Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 34 / 4

The stochastic exponential of $M$ isItô's formula gives $d\mathcal{E}(M{)}_t=\mathcal{E}(M{)}_{t}d{M}_t$, so it is a positive continuous local martingale starting from one. For $p>1$,Every nonnegative local martingale is a supermartingale. Hence, under $[M{]}_{\infty }\le C$,The bound applies also to stopped versions. Localize the exponential to true martingales. Their uniform $L^p$ bound permits $L^1$ passage to the limit, proving that the unstopped exponential is a true martingale of mean one. The same bound makes its entire time-indexed family uniformly integrable. This proves the bounded-bracket criterion for a stochastic exponential, without needing an additional exponential-moment assumption.

One continuous Girsanov theorem formulation is as follows. Let $L$ be a zero-starting continuous local martingale such that $Z=\mathcal{E}(L)$ is a strictly positive uniformly integrable martingale, and define $d\mathbb{Q}=Z_{\infty }d\mathbb{P}$. For every continuous $\mathbb{P}$-local martingale $N$,On a finite horizon the same assertion holds with terminal density $Z_T$, provided $Z$ is a true martingale through $T$. In particular, if $L_t={\int }_0^t{\theta }_{s}d{B}_s$, thenis a Brownian motion under $\mathbb{Q}$, on the specified horizon. Its quadratic variation remains $t$ under this absolutely continuous measure (equivalent on every finite-horizon sigma-algebra), and the Lévy characterization of Brownian motion identifies it as Brownian. The density sign and drift subtraction are paired: choosing the exponential of $-\int \theta dB$ instead produces $B+\int \theta ds$.

Use the predictable processThe value at zero is irrelevant to the Brownian integral. The numerator is bounded and ${\sigma }_s\ge \delta >0$, so $\theta$ is bounded and $[L{]}_{\infty }={\int }_0^T{\theta }_s^{2}ds$ has a deterministic bound. Part (a) supplies a uniformly integrable density martingale. Define a measure on the entire given sigma-algebra byIts density is positive and has expectation one, so $\mathbb{Q}$ is a probability measure equivalent to $\mathbb{P}$. The Girsanov theorem makes ${\tilde{B}}_t=B_t-{\int }_0^{t\wedge T}({\nu }_s-{\mu }_s)/{\sigma }_{s}ds$ a Brownian motion under $\mathbb{Q}$. For $t\le T$, substitute $dB=d\tilde{B}+\theta dt$:Continuity of $\sigma$ gives pathwise local square integrability, even though it is not assumed bounded. Equivalence preserves this property. This is finite-horizon drift replacement by a change of measure.