{c}

For continuous local martingales, the <total-variation process> of their <quadratic covariation> satisfies
$$
V_t([M,N])\leq[M]_t^{1/2}[N]_t^{1/2}.
$$
It is the <Cauchy-Schwarz inequality> for the matrix-valued measure formed by their <quadratic variations> and <quadratic covariation>.

Kunita-Watanabe inequality

With the notation from part (i), the <total-variation process> of $[M,N]$ is
$$
V_t=\int_0^t|c|\,dC.
$$
For the centered <bivariate normal distribution> $(U,V)$ used there, $c=\mathbb E[UV]$. The <integral triangle inequality> gives
$$
|c|=|\mathbb E[UV]|\leq\mathbb E|UV|.
$$
Integrating this pointwise inequality against $dC$ proves $V_t\leq\widetilde V_t$ for every $t\geq0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 202 / 2 / b / ii / Solution

{title2=$V_t(A)$}

For a path of <finite variation>, its total-variation process is
$$
V_t(A)=\sup_\pi\sum_{[u,v]\in\pi}|A_v-A_u|,
$$
where the <supremum> is over every <partition of an interval> of $[0,t]$. It is an increasing process, and the signed measure $dA$ satisfies $|dA|=dV(A)$.

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