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In recent few days I used to ask lots of questions, but I found them really interesting (from my perspective) and want to share with larger community.

Let $(X_t,Y_t)$ be two-dimensional Brownian Motion and denote $\mathscr{A}_t=\int_0^t X_sdY_s-\int_0^t Y_sdX_s$ Levy's area. How to prove that $\mathscr{A}_t$ is a true $L^2$-bounded martingale?

Given that $\mathscr{A}_0=0$, it is known that $\mathscr{A}_t$ will be $L^2$-bounded martingale if and only if $\mathbb{E}\langle \mathscr{A},\mathscr{A}\rangle_\infty<\infty$. Using this criteria of $L^2$-boundedness, we have $$ \langle\mathscr{A},\mathscr{A}\rangle_t=\langle \int_0^\cdot X_sdY_s-\int_0^\cdot Y_sdX_s\rangle_t= $$ $$ =\langle \int_0^\cdot X_sdY_s, \int_0^\cdot X_sdY_s\rangle_t + \langle \int_0^\cdot Y_sdX_s, \int_0^\cdot Y_sdX_s\rangle_t-2\langle \int_0^\cdot X_sdY_s, \int_0^\cdot Y_sdX_s\rangle_t $$ $$ =\int_0^t X_s^2d\langle Y, Y\rangle_s +\int_0^t Y_s^2d\langle X, X\rangle_s - 2\int_0^t X_sY_sd\langle X, Y\rangle_s $$ It is well-known that as $X_t,Y_t$ are independent Brownian Motions, they are orthogonal in a sense $\langle X, Y\rangle_s=0$ for any $s$ which gives us $$ \langle\mathscr{A},\mathscr{A}\rangle_\infty=\int_0^\infty X_s^2ds +\int_0^\infty Y_s^2ds $$ so $$ \mathbb{E} \langle\mathscr{A},\mathscr{A}\rangle_\infty=\int_0^\infty \mathbb{E}X_s^2ds +\int_0^\infty \mathbb{E}Y_s^2ds=2\int_0^\infty sds $$ This means that $\mathbb{E} \langle\mathscr{A},\mathscr{A}\rangle_\infty=\infty$ and $\mathscr{A}_t$ therefore can't be $L^2$-bounded martingale. Is this logic correct?

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Yes, your logic is correct. $\mathscr{A}_t$ is not $L^2$-bounded.

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  • $\begingroup$ Dear Nate! Thanks a lot for the confirmation. It seems I made a too strong statement that $\mathscr{A}_t$ is $L^2$-bounded. In fact, it seems to be square integrable only. $\endgroup$ – Mushtandoid Jun 5 '17 at 15:42

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