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I have to show that $e^{B_t^1}\cos(B_t^2)$ is a martingale ($B=(B^1,B^2)$ is a two-dimensional Brownian Motion). I used Ito's formula and got $e^{B_t^1}\cos(B_t^2)=1+\int_0^t e^{B_s^1}\cos(B_s^2)dB_s^1-\int_0^t e^{B_s^1}\sin(B_t^2)dBs^2$. From the right side I know that these are local martingales, but are they martingales, and how can I show that? Do I have to show that it is bounded? Please help me!!!

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Apply Ito's formula for: $$ f(x_1,x_2)=e^{x_1}\cos(x_2) $$ $$ f_{x_1}(x_1,x_2)=e^{x_1}\cos(x_2)\quad f_{x_2}(x_1,x_2)=-e^{x_1}\sin(x_2) $$ $$ f_{x_1x_1}(x_1,x_2)=e^{x_1}\cos(x_2)\quad f_{x_1x_2}(x_1,x_2)=-e^{x_1}\sin(x_2) $$ $$ f_{x_2x_2}(x_1,x_2)=e^{x_1}\cos(x_2) $$ Then: $$ \mathrm{d}f(B_1,B_2)=f_{x_1}(B_1,B_2)\mathrm{d}B_1+f_{x_2}(B_1,B_2)\mathrm{d}B_2+ $$ $$ +\frac{f_{x_1x_1}(B_1,B_2)+2{f_{x_1x_2}(B_1,B_2)}+{f_{x_2x_2}(B_1,B_2)}{}}{2}=... $$

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  • $\begingroup$ but how do I know that this is a real martingale? $\endgroup$ – Mathfreak Apr 25 '16 at 20:25
  • $\begingroup$ $\mathbb{E}(|e^{B_1(t)}\cos(B_2(t))|)\leq\mathbb{E}e^{B_1(t)}$. Since $B_1(t)$ has distribution $\mathcal{N}(0,\sqrt{t})$, this is the moment generating function of a normal at $u=1$, and since $\mathbb{E}(e^{uX})=\exp({u\mu+u^2\sigma^2/2})$. $\endgroup$ – Ákos Somogyi Apr 25 '16 at 20:31
  • $\begingroup$ do you know the theorem or a link? $\endgroup$ – Mathfreak Apr 25 '16 at 20:36
  • $\begingroup$ le.ac.uk/users/dsgp1/COURSES/MATHSTAT/6normgf.pdf $\endgroup$ – Ákos Somogyi Apr 25 '16 at 20:39
  • $\begingroup$ i meant something like if $\mathbb{E}[|X_t|]<\infty$ a local martingale is a martingale. Thanks for the rest. :) $\endgroup$ – Mathfreak Apr 25 '16 at 20:59

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