Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 2 / 28J (Stochastic Financial Models)

The independent Gaussian increments of Brownian motion give $E[Y_t\mid {\mathcal{F}}_s]=Y_s{e}^{\mu (t-s)}$; the exponential is integrable at every finite time. ThereforeSimilarly $E[Y_t^{-1}\mid {\mathcal{F}}_s]=Y_s^{-1}{e}^{({\sigma }^2-\mu )(t-s)}$, so $\boxed{Y^{-1}\text{ is a martingale}⟺\mu ={\sigma }^2}$. This uses actual conditional expectations, not only the local stochastic differential equation.

The Dollar bank account is a tradable asset whose Euro value is $Y_t{e}^{r_{D}t}$. Under the Euro equivalent martingale measure, discount this value by $e^{r_{E}t}$. The result is a martingale precisely when $\mu +r_D-r_E=0$, givingWhen the rates coincide, $\mu =0$ and part (a) proves that the exchange rate itself is a martingale under $Q_E$.

For a Dollar investor, the Euro bank account has Dollar value $e^{r_{E}t}/Y_t$. After Dollar discounting, it is $e^{(r_E-r_D)t}/Y_t$. Its conditional mean growth exponent is $r_E-r_D+{\sigma }^2-\mu$, so the martingale requirement givesWith equal rates, $Y^{-1}$ is a martingale under $Q_D$ by part (a), whereas $Y$ has positive drift ${\sigma }^2$ and is not a martingale when $\sigma \ne 0$. In the degenerate zero-volatility case, both exchange rates are constant and both are martingales; the asserted nonmartingale conclusion implicitly requires a genuinely risky exchange rate.

For equal rates, the change of numeraire from the Euro bank account to the Dollar account has density process $Z_t=Y_t/Y_0$, a strictly positive mean-one martingale under $Q_E$ by part (b). Conditional change of measure givesThus the reciprocal is a genuine martingale under $Q_D$, directly from the Euro-measure martingale property, without recalculating its stochastic differential equation.