How do I differentiate this stochastic process?

Question

I have the following stochastic equation:

$X(t)=e^{-t}W(e^{2t})$

How do I now find $dX$? I have tried using Ito's equations but I seem to get lost in the calculus ($W$ is a Brownian motion).

Answer 1

Basically copy-paste from https://en.wikipedia.org/wiki/It%C3%B4%27s_lemma

Set $$h(t,x)=e^{-t}x$$ so $$X(t)=h\left(t,W(e^{2t})\right)$$ and consider $${\rm d}X(t)= \frac{\partial h}{\partial t}\left(t,W(e^{2t})\right){\rm d}t + \frac{\partial h}{\partial x}\left(t,W(e^{2t})\right) {\rm d}(W(e^{2t})) + \frac{1}{2}\frac{\partial^2 h}{\partial x^2} \left(t,W(e^{2t})\right) \left({\rm d}(W(e^{2t}))\right)^2 \\ =-e^{-t}W(e^{2t}) \, {\rm d}t + e^{-t} {\rm d}(W(e^{2t})) + 0 \\ =-e^{-t} W(e^{2t}) \, {\rm d}t + \sqrt{2} \, {\rm d}W(e^{2t})$$ where ${\rm d}W$ has variance ${\rm d}t$.

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I think in the last line, second term it should be $\sqrt{2}$ instead of $2e^{t}$, as $${\rm d}(W(e^{2t}))=W(e^{2(t+{\rm d}t)})-W(e^{2t}) = W(e^{2t}+2e^{2t}{\rm d}t)-W(e^{2t}) \sim \mathcal{N}(0,2e^{2t}{\rm d}t)$$ for infinitesimal ${\rm d}t$, while $${\rm d}W(e^{2t}) \sim \mathcal{N}(0,{\rm d}t)$$ i.e. $$({\rm d}(W(e^{2t})))^2=2e^{2t}({\rm d}W(e^{2t}))^2 \, .$$

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Could you add some further information on how you obtained this motion and what you would expect and why?