Applying Itô’s Lemma to the Reciprocal of an FX Rate

Summary

The document examines whether the reciprocal of an exchange rate following geometric Brownian motion is itself a Brownian motion. It applies Itô’s lemma to the function that takes a rate to its reciprocal, accounting for both the first derivative term and the quadratic-variation correction. The resulting process has a proportional stochastic term and a drift term, so the reciprocal remains a geometric Brownian motion rather than a standard Brownian motion.

The explanation is a compact calculation, not an empirical study. Its conclusion depends on the assumed exchange-rate dynamics and notation in the question; it does not address alternative models, boundary behavior, or how the result would be used in a trading strategy. The distinction matters because transforming a stochastic process can change its drift even when the transformed process still contains Brownian noise.

Key ideas

• Itô’s lemma includes a quadratic-variation correction when applied to the reciprocal function.
• The reciprocal of a geometric Brownian exchange rate has proportional diffusion and drift.
• A process with this form is geometric Brownian motion, not standard Brownian motion.
• The conclusion relies on the exchange rate following the stated stochastic model.

Tags

# What are the dynamics of the reverse of this FX process?


# What are the dynamics of the reverse of this FX process?












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?

## Answer by Drew (score 3)

https://quant.stackexchange.com/a/15553

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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