Angle bracket (quadratic variation) process for martingales

Question

Probability with Martingales:

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What is the relation between $\langle M_{S(k) \wedge n}\rangle \ = A_{S(k) \wedge n}$ and $\{N_n\}, \{ N_{ S(k) \wedge n } \}$ being martingales?

It seems that $$\langle M_{S(k) \wedge n}\rangle \ = A_{S(k) \wedge n}$$ is supposed to be true by defintion. Am I wrong?

Answer 1

Since $M^2-A=N$ is a martingale, the stopped process $(M^2-A)^{S(k)}$ is also a martingale. Also, $A^{S(k)}$ is previsible. However, the Doob decomposition of an $\mathcal{L}^1$ adapted process into a martingale and a previsible process is essentially unique. Hence, $\left<M^{S(k)}\right>=A^{S(k)}$.

Answer 2

You are confusing $\langle M^{S(k)}\rangle,$ the angle bracket of the stopped process with $\langle M \rangle^{S(k)}.$ Only the latter is equal to $A^{S(k)}$ by definition.

The fact that $$(M^{S(k)})^2-A^{S(k)}$$ is a martingale demonstrates that the first is also equal, by the uniqueness of the Doob-Meyer decompostion.