63.2. THE QUADRATIC VARIATION 2119

= E(E((Mτ (tk+1)−Mτ (tk))

(Mτ(t j+1

)−Mτ (t j)

)|Ftk

))= E

((Mτ(t j+1

)−Mτ (t j)

)E((Mτ (tk+1)−Mτ (tk)) |Ftk

))= E

((Mτ(t j+1

)−Mτ (t j)

)(Mτ (tk)−Mτ (tk))

)= 0

It follows

E(

Mτ (l)2)

= E

(n−1

∑k=0

(Mτ (tk+1)−Mτ (tk))2

)

≤ E

(n−1

∑k=0

Mτ (Pn) |Mτ (tk+1)−Mτ (tk)|)

≤ E

(n−1

∑k=0

Mτ (Pn)(|Aτ (tk+1)−Aτ (tk)|+ |Bτ (tk+1)−Bτ (tk)|)

)

≤ E

(Mτ (Pn)

n−1

∑k=0

(|Aτ (tk+1)−Aτ (tk)|+ |Bτ (tk+1)−Bτ (tk)|)

)≤ E (Mτ (Pn)2C)

the last step holding because A and B are increasing. Now letting n→ ∞, the right sideconverges to 0 by the dominated convergence theorem and the observation that for a.e. ω,

limn→∞

Mτ (Pn)(ω) = 0

because of continuity of M. Thus for τ = τC given above,

M (l∧ τC) = 0 a.e.

Now let C ∈ N and let NC be the exceptional set off which M (l∧ τC) = 0. Then letting Nldenote the union of all these exceptional sets for C ∈N, it is also a set of measure zero andfor ω not in this set, M (l∧ τC) = 0 for all C. Since the martingale is continuous, it followsfor each such ω, eventually τC > l and so M (l) = 0. Thus for ω /∈ Nl ,

M (l)(ω) = 0

Now let N = ∪l∈Q∩[0,∞)Nl . Then for ω /∈ N,M (l)(ω) = 0 for all l ∈ Q∩ [0,∞) and so bycontinuity, this is true for all positive l.

Note this shows a continuous martingale is not of bounded variation unless it is aconstant.

63.2 The Quadratic VariationThis section is on the quadratic variation of a martingale. Actually, you can also considerthe quadratic variation of a local martingale which is more general. Therefore, this conceptis defined first. We will generally assume M (0) = 0 since there is no real loss of generalityin doing so. One can simply subtract M (0) otherwise.