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?

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.
