Geometric Brownian Motion and Long-Run Exchange Rate Behavior

# Why would exchange rates follow a geometric brownian motion?


# Why would exchange rates follow a geometric brownian motion?












I'm reading Shreve's Stochastic Calculus for Finance.

On page 382, he begins talking about exchange rates:

> Finally, there is an exchange rate $Q(t)$, which gives units of domestic currency per unit of foreign currency. We assume this satisfies $$\mathrm{d}Q(t) = \gamma(t)Q(t)\mathrm{d}t + \sigma_2(t)Q(t)\Big[\rho(t)\mathrm{d}W_1(u) + \sqrt{1-\rho^2(t)} \mathrm{d}W_2(t) \Big]\text{.}\tag{9.3.2} $$ We define $$ W_3(t) = \int_0^t \rho(u) \mathrm{d}W_1(u) + \int_0^t \sqrt{1-\rho^2(t)} \mathrm{d}W_2(t)\text{.}\tag{9.3.3}$$ By Lévy's Theorem, Theorem 4.6.4, $W_3(t)$ is a Brownian motion under $\mathbb{P}$. We may rewrite (9.3.2) as $$\mathrm{d}Q(t) = \gamma(t)Q(t)\mathrm{d}t + \sigma_2(t)Q(t) \mathrm{d}W_3(t)\text{,}\tag{9.3.4}$$ from which we see that $Q(t)$ has volatility $\sigma_2(t)$.

Why would it make sense to model exchange rates as in (9.3.4)? Why would exchange rates be compounding? Wouldn't that result in every increasing (or decreasing) exchange rates?

If $\gamma(t)$ is chosen to prevent that, how are we meant to choose $\gamma(t)$?

## Answer by fes (score 2)

https://quant.stackexchange.com/a/69068

You are right that for long horizons this may be a strange model for FX dynamics. However, it doesn't always result in the FX rate tending to zero or infinity. For constant parameters the GBM has the well known solution.

$$Q(t)=Q(0)\exp\left((\gamma-\frac{1}{2}\sigma^2)t+\sigma W_{3}(t)\right)$$

For example if $$\gamma-\frac{1}{2}\sigma^2>0$$

the rate tends to infinity almost surely. But if you set $\gamma-\frac{1}{2}\sigma^2=0$ is does not tend to anything.

If we allow the coefficients to depend on $Q$, we can also set $\gamma(t)=0$ and $\sigma(t)=\sigma \frac{1}{Q(t)}$. This gives the zero mean arithmetic Brownian motion which also does not tend to anything:

$$\frac{dQ(t)}{Q(t)}=\sigma dW_3(t)$$

## Answer by jdaw1 (score 0)

https://quant.stackexchange.com/a/69071

Whichever distribution is chosen, is must be able to cope with the Argentine peso, which has had multiple redenominations. Wikipedia: “After the various changes of currency and dropping of zeros, one peso convertible was equivalent to 10 trillion pesos moneda nacional.”

Whichever distribution is chosen, is must be able to cope with the US dollar, which has gone from being worth a handful of pesos moneda nacional to being worth about a quadrillion.