i am trying to construct a brownian motion on the sphere using the method given in Price and williams paper.$\partial$ represents the SDE of stratonovich type which is converted to ito form in last expression $$\partial X=n(X) \times \partial B $$ where the $\times$ is a vector product and $n(X)$ is a unit normal vector on the surface the $$n(X)=\begin{pmatrix} -\cos(\theta)\cos(\phi) &&0&&0 \\ 0&& \sin(\theta)\sin(\phi) && 0\\0&& 0&& -\cos(\phi)\end {pmatrix}$$

equation that i formed is $$\partial \begin{pmatrix}\cos(\theta)\sin(\phi)\\\sin(\theta)\sin(\phi)\\\cos (\phi)\end{pmatrix} =\begin{pmatrix} -\cos(\theta)\cos(\phi) &&0&&0 \\ 0&& \sin(\theta)\sin(\phi) && 0\\0&& 0&& -\cos(\phi)\end {pmatrix}\times \begin {pmatrix} \partial B_1\\\partial B_2\\\partial B_3\end {pmatrix}$$ and subsequently changing the equation back to ito form i get $$d\phi=\cot(\phi)dB_3+\frac{1}{2}dt $$ but the result in the book is given by $$d\phi=dB_3+\frac{1}{2}\cot(\phi) dt $$

i cant find where i am wrong please help

the text i am referring is this the

  • $\begingroup$ Shouldn't it be $\partial X = X \times \partial B$? If $X$ is a position vector of a point on the sphere, then the unit normal vector at that point is also $X$. (Also, why does your normal vector look like a matrix?) $\endgroup$ – Lukas Geyer Apr 14 '16 at 19:40
  • $\begingroup$ i thought of $n(X)$ as the unit normal vector which i represented as a matrix @LukasGeyer $\endgroup$ – Nebo Alex Apr 14 '16 at 19:53
  • $\begingroup$ Sorry, I don't know what that means, or what a cross product between a matrix and a vector is. But most importantly, where do you get the entries in your $n(X)$ from? Shouldn't they be the same as in $X$? $\endgroup$ – Lukas Geyer Apr 14 '16 at 20:05
  • $\begingroup$ it is a normal by using the $n(X)=\frac{\Phi_{\theta}\times\Phi_{\phi}}{|\Phi_{\theta}\times\Phi_{\phi}|}$ $\endgroup$ – Nebo Alex Apr 15 '16 at 9:34
  • $\begingroup$ i think that i attempted this question wrongly @lukasGeyer $\endgroup$ – Nebo Alex Apr 15 '16 at 17:25

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