When a Driftless Exchange-Rate SDE Is a Martingale

Summary

The document asks why an exchange-rate process governed under a domestic risk-neutral measure by a proportional stochastic differential equation with constant volatility and no drift is a martingale. One response invokes the stochastic exponential solution and checks its expectation using the normal distribution of Brownian motion, concluding that the process retains its initial expected value.

Another response gives the intuition that Brownian motion is itself a martingale and that this simple driftless model has the expected value required for a martingale. It also warns that a driftless stochastic differential equation does not in general guarantee a true martingale: some processes are only local martingales or supermartingales. The discussion is limited to the stated model and contains a likely formula error in the displayed solution and expectation calculation, so its derivation should not be relied on without correction. For the stated SDE, the usual stochastic exponential has a quadratic-variation correction proportional to time, not time squared.

Key ideas

• A martingale has conditional expected future value equal to its current value.
• The document explains the stated driftless exchange-rate model through its stochastic exponential solution.
• Brownian motion is a martingale, providing intuition for the model's behavior.
• A driftless SDE is not automatically a true martingale in every setting.
• The displayed derivation contains a time-dependence error and needs correction before use.

Tags

# Exchange rate model and Martingales


# Exchange rate model and Martingales












In exchange rate model explanation,

"...If under the domestic risk neutral measure $Q_d$, the process $X(t)$ satisfies

$\displaystyle \frac{dX(t)}{X(t)}=\sigma dZ_d(t)$

Since $Z_d(t)$ is $Q_d$-Brownian motion, so $X(t)$ is a $Q_d$-martingale."

How does this last line work? I cannot see what theorem has been applied

## Answer by emcor (score 1, accepted)

https://quant.stackexchange.com/a/14319

The solution of your SDE is known as the Stochastic Exponential:

$$X_t=X_0e^{\sigma Z_t-\sigma^2t^2/2}$$

(You can check this solution by applying Ito to the function $f(t,Z_t)=\ln X_t$.)

Taking the expectation of $X_t$ to check its martingale property, since $Z_t\sim N(0,t)$ then $E(e^{\sigma Z_t})=e^{\sigma^2t^2/2}$:

$$E(X_t)=E(X_0e^{\sigma Z_t-\sigma^2t^2/2})=X_0e^{-\sigma^2t^2}E(e^{\sigma Z_t})=X_0$$

Hence $X_t$ is martingale for $Z_t$ Brownian Motion (and Brownian Motions are Martingales).

## Answer by SmallChess (score 4)

https://quant.stackexchange.com/a/14317

Most of the time, when you have a simple SDE without a drift, it's a martingale because the Wiener process itself is a martingale. In your example, you have a constant with the Wiener process, therefore the whole process must also be a martingale because the expectation is clearly X(t).

However, we can't conclude a driftless SDE is always a martingale. There're cases that a driftless SDE is a local-martingale or super-martingale. What we can conclude in your example that it's an obvious martingale.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.