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Stat 761 Winter 2015
Stochastic Processes
Instructor: A. Swishchuk
Lecture 11: Martingales, Sub- and
Supermartingales, Doob-Meyer
Decomposition: Discrete Time
Outline
⇒ Martingales: Definition and Examples
⇒ Martingale Transformation
⇒ Sub-and Supermartingales: Definitions and Examples
⇒ Doob-Meyer Decomposition
⇒ Doob Inequalities
1
Martingales in Discrete Time
1.1
Definition and Examples
Let (Ω, F, P ) be a probability space and (Fn )n≥0 be a filtration. A sequence
of r.v. mn is an Fn -martingale, if
1) mn is Fn -measurable;
2)
E(mm /Fn ) = mn , m > n.
For 2) to be fulfilled it is sufficient to check that
E(mn+1 /Fn ) = mn ,
because, for example,
E(mn+2 /Fn ) = E(E(mn+2 /Fn+1 )/Fn ) = E(mn+1 /Fn ) = mn .
We’ve just used Tower Law.
We note, that linear combination of martingales is a martingale.
Examples.
1. If mn is a martingale, then Emn = Em0 , since E(mn+k |Fk ) = mn and
for n = 0 we have E(mk |Fk ) = mk = m0 (mk is Fk -measurable).
2. Let Xn be Bernoulli r.v., i.e., independent r.v. such that P (Xn =
1) = P (Xn = −1) = 1/2. Then the sequence Sn := X1 + X2 + ... + Xn
is a martingale wrt algebra Dn = DX1 ...Xn generated by r.v. X1 , X2 , ..., Xn .
Here, D1 = (D+ , D− ), where D+ := (ω : X1 = +1) and D− := (ω : X1 =
−1); D2 = (D++ , D+− , D−+ , D−− ), where, for example, D++ := (ω : X1 =
+1, X2 = +1), and so on. Indeed, Sn is Dn -measurable by construction and
E(Sn+1 /Dn ) = E(Sn +Xn+1 /Dn ) = E(Sn /Dn )+E(Xn+1 Dn ) = Sn +EXn+1 = Sn ,
because EXn+1 = 1 × (1/2) + (−1) × (1/2) = 0.
3. P
Let Xn be a sequence of independent r.v. such that EXn = 0 and
Sn := ni=1 Xi . Then (Sn , Dn ) is a martingale, where Dn generated by r.v.
X1 , X2 , ..., Xn :
E(Sn+1 /Dn ) = E(Sn + Xn+1 /Dn ) = Sn + E(Xn+1 /Fn ) = Sn + EXn+1 = Sn .
1.2
Martingale Transformation
Let mn be an Fn -martingale, Hn be a predictable sequence (i.e., Hn is Fn−1
measurable), X0 := 0,
Xn := H1 ∆m1 + ... + Hn ∆mn
and ∆mn := mn − mn−1 .
Xn is called a martingale transformation of mn by Hn .
Lemma on Martingale Transformation. Xn is an Fn -martingale.
Proof. We note that Xn is Fn -adapted sequence. Then
E(Xn+1 − Xn |Fn ) = E(Hn (mn+1 − mn )|Fn )
= Hn+1 E(mn+1 − mn )|Fn ) = 0
,
since mn is Fn -martingale. Hence,
E(Xn+1 − Xn |Fn ) = 0,
and
E(Xn+1 |Fn ) = Xn ,
and Xn is an Fn -martingale.
Lemma on Martingale Criterion.
Adapted process (sequence) mn is an Fn -martingale iff for any predictable
sequence Hn
N
X
E(
Hn ∆mn ) = 0.
n=1
2
Sub- and Supermartingales, Examples
Let (Ω, F, P ) be a probability space and (Fn )n≥0 be a filtration.
A sequence of r.v. mn is an Fn -submartingale, if
1) mn is Fn -measurable;
2)
E(mm /Fn ) ≥ mn , m > n.
For 2) to be fulfilled it is sufficient to check that
E(mn+1 /Fn ) ≥ mn ,
Examples.
1. If mn is a martingale, them |mn | is a submartingale:
E(|mm |/Fn ) ≥ |E(mm /Fn )| = |mn |,
m > n.
2. If mn is a martingale, them m2n is a submartingale:
E(m2m /Fn ) ≥ (E(mm /Fn ))2 ≥ m2n ,
m > n.
Jensen Inequality.
Let g(x) be a convex function and E|X| < +∞. Then
g(EX) ≤ Eg(X).
If mn is a martingale, then g(mn ) is a submartingale for every convex function
g, that follows from Jensen inequality.
Let (Ω, F, P ) be a probability space and (Fn )n≥0 be a filtration.
A sequence of r.v. mn is an Fn -supermartingale, if
1) mn is Fn -measurable;
2)
E(mm /Fn ) ≤ mn , m > n.
For 2) to be fulfilled it is sufficient to check that
E(mn+1 /Fn ) ≤ mn ,
Examples.
1. If mn is a positive martingale and An is a positive decreasing a.s. sequence such that An is Fn−1 -measurable for any n = 1, 2, 3, ..., then sequence
Sn := mn + An is a supermartingale:
E(Sn+1 /Fn ) = E(mn+1 + An+1 /Fn ) ≤ mn + An = Sn ,
n ≥ 1.
3
Doob-Meyer Decomposition, Quadratic Variation, Covariance
A sequence Xn is predictable, if it is Fn−1 -measurable.
Doob-Meyer Decomposition (DMD). Let (Xn , Fn ) be a submartingale. Then there exists a martingale (mn , Fn ) and a predictable increasing
sequence (An , Fn ) such that for any n ≥ 0
Xn = mn + An .
This decomposition is unique.
Proof. Set m0 = X0 , A0 = 0,
mn := m0 +
n−1
X
[Xi+1 − E(Xi+1 /Fi )]
i=0
and
An :=
n−1
X
[E(Xi+1 /Fi ) − Xi ],
i=0
and decomposition is proved. Uniqueness follows from standard arguments.
Q.E.D.
A similar result may be proved for a supermartingales. Let (Xn , Fn ) be a
supermartingale. Then there exists a martingale (mn , Fn ) and a predictable
decreasing sequence (An , Fn ) such that for any n ≥ 0
Xn = mn + An .
This decomposition is unique.
From the Doob-Meyer decomposition follows that An compensates submartingale Xn to the martingale mn .
The sequence An in Doob-Meyer decomposition is called the compensator
of the submartingale Xn .
Doob-Meyer decomposition plays a crucial role in the study of squareintegrable martingale Mn , EMn2 < +∞, because Mn2 is a submartingale and
hence
Mn2 = mn + < M >n ,
where mn is a martingale and < M >n is a predictable increasing sequence.
The sequence < M >n is called the quadratic variation of the martingale
Mn . From the representation mn in DMD we have
< M >n =
n
X
E[(∆Mi )2 /Fi−1 ].
i=1
If Xn and Yn are square-integrable martingales, then
1
< X, Y >n := [< X + Y >n − < X − Y >n ]
4
is called a covariation of X and Y.
4
Doob Inequalities
Let Xn be a non-negative sequence of r.v..
1. If Xn , 0 ≤ n ≤ N, is a submartingale, then
aP ( max Xn ≥ a) ≤ E[XN ],
a ∈ R.
0≤n≤N
2. If Xn ,
0 ≤ n ≤ N, is a supermartingale, then
aP ( max Xn ≥ a) ≤ E[X0 ],
0≤n≤N
3. If Xn ,
a ∈ R.
0 ≤ n ≤ N, is a martingale, then
aP ( max |Xn | ≥ a) ≤ E[XN ],
0≤n≤N
4. If Mn ,
a ∈ R.
0 ≤ n ≤ N, is a martingale, then
E[ max Mn2 ] ≤ 4E[MN2 ].
0≤n≤N
Recommended Textbook: ’A First Course in Stochastic Processes’ by
S. Karlin and H. Taylor, Academic Press, 2nd ed., 1975, p.238-253.
Recommended Exercises:
1. Q
Let Xn be a sequence of independent r.v. such that EXn = 1 and
Sn := ni=1 Xi . Then (Sn , Dn ) is a martingale, where Dn generated by r.v.
X1 , X2 , ..., Xn .
2. Let Xn be Bernoulli r.v., i.e., independent r.v. such that P (Xn = 1) =
p, P (Xn = −1) = q, p + q = 1, p 6= q. Then the sequence ηn := ( pq )Sn
and ξn := Sn − n(p − q), where Sn := X1 + X2 + ... + Xn are martingales wrt
algebra Dn = DX1 ...Xn generated by r.v. X1 , X2 , ..., Xn .
3. If X is a r.v., then Sn := E(X/Dn ) is a martingale wrt Dn generated
by r.v. X1 , X2 , ..., Xn .
4. Let Xn be a sequence of independent identically distributed r.v.
and Sn := X1 + X2 + ... + Xn . Then the sequence m1 := Snn , m2 :=
Sn−1
n+1 −k
, ..., mk := Sn+1−k
, mn = S1 is a martingael wrt to Dn generated
n−1
by X1 , X2 , ..., Xn .
5. Prove Doob inequality 2.