local martingale and submartingale 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?
 A: 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.
A: 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.
