Suppose we have that $dp=u dt+s dz$, where $dz$ is brownian motion. What would be $d^2 p$? that is, the second derivative under ito calculus.
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you are right. possibly you can proceed by analogy with 2D, taking a total differential of a sorts to get $d^2 p = u (dt)^2 + s (dz)^2$ and maybe using $dz^2 = dt$, but not sure how. – gt6989b Feb 18 '16 at 19:28
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That did not answer the question. Does anyone know the answer to the question? – he wei Feb 19 '16 at 20:20
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I'd recommend you refrain from commenting if you don't know the answer, especially you asked me several questions that was not productive at all and was taking up other people's time. – he wei Feb 19 '16 at 22:21
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when i posted an answer, i made a mistake in calculations, so i thought i well knew what the answer was. as for this comment, it is not an answer, just a suggestion along which lines i would think... – gt6989b Feb 22 '16 at 19:05
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That's why I asked you to post the answer first right off the bat. – he wei Mar 05 '16 at 06:50