Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 37 / 3

Set $N_t={\int }_0^t{Z}_s^{-1}d{Z}_s$, the stochastic logarithm of $Z$. On each finite horizon, a strictly positive continuous path has its reciprocal bounded; stopping at levels of $Z$, $Z^{-1}$ and $[Z]$ rigorously localizes this stochastic integral. Thus $N$ is a zero-start continuous local martingale, withThe Itô formula for the ordinary logarithm givesAccordingly, with the exponential convention printed in this question,The initial constant in $M$ contributes no quadratic variation. Under the standard normalized Doléans-Dade exponential convention, which subtracts the driver's initial value in the exponent, the same formula is written $Z=Z_0\mathcal{E}(N)$.

For uniqueness, suppose another continuous driver $L$ gives the printed exponential. Evaluating at zero forces $L_0=\log Z_0$. Applying the Itô formula to its exponential gives $d{Z}_t=Z_{t}d{L}_t$. Since $Z_t>0$, associativity of stochastic integration yields $d{L}_t=Z_t^{-1}d{Z}_t$. Thus $L_t=L_0+N_t=M_t$, up to indistinguishability of stochastic processes.

There is an initial-integrability qualification if “local martingale” is used with the usual integrable-initial-value definition. The displayed $M$ is a continuous local martingale when $\log Z_0\in L^1$, in particular when $Z_0$ is a positive deterministic number. An arbitrary positive integrable $Z_0$ does not ensure this. For example, take an ${\mathcal{F}}_0$-measurable positive integer $W$ with $\mathbb{P}(W=n)=6/({\pi }^2{n}^2)$ and put $Z_t=e^{-W}$ for all $t$. This is a bounded strictly positive martingale, but any representing driver must have $M_0=-W$, which is not integrable. Thus the unqualified claim is false under that definition. For general $Z_0$, the always-valid local-martingale statement is the zero-start representation $Z=Z_0\mathcal{E}(N)$; the printed representation additionally requires the stated integrability or the convention that allows arbitrary finite initial constants.

Let $D=d\mathbb{Q}/d\mathbb{P}$ on $\mathcal{F}$. The Radon-Nikodym derivative is nonnegative and ${\mathbb{E}}_{\mathbb{P}}D=1$. For $A\in {\mathcal{F}}_t$,Uniqueness of the restricted Radon-Nikodym derivative therefore identifiesFor $s\le t$, the tower property of conditional expectation gives ${\mathbb{E}}_{\mathbb{P}}(Z_t\mid {\mathcal{F}}_s)=Z_s$, and ${\mathbb{E}}_{\mathbb{P}}{Z}_t=1$. Thus $Z$ is a nonnegative true martingale, indeed a uniformly integrable martingale, since it consists of conditional expectations of one integrable random variable. This is the Radon-Nikodym density martingale. Under the usual conditions for a filtration, it has the standard càdlàg version; continuity is the additional assumption in part (c), not a consequence of absolute continuity alone.

On the finite horizon $[0,T]$, strict positivity gives equivalence of the restricted probability measures: for $A\in {\mathcal{F}}_T$, $\mathbb{Q}(A)={\mathbb{E}}_{\mathbb{P}}(Z_T{1}_A)$ vanishes if and only if $\mathbb{P}(A)$ vanishes. The standard change-of-measure theorem gives semimartingale invariance under equivalent measures: a process is a semimartingale under $\mathbb{P}$ on this horizon if and only if it is one under $\mathbb{Q}$. The sample-path quadratic variation is unchanged, but the local-martingale and finite-variation parts generally change.

More explicitly, set $N=\int Z^{-1}dZ$, the zero-start stochastic logarithm of the density. For every continuous $\mathbb{P}$-local martingale $L$, the Girsanov theorem states thatis a continuous $\mathbb{Q}$-local martingale. Consequently, if the continuous semimartingale decomposition under $\mathbb{P}$ is $X=X_0+L+V$, its decomposition under $\mathbb{Q}$ isThe covariation correction is a continuous finite-variation process, locally well-defined because $Z$ stays strictly positive. In the Brownian case, if $d{Z}_t=Z_t{\theta }_{t}d{B}_t$, thenis a Brownian motion under $\mathbb{Q}$. The inverse finite-horizon density is $1/Z$, giving the reverse change of measure. These assertions apply on each horizon on which the density is strictly positive; they do not assert equivalence on a larger terminal sigma-algebra where the full density $D$ may vanish.