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

For every integrable real or complex function,
$$
\left|\int f\,d\mu\right|\leq\int|f|\,d\mu.
$$


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