Quadratic Variation of Increasing Process?

I am looking through my notes and I came across the following statement:

Let Xs be a positive local martingale and let Mt=max0stXs. Then since Mt is an increasing process, [X,M]=[M]=0. Why is this true?

On a similar note, the claim is made that in terms of stochastic integrals, dMs is 0 unless Xs=Ms. I’m not sure why this holds either.


Let (πn)n0 be a sequence of partitions of [0,T] whose mesh is going to zero (that is, if the partition is πi:0=t0t1...tN(i)=T, then mesh(πi)=maxp(tp+1tp)).

Then a/the definition of quadratic covariation of processes X, Y over [0,T] is:
(see Revuz & Yor, Continuous Martingales and Stochastic Calculus, Chapter IV on Stochastic Integration, Theorem 1.9 for the equivalence of this definition with the other main definition; that quadratic covariation is the unique increasing stochastic process such that (XtYt[X,Y]t)t0 is a martingale) and [X]:=[X,X].

Then if M is some general increasing process, and X some general continuous process, then we prove that [X,M]T=0 for all T:
since M is increasing, and then as the sum is telescoping:
since X is continuous, and the mesh of the partition is tending to zero, so maxj|Xtj+1Xtj|0

Since our proof here is for general processes X, M, with M increasing and X increasing, we thus have [X,M]=[M](=[M,M])=0.

This will not in general be true if X is not a continuous process. If X has a jump discontinuity which causes M to have a jump discontinuity, say at t(ti,ti+1), then |Xti+1Xti||Mti+1Mti| can never go to zero; it will always equal at least the product of the two jumps. For a partial discussion on this see https://almostsuremath.com/2010/01/19/properties-of-quadratic-variations/.

As for your second question, imagine drawing sample paths for X and M on the same graph. So at every point, M is either constant or increasing. But by the definition of M, M can only increase, when X is reaching a new, higher maximum. But, whenever X is reaching a new maximum, say at Xt, Mt=Xt by definition of M being the maximum X up to that point. Thus the only case when dMs=0 is when Ms=Xs. I hope that makes sense – try drawing the sample paths of X and M for yourself; this should elucidate what I’m trying to explain.

Source : Link , Question Author : Eric , Answer Author : Jacob

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