Question

$\mathbb{E}[(\int_{0}^{t} V_s^2: d\langle M^{c}\rangle_{s})^{\frac{1}{2}}]<\infty$ implies $N_t:= \int_0^t V_s \cdot d M_{s}^{c}$ is a martingale.

score 33 · Answer 1 · 2026-04-12 15:22:57

33

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$\mathbb{E}[(\int_{0}^{t} V_s^2: d\langle M^{c}\rangle_{s})^{\frac{1}{2}}]<\infty$ implies $N_t:= \int_0^t V_s \cdot d M_{s}^{c}$ is a martingale.

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174

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List Question

$\mathbb{E}[(\int_{0}^{t} V_s^2: d\langle M^{c}\rangle_{s})^{\frac{1}{2}}]<\infty$ implies $N_t:= \int_0^t V_s \cdot d M_{s}^{c}$ is a martingale.

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How to show that $E[|Y_n|]<\infty$ in martingale?

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