Hyoung-Goo Kang
Abstract
Students will understand preliminary denitions and concepts.
1. Martingales
Preliminary denitions and examples
Denition 1. A stochastic process {Xn|n= 0,1,2, ...} is a martingale if for n = 0,1,2, ...
(i)E[|Xn|] < ∞ (ii)E[Xn+1|X0, ..., Xn] = Xn
Denition 2. Let{Xn|n= 0,1,2, ...}and {Yn|n = 0,1,2, ...}be stochastic process.
We say {Xn} is martingale with respect to{Yn} if, for n= 0,1, ...
(i)E[|Xn|] < ∞ (ii)E[Xn+1|Y0, ..., Yn] = Xn
It is useful to think of (Y0, ..., Yn) as the information or history up to time or
stage n.
The history determines Xnin the sense that Xn is a function of (Y0, ..., Yn), i.e.
knowledge of (Y0, ..., Yn) determine the value ofXn.
E[Xn|Y0, ..., Yn] = Xn.
Using the law of iterative expectations, we have:
E[Xn+1] =E(E(Xn+1|Y0, ..., Yn)) = E(Xn).
To review the law of iterative expection: If G⊂H, thenE(X|G) =E(E(X|G)|H).
Example 3. Sum of independent random variables.
Let Y0 = 0 and Y1, Y2, ... be independent random variables with E(|Yn|) < ∞
and E(Yn) = 0 for all n. If X0 = 0 and Xn =Y1 +...+Yn for n ≥ 1, then {Xn} is
martingale w.r.t {Yn}. We can check the conditions as
(1)E(|Xn|) =E(|Y1+...+Yn|)≤E(|Y1|+...+|Yn|)<∞ (2)E(Xn+1|Y0, ..., Yn) = E(Xn+Yn+1|Y0, ..., Yn)
=E(Xn|Y0, ..., Yn) +E(Yn+1|Y0, ..., Yn) =Xn+E(Yn+1) =Xn.
Example 4. The variance of a sum as a martingale
Let Y0 = 0 and Y1, Y2, ... be independent and identically distributed random
variables with E(Yk) = 0 and E(Yk2) =σ2, k = 1,2, .... Let X0 = 0 and
Xn= n
X
k=1
Yk
!2
−nσ2.
Check:
(1) E|Xn|=E|(Pnk=1Yk)
2
−nσ2| ≤E(Pn
k=1Yk)
2
+nσ2
=E(Pn
k=1Yk2) +nσ2 = 2nσ2 <∞.
(2) E(Xn+1|Y0, ..., Yn) = E
(Yn+1+Pnk=1Yk)2−(n+ 1)σ2|Y0, ..., Yn
=EYn2+1+ 2Yn+1Pnk=1Yk+ (Pnk=1Yk)
2
−(n+ 1)σ2|Y0, ..., Yn
=EY2
n+1|•
+ 2E(Yn+1|•)Pnk=1Yk+Xn−σ2 =Xn.
Example 5. Doob's martingale process
Let Y1, Y2, ... be an arbitrary sequence of random variables. Suppose X is a
random variable satisying E(|X|) < ∞. Then, Xn ≡ E(X|Y0, Y1, ..., Yn) forms a
martingale w.r.t Yn. Xn is called Doob's process.
(1) E(|Xn|)= E[|E(X|Y0, ..., Yn)|]≤ E[E(|X||Y0, ..., Yn)] =E(|X|) <∞.
(2) E(Xn+1|Y0, ..., Yn) =E[E(X|Y0, ..., Yn+1)|Y0, ..., Yn] =E(X|Y0, ..., Yn) =Xn.
2. Martingales with respect to σ-eld
Denition 6. (Ω,F, P) is called a probability space
(1) The sample space: a set Ω whose elements ω correspond to the possible outcomes of an experiment
(2) The family of events: a collectionF of subsets Aof Ω. We say that the event
A occurs if the outcome ω of the experiment is an element of A
(3) The probability measure: a functionP dened on F and satisfying (a) 0 =P(φ)≤P(A)≤P(Ω)≤1
(b) P(A1∪A1) = P(A1) + P(A2) − P(A1∩A2) for Ai ∈ F, i= 1,2
(c)P(∪∞
Denition 7. σ-eld: A σ-eld of Ω, denoted byF, satises (i) A∈ F ⇒Ac∈ F where Ac denotes the complements of A
(ii) If A1 and A2 are inF, A1∩A2 is inF. For any sequence A1, A2, ...inF, the
unition A1∪A2∪...is in F
Example 8. Ω ={ω1, ω2, ω3, ω4},
F ={{ω1, ω2},{ω3},{ω4},{ω1, ω2, ω3},{ω1, ω2, ω4},{ω3, ω4},{ω1, ω2, ω3, ω4}, φ},
P({ω1, ω2}) = 0, P({ω3}) = P({(ω4}) = 1/2.
Note that (a) one cannot distinguish between states ω1 and ω2; (b) ω1 and ω2
can only occur with zero probability (Null set: {ω1}, {ω2}). To check, let A1 =
{ω1, ω2} ∈ F. Ac1 ={{ω3},{ω4},{ω3, ω4}, φ} ∈ F. Take A1 ={ω1, ω2}, A2 ={ω3}.
A1∪A2 ={ω1, ω2, ω3} ∈ F.
Denition 9. A random variable is a function X : Ω−→ R with the property: For any a ∈ R, the set {ω ∈ Ω : X(ω) = a} is in F. Intuitively, X is a r.v. if, for any possible outcome a, we will know whether X has this outcome from knowing the outcomes of events in F. If X is a r.v. w.r.t (Ω,F), we also say that X is F-measurable. Theσ-eld generated by a r.v. X is dened to be the smallestσ-eld w.r.t which X is measurable. It is denoted by F(X).
Example 10. If X has only denumerably many possible values a1, a2, ... and if
X(ω) = ω. Ai = {ω|X(ω) =ω =ai} which means that ω can only take on ai.
Dene Ω = ∪∞
i=1Ai. Note that Ai∩Aj = φ. What is the σ-eld generated X(ω)?
F(X) = {φ,Ω, every set that is the unition of some of the A0is}. Much smaller than F(ω).
Denition 11. Filtration. In order to facilitate our analysis of dynamic securities markets, we need to nd a way to model the information structure of the economy. A ltration is a family of sub σ-eld of F, Ft = {Ft : t ∈ [0, T]}, Fs ∈ Ft, ∀s < t.
Ft is interpreted as the information set that contains events observable at or before
time t. In a discrete time and nite space model, a ltration can be representated by an event tree, and the family of sub-F-elds can be generated from the partitions (unions) of the events.
Martingale w.r.t an increasing family of σ-elds
Let{Xn}∞0 be a sequence of real r.v.s on a probability space(Ω,F, P). Let{Fn}
be a sequence of sub σ-elds of F with F0 ⊂ F1... ⊂ Fn...⊂ F. We say that Xn is
adapted toFnif, for everyn,XnisFn-measurable. We think ofFn as containing the
information available at stage n. Fn ⊂ Fn+1 expresses the increases in information
Example 12. Let Ω = {ω1, ω2, ..., ω6} and P(ωi) > 0,∀i. Let F0 = {Ω, φ}, F1 =
{{ω1, ω2, ω3},{ω4, ω5, ω6},Ω, φ}, F2 = {all subsets ofΩ}. Check F1 is a σ-eld.
A1 ={ω1, ω2, ω3}, Ac1 ={ω4, ω5, ω6} ∈ F1. A1∪A2 = Ω. Done.
(1) Draw decision tree for this process) (2) LetX be a stochastic process dened by X(ωi,0) = 0, i= 1, ...,6
X(ω1,1) =X(ω2,1) = 1, X(ω3,1) =X(ω4,1) = 2,X(ω5,1) =X(ω6,3) = 1
X(ωi,2) =i, i= 1, ...,6
Is X adapted to F?
Note that adaptedness is simply an informational constraint: the value of the process at any time t can only depend on the information revealed on or before time t. In order to check measurability: For anyai ∈ R, the set {ω ∈ Ω|X(ω) = ai is in
F. Here, a1 = 1, a2 = 2, a3 = 3.
When X(ω) = a1 = 1, the set is (ω1, ω2)∈ F/ 1.
When X(ω) = a2 = 2, the set is (ω3, ω4) ∈ F/ 1. Thus, not measurable, not
adapted. Intuitively, at time 1, one cannot distinguish amongω1, ω2, ω3 according to
mathcalF. The value of X1, however, vary across {ω1, ω2}andω3}.
In nance, we always assume this property.
Denition 13. Martingale: A process X is called a martingale adapted toF if (i) Xt is measurable wrtFt
(ii) E(Xs|Ft) =Xt
(iii) E|Xt|<∞
Denition 14. A standard Brownian Motion is a process B dened by the proper-ties:
(a) B0 = 0 a.s.
(b) For times t and s > t, Bs−Bt is normally distributed with mean zero and
variance s−t.
(c) For timest0, ..., tnsuch that 0≤t0 < t1 < ... < tn <∞, the random variables
B(t0), B(t1)−B(t0), ..., B(tn)−B(tn−1) are independently distributed; and
(d) For each ω∈Ω, the sample patht →B(ω, t)is continuous
The following property regarding a BM will be useful (Homework).
• {Bt} is a martingale
• {B2
t −t}is a martingale, and E[(Bt−Bs)2|Fs] =t−s, t≥s
3. Ito Integrals
´T
0 f(t)dBt: If E
´T
0 |f(ω, t)|
2dt < ∞, it is a martingale (H2 − space). If
P ´ |f(ω, t)|2dt <∞ = 1, the integral is also well-dened, but it's no longer a
martingale. It's called a local martingale (L2 −space). L2 −space is larger. For
example, if E´0T |f(ω, t)|2dt<∞, then´ |f(ω, t)|2dt < ∞where p(ω)<∞.
An Ito process is an Ito integral plus an adapted process X(t) = X(0) + ´0ta(s)ds +´0tf(s)dBs, t ∈ [0, T], where
´T
0 |a(s)|ds < ∞ a.s.
´T
0 |f(s)|
2ds <∞ a.s. Very often, we rewrite an Ito process in terms of a stochastic
dierential equation: dX(t) =a(t)dt+f(t)dBt.
Example 15. Ito's Lemma
Letf be twice continously dierntiable w.r.t the rst argument, and continuously dierentialble w.r.t the second argument. Let X be an Ito process: X(t) =X(0) +
´t
0a(s)ds+
´t
0 σ(s)dBs. Then,
f(X(t), t) =f(X(0),0) +
ˆ t
0
1 2fxxσ
2
(s)ds+
ˆ t
0
fxa(s)ds+
ˆ t
0
ftds+
ˆ t
0
fXdBs
In dierential form, we have
df(X(t), t) =
1
2fXXσ
2
+fXa(t) +ft
dt+fXσdBt
To illustrate,
• d[Bm
t ] =mB m−1
t dBt+ m(m
−1)
2 B
m−1
t dt
• Geometric BM: dS = µSdt+σSdBt. Let X = log(S), then dX/dS = S−1,
d2X/dS2 =−S−2,dX = (−1 2S
−2σ2S2+S−1µS)dt+S−1σSdB
t=(µ−12σ2)dt+
σdBt. Let X0 ≡X(0) =logS0 then Xt=X0+ (µ− 12σ2)t+σ(Bt−B0).
• Let Z(t)≡exp´0tf(s)dBs− 12
´t
0 f
2(s)ds. Homework: dZ(t)=?
• ´0tBn(s)dB
s: Tryd[Btn+1] = (n+ 1)BntdBt+12n(n+ 1)Btn−1dt. Thus, BtndBt=
1
n+1dB
n+1
t −12nB
n−1
t dt
Homeworks
1. dWt = (a(t)Wt+b(t))dt+c(t)dBt. (1)
´T
0 e
´T
t a(u)du(dWt−a(t)Wtdt) =? (2)
F ≡exp−´0tβ1(u, Wu)du−
´t
0β2(u, Wu)dWu
V(t, W) where V(t, W)is well
2. dYt/Yt=µdt+σdBt;u(x, t) = e−δt x
1−b
1−b,δ > 0and b <0.
P(Y, t) =Et
ˆ ∞
t
ux(Ys, s)
ux(Yt, t)
Ysds
!
=?
