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Consider a local continuous martingale $M_t$ for $t \in [0,T]$. Taking a localizing $\{\tau_k\}$ yields that $(M^2 - \langle M \rangle)(\tau_k \wedge t)$ is martingale. How can I show that $M^2(\tau_k \wedge t)$ is a submartingale then.

I know that the quadratic variation is increasing and it holds:

$$\mathbb{E}[(M^2 - \langle M \rangle)(\tau_k \wedge t) | F_t]= (M^2 - \langle M \rangle)(\tau_k \wedge s) $$ How can I isolate $M^2$ from there?

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2 Answers 2

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By Jensen's inequality and the fact that $(M_t)_{t \in \mathbb{R}^+}$ is a local martingale: $$E[M^2_{t \wedge \tau_n}|\mathcal{F}_s]\geq (E[M_{t \wedge \tau_n}|\mathcal{F}_s])^2=M_{s \wedge \tau_n}^2$$ The claim follows.

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You have $$ \Bbb E[M^2_{\tau_k\wedge t}-\langle M\rangle_{\tau_k\wedge t}\mid\mathcal F_s]= M^2_{\tau_k\wedge s}-\langle M\rangle_{\tau_k\wedge s}. $$ Now move the $\Bbb E[\langle M\rangle_{\tau_k\wedge t}\mid\mathcal F_s]$ term to the right side: $$ \eqalign{ \Bbb E[M^2_{\tau_k\wedge t}\mid\mathcal F_s] &= M^2_{\tau_k\wedge s}+\Bbb E[\langle M\rangle_{\tau_k\wedge t}\mid\mathcal F_s]-\langle M\rangle_{\tau_k\wedge s}\cr &= M^2_{\tau_k\wedge s}+\Bbb E[\langle M\rangle_{\tau_k\wedge t}-\langle M\rangle_{\tau_k\wedge s}\mid\mathcal F_s]\cr &\ge M^2_{\tau_k\wedge s},\cr } $$ where the second equality is true because $\langle M\rangle_{\tau_k\wedge s}$ is $\mathcal F_s$-measurable, and the final inequality is true because $u\mapsto\langle M\rangle_{\tau_k\wedge u}$ is non-decreasing.

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