Solution
= Solution
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 $t\geq0$.