It's from Brownian Motion and Stochastic Calculus by I. Karatzas chapter 1, problem 5.7 property (iv) trying to show an inequality wrt total variation and cross variation of martingales.
$$\check{\xi}_t – \check{\xi}_s\le \frac{1}{2}[\langle X\rangle_t+\langle Y\rangle_t-(\langle X\rangle_s+\langle Y\rangle_s)]$$
where $X$ and $Y$ are two martingales and $\check{\xi}_t$ is the total variation of the cross variation $\langle X,Y\rangle_t$ on $[0,t]$.
For now, I know how to show that $\check{\xi}_t\le \frac{1}{2}[\langle X\rangle_t+\langle Y\rangle_t]$ using $\langle X-Y\rangle\ge0$ but as for the $s$-term, I don't know how to deal with it.
It is said to be the final step to verify that cross variation is a bilinear function but I really don't understand why aren't the first three properties sufficient to show the result. I'll appreciate help of any kind. Thank you so much!
Best Answer
Properties (i) and (ii) show that $\langle\cdot,\cdot\rangle$ is a bilinear form. Properties (iii) and (iv) seem to be extra properties.
Let $0\le s< t$.
Observation 1. Suppose $h$ is given by the difference between two nondecreasing functions $f$ and $g$: $h=f-g$. Then $\check h(t)-\check h(s)\le f(t)-f(s)$.
Proof. $|f(t_2)-g(t_2)-f(t_1)+g(t_1)|\le f(t_2)-f(t_1)$, so $\sum_i|h(t_{i+1})-h(t_i)|\le f(t)-f(s)$ for any partition.
Observation 2. For any function $f$ of bounded variation, $f(t)-f(s)\le \check f(t)-\check f(s)$.
Now Property (iv): \begin{align} \check\xi_t-\check\xi_s &\le \frac{1}{4}\left(\langle X+Y\rangle_t-\langle X+Y\rangle_s\right),\quad\text{Observation 1},\\ &=\frac{1}{4}( \langle X\rangle_t-\langle X\rangle_s+\langle Y\rangle_t-\langle Y\rangle_s +2\langle X,Y\rangle_t-2\langle X,Y\rangle_s ),\quad\text{Properties (i) and (ii)},\\ &\le \frac{1}{4}( \langle X\rangle_t-\langle X\rangle_s+\langle Y\rangle_t-\langle Y\rangle_s +2(\check\xi_t-\check\xi_s) ),\quad\text{Observation 2}, \end{align} which, rearranged, yields the desired inequality.