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This question pertains to the topic titled: Is this local martingale a true martingale?

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"Using the Ito's formula I have shown that $X_t$ is a local martingale, because $\mathrm{d}X_t = \underline{\phantom{b_t}} \mathrm{d} B_t$",

Can someone please explain this opening phrase to the original question to me further? I can not get my head around why $X_t$ would not indeed be a true martingale if it is written in this form (all Ito integrals are martingales! (?)).

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In order to make any statements about $\mathsf{E}\left(X_t| \mathcal{F}_s\right)$ one needs to make sure that the expectation is defined. –  Sasha Aug 31 '12 at 5:04
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2 Answers

In fact, not ALL Ito integrals are martingales. For instance, an Ito integral $\int_0^T f(t,\omega)dW_t$ is well defined for the class of stochastic processes $\mathcal P=\{f=f(t,\omega)\mid f \text{ is non-anticipating, jointly measurable and} \mathsf P(\int_0^Tf^2(t,\omega)dt<\infty)=1\}$ and in this case $\int_0^t f(s,\omega)dW_s, t \in [0,T]$ is local martingale, which will be square integrable martingale iff $\mathsf E \int_0^T f^2(s,\omega)\, ds<\infty$.

In the case, mentioned above, that process is a square integrable martingale.

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In order to make $\left(\int_0^t \sigma_s dB_s\right)_{t \ge 0}$ a true martingale, one has to show e.g. that its quadratic variation is integrable for any $t$. Use the fact that $$ \langle \int_0^. \sigma_s dB_s\rangle_t = \int_0^t \sigma_s^2 ds $$ and show that $$ \mathbb{E}\left(\int_0^t \sigma_s^2 ds \right) < \infty $$ Keep in mind that it is not as trivial as it may seem, but often true in practice.

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