EP [M ] = 1. Take
M to be the likelihood ratio of Y vs X
M = exp
(
−µ Y
σ2− µ2
2 σ2
)
= exp
(
−µ Z
σ− µ2
2 σ2
)
Q(X ≤ x) = EP [M 1(X ≤ x)]
=
∫
1(y + µ ≤ x) exp
(
−µ y
σ2− µ2
2 σ2
)
1√2 π σ
exp
(
y2
2 σ2
)
dy
=
∫
1(y + µ ≤ x)1√
2 π σexp
(
−(y + µ)2
2 σ2
)
dy
=
∫
1(u ≤ x)1√
2 π σexp
(
− u2
2 σ2
)
du
=
∫ x
−∞
1
σφ
(u
σ
)
du
= Φ(y
σ
)
,
29
that is, under Q, X ∼ N(0, σ2). Similarly we can show that
Q(Y ≤ y) = Φ
(
y − µ
σ
)
that is, under Q, Y ∼ N(−µ, σ2).
Now we work on Bnt on [0, 1]. Since
Bntk
=
√
1
n
k∑
`=1
ε`,
Then
Bntk
+
k∑
`=1
µt` n−1 =
√
1
n
k∑
`=1
[ε` + µt` n−1/2]
follows N(∑k
`=1 µt` n−1, tk). Take
Mn = exp
(
−n−1/2n
∑
`=1
µt` ε` −n
∑
`=1
µ2t`
2 n
)
EQn[H ] = EP [Mn H ],
Then under Qn, ε` + µt`/√
n ∼ N(0, 1), and Bntk
+∑k
`=1 µt` n−1 ∼
N(0, tk). As Bntk
+∑k
`=1 µt` n−1 converges to Bt +∫ t
0 µs ds,
Mn → M = exp
(
−∫ t
0
µs dBs −∫ t
0 µ2s ds
2
)
EQ[H ] = EP [M H ],
Then under Q, Bt +∫ t
0 µs ds is a Brownian motion.
30
Girsanov theorem Suppose
Xt = Bt +
∫ t
0
µ(Xs) ds, or dXt = dBt + µ(Xt)dt,
and Bt is a Brownian motion with respect to probability P . Let
Mt = exp
(
−∫ t
0
µ(Xt)dt − 1
2
∫ t
0
µ2(Xt)dBt
)
.
By Ito lemma, Mt is a martingale with M0 = 1. Define a new
probability Q
Q(A) = E[MT ; A],
that is, Q is absolutely continuous with respect to P and MT is the
Radon-Nikodym derivative of Q with respect to P ,
dQ
dP= MT .
Then under Q, Xt itself is a Brownian motion.
31
Martingale Representation theorem Suppose Mt is a mar-
tingale with respect to the σ-fields generated by a Brownian motion
B. Then There exists Hs such that
Mt = M0 +
∫ t
0
Hs dBs.
If V is an option, and
V = constant +
∫ t
0
Hs dSs
Any option can be replicated by buying and selling the stock.
Fundamental theorems Risk-neutral measure ⇐⇒ martingale
measure.
Under the risk-neutral measure, discounted stock price is a mar-
tingale ⇐⇒ no arbitrage exists.
Market is complete ⇐⇒ martingale measure is unique.