Are all continuous time $L^2$-martingales i.e martingales such that $\mathbb{E}X_{t}^2<\infty$ for all $t$ also martingales in the sense that $\mathbb{E}\mid X_{t} \mid<\infty$ for all $t$, which is the common definition
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Yes, they are. Martingales have finite first moment and $L^{2}$ martingales are martingales with the extra condition that second moments are also finite.
Kavi Rama Murthy
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This follows also from the Lp inclusion and the fact that $\mathbb{E}\mid X_{t} \mid<\infty$ is equivallent to $\mathbb{E} X_{t} <\infty$, right? – Second Oct 28 '19 at 10:10
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@Probablynot Yes, $L^{2} \subset L^{1}$. – Kavi Rama Murthy Oct 28 '19 at 10:11
