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


2 Answers 2


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).


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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