# Quadratic Variation of Increasing Process?

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

Let $X_s$ be a positive local martingale and let $M_t = max_{0 \le s \le t} X_s$. Then since $M_t$ 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, $dM_s$ is $0$ unless $X_s = M_s$. I’m not sure why this holds either.

Let $$(πn)n≥0(\pi_n)_{n\geq 0}$$ be a sequence of partitions of $$[0,T][0,T]$$ whose mesh is going to zero (that is, if the partition is $$πi:0=t0≤t1≤...tN(i)=T\pi_i:0=t_0\leq t_1 \leq ... t_{N(i)} = T$$, then $$mesh(πi)=maxp(tp+1−tp)mesh(\pi_i)=max_p(t_{p+1}-t_p)$$).

Then a/the definition of quadratic covariation of processes X, Y over $$[0,T][0,T]$$ is:
$$[X,Y]T=limn→∞N(n)−1∑i=0|Xti+1−Xti||Yti+1−Yti| [X,Y]_T = lim_{n \to \infty} \sum_{i=0}^{N(n)-1}|X_{t_{i+1}}-X_{t_i}||Y_{t_{i+1}}-Y_{t_i}|$$
(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 $$(XtYt−[X,Y]t)t≥0(X_tY_t-[X,Y]_t)_{t\geq 0}$$ is a martingale) and $$[X]:=[X,X][X]:=[X,X]$$.

Then if M is some general increasing process, and X some general continuous process, then we prove that $$[X,M]T=0[X,M]_T=0$$ for all T:
$$[X,M]T=limn→∞N(n)−1∑i=0|Xti+1−Xti||Mti+1−Mti|≤limn→∞N(n)−1∑i=0(maxj|Xtj+1−Xtj|)|Mti+1−Mti|=limn→∞(maxj|Xtj+1−Xtj|)N(n)−1∑i=0|Mti+1−Mti|=limn→∞(maxj|Xtj+1−Xtj|)N(n)−1∑i=0(Mti+1−Mti) [X,M]_T = lim_{n \to \infty} \sum_{i=0}^{N(n)-1}|X_{t_{i+1}}-X_{t_i}||M_{t_{i+1}}-M_{t_i}| \\ \leq lim_{n \to \infty} \sum_{i=0}^{N(n)-1}(max_j|X_{t_{j+1}}-X_{t_j}|)|M_{t_{i+1}}-M_{t_i}| \\ = lim_{n \to \infty} (max_j|X_{t_{j+1}}-X_{t_j}|)\sum_{i=0}^{N(n)-1}|M_{t_{i+1}}-M_{t_i}|\\ = lim_{n \to \infty} (max_j|X_{t_{j+1}}-X_{t_j}|)\sum_{i=0}^{N(n)-1}(M_{t_{i+1}}-M_{t_i})$$
since M is increasing, and then as the sum is telescoping:
$$=limn→∞(maxj|Xtj+1−Xtj|)(MT−M0)=0 = lim_{n \to \infty} (max_j|X_{t_{j+1}}-X_{t_j}|)(M_T-M_0)=0$$
since X is continuous, and the mesh of the partition is tending to zero, so $$maxj|Xtj+1−Xtj|→0max_j|X_{t_{j+1}}-X_{t_j}|\to 0$$

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

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 $$t∈(ti,ti+1)t\in (t_i, t_{i+1})$$, then $$|Xti+1−Xti||Mti+1−Mti||X_{t_{i+1}}-X_{t_i}||M_{t_{i+1}}-M_{t_i}|$$ 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 $$XtX_t$$, $$Mt=XtM_t = X_t$$ by definition of M being the maximum X up to that point. Thus the only case when $$dMs=0dM_s=0$$ is when $$Ms=XsM_s=X_s$$. 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.