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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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  • $\begingroup$ Plese write down your model for $FX$ - then on can look at $1/FX$. $\endgroup$
    – Richi Wa
    Nov 20, 2014 at 14:18
  • $\begingroup$ 1/Fx is supposed to be brownian process. How to prove it. $\endgroup$
    – user13524
    Nov 20, 2014 at 16:45
  • $\begingroup$ Its a GBM Process not a brownian. See below. $\endgroup$
    – Drew
    Nov 20, 2014 at 19:30
  • $\begingroup$ 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. $\endgroup$
    – SRKX
    Nov 21, 2014 at 1:00

1 Answer 1

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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 $H \bullet X = \int H dX_s$

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  • $\begingroup$ 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. $\endgroup$
    – Richi Wa
    Nov 21, 2014 at 7:24
  • $\begingroup$ I changed it to $\Delta r$ on seeing the change made above. $\endgroup$
    – Drew
    Nov 21, 2014 at 15:13
  • $\begingroup$ Your quadratic appears missing the fraction $\frac{1}{2}$. $\endgroup$
    – Gordon
    Apr 1, 2015 at 15:54

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