Angle bracket (quadratic variation) process for martingales Probability with Martingales:



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?

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, $\left<M^{S(k)}\right>=A^{S(k)}$.
