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In the book "Brownian Motion and Stochastic Calculus" by Karatzas and Shreve it is said that if $X\in\mathcal{M}_2^c$ (i.e. $X$ is a continuous square integrable martingale) the predictable quadratic variation $\left<X\right>_t$ (defined, through the Doob-Meyer decomposition, as the a.s. unique natural increasing process $Y$ such that $X^2-Y$ is a martingale) is its own first variation, that is

$$p-\lim_{n\rightarrow\infty}\sum_{k=1}^n\left| Y_{t_{k}}-Y_{t_{k-1}} \right|=\left<X\right>_t$$

where $0=t_1<t_2<\cdots<t_n=t$ is a partition of $[0,t]$. Although this should be obvious, I can't figure out why the identity holds.

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Since $Y$ is increasing, $|Y_{t_k} - Y_{t_{k-1}}| = Y_{t_k}-Y_{t_{k-1}}$, so this is a telescoping sum: \begin{align*} \sum_{k=1}^n |Y_{t_k} - Y_{t_{k-1}}| &= \sum_{k=1}^n (Y_{t_k}-Y_{t_{k-1}}) \\ &= Y_{t_n} - Y_{t_1} \\ &= Y_t = \langle X\rangle_t. \end{align*} The fact that $Y_{t_1}=Y_0 = 0$ is because part of the definition of the quadratic variation is $\langle X\rangle_0 = 0$.

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