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I am trying to solve the following problem:

Give the Stochastic Differential Equations for $X_t = \cos B_t$ and $Y_t = \sin B_t$, where $B_t$ is the standard Brownian Motion.

I have used the following Ito formula:

$$dg(t,B_t) = \frac{\partial g}{\partial t}(t, B_t)\,dt + \frac{\partial g}{\partial x}(t, B_t)\,dB_t + \frac12\frac{\partial^2 g}{\partial x^2}(t, B_t)\,dt$$ where $g(t, B_t) = \cos B_t$ and $g(t, B_t) = \sin(t, B_t)$ for $X_t$ and $Y_t$ respectively.

This results in:

$$dX_t = -\frac12\cos B_t\,dt - \sin B_t\,dB_t\,,$$

$$dY_t = -\frac12\sin B_\,dt + \cos B_t\,dB_t\,.$$

But I wasn't sure if I was missing anything.

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  • $\begingroup$ This question doesn't make any sense [yet]. What is an SDE? What is $B(t)$? $\endgroup$ Oct 3, 2023 at 11:41
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    $\begingroup$ Hint: what are $dX$ and $dY\,?$ $\endgroup$
    – Kurt G.
    Oct 3, 2023 at 11:56
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    $\begingroup$ Hint: Ito's lemma. $\endgroup$
    – Kurt G.
    Oct 3, 2023 at 12:59

1 Answer 1

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From \begin{align} X_t&=\cos B_t\,,&Y_t&=\sin B_t \end{align} you got by Ito's formula \begin{align} dX_t&=-\sin B_t\,dB_t-\frac12\cos B_t\,dt\,,&dY_t&=\cos B_t\,dB_t-\frac12\sin B_t\,dt \end{align} which can be written as a coupled SDE for $X$ and $Y\,,$ \begin{align} dX_t&=-Y_t\,dB_t-\frac12X_t\,dt\,,&dY_t=X_t\,dB_t-\frac12Y_t\,dt\,, \end{align} or as two uncoupled SDEs \begin{align} dX_t&=-\sin(\arccos X_t)\,dB_t-\frac12X_t\,dt\,,&dY_t=\cos(\arcsin Y_t)\,dB_t-\frac12Y_t\,dt\,. \end{align}

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