Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 29 / 4 / a / Solution

Put $U=M-\frac{1}{2}[M]$. Since the bracket is continuous finite variation, $[U]=[M]$. The Itô formula applied to $Z=e^U$ givesThereforeWith the printed exponential formula, $Z_0=e^{M_0}$. The usual unit-initial-value stochastic exponential is $\exp (M_t-M_0-\frac{1}{2}[M{]}_t)$; the two coincide when $M_0=0$, as in part (c).

For any strictly positive solution of this stochastic equation, its bracket is $d[Z{]}_t=Z_t^{2}d[M{]}_t$. Applying Itô to the logarithm, after localizing away from zero if necessary, givesThe same identity holds for $Z^{\prime }$, so $\log Z_t^{\prime }-\log Z_t$ is constant in time and is zero initially. Their continuous versions agree at all times on one full-probability event. Thus pathwise uniqueness for a multiplicative martingale equation givesThis is indistinguishability of stochastic processes, not just equality almost surely at each separately chosen time.