Assuming the dynamics of the exchange rate between two currencies at time $t$ is given by:
$$ dX_t=\Delta r X_t dt+ σ X_t dW_t$$
Is the FX Reverse process $\frac{1}{X_t}$ a brownian motion?
How can Ito's Lemma be applied to prove that?
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Plese write down your model for $FX$ - then on can look at $1/FX$. – Richi Wa Nov 20 '14 at 14:18
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1/Fx is supposed to be brownian process. How to prove it. – user13524 Nov 20 '14 at 16:45
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Its a GBM Process not a brownian. See below. – Drew Nov 20 '14 at 19:30
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I tried to keep you notation but frankly the way you expressed your question and added details in comments were not very community-friendly. Please pay some attention to formatting and clarity next time. – SRKX Nov 21 '14 at 01:00
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Well, if you assume Fx is a Brownian Motion $W_t$ then $\frac{1}{X_t} = -\frac{1}{X^2_t} \bullet X_t + \frac{1}{X^3_t} \bullet \langle X\rangle_t = -\frac{1}{X^2_t} \bullet X_t + \frac{1}{X^3_t} \bullet \sigma^2 X^2_t t$.
So $d\bigl(\frac{1}{X_t}\bigr) = -\frac{1}{X_t} [(\Delta r + \sigma^2 ) dt + \sigma dW_t ]$
Setting $\frac{1}{X_t} = M_t$, we see it is not a Brownian Motion, but a Geometric Brownian Motion. I think thats the proof you were asked.
This is Yor's notation where $H \bullet X = \int H dX_s$
Drew
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Nice solution! I also think that it is important that it is GBM, which is a positive process. With BM we could run into serious problems at zero. – Richi Wa Nov 21 '14 at 07:24
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