= Solution
\b[The printed assumptions omit the orthogonality needed for the claimed local martingales.] Merely placing two <Brownian motions> on the same <probability space> does not make them independent. Interpret both as <Brownian motions> relative to a common <filtration>, and put $C_t=[B,\vartheta]_t$. For $X=e^B\cos\vartheta$ and $Y=e^B\sin\vartheta$, the two-variable <Itô formula> gives
$$
\boxed{dX=X\,dB-Y\,d\vartheta-Y\,dC,\qquad dY=Y\,dB+X\,d\vartheta+X\,dC.}
$$
The pure second derivatives cancel because $[B]_t=[\vartheta]_t=t$, while the mixed <partial derivatives> are $X_{b\vartheta}=-Y$ and $Y_{b\vartheta}=X$. Thus the mixed <quadratic covariation> term cannot be omitted.
For a counterexample allowed by the PDF, take $\vartheta=B$. Then $C_t=t$ and $X$ has drift $-Y\,dt$, while $Y$ has drift $X\,dt$. In particular $X_0=1$ makes the latter drift nonzero near zero, so $Y$ is not a <local martingale>. Nor is $X$: $Y$ cannot be identically zero, so its accumulated continuous drift is not identically zero. Uniqueness of the <continuous semimartingale decomposition> excludes cancellation by another <local martingale> term.
The corrected hypothesis is $[B,\vartheta]\equiv0$, for example independent <Brownian motions> in their joint <filtration>. Conversely, if both were <local martingales>, uniqueness of the <continuous semimartingale decomposition> would force $Y\,dC=X\,dC=0$; since $X^2+Y^2=e^{2B}>0$, this forces $dC=0$. Thus the correction is exactly the condition needed for both <local martingales>. Under this hypothesis the drift terms vanish and the intended <stochastic differential equations> are $dX=X\,dB-Y\,d\vartheta$ and $dY=Y\,dB+X\,d\vartheta$. The continuous integrands are locally bounded, so localization makes these <Itô integrals> <continuous local martingales>. Subsequent intended conclusions are proved under this explicitly stated correction.
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