= Bounded Girsanov density for exit of an unstable linear diffusion
{title2=$Z_t=\exp\!\left(\tfrac12X_{t\wedge T}^2-\tfrac12(t\wedge T)-\tfrac12\int_0^{t\wedge T}X_s^2ds\right)$}
Under $P$, let $X$ be standard <Brownian motion> started at zero and let $T$ be its first exit from $(-1,1)$. The <Itô formula> makes
$$
Z_t=\exp\!\left(\frac{X_{t\wedge T}^2}{2}-\frac{t\wedge T}{2}-\frac12\int_0^{t\wedge T}X_s^2ds\right)
$$
the <stochastic exponential> with integrand $X\mathbf1_{[0,T]}$. Since $Z_t\leq e^{1/2}$, it is a <uniformly integrable martingale> and defines $dQ=Z_TdP$ on the whole original <sigma-algebra>. The <Girsanov theorem> makes $X_t-\int_0^{t\wedge T}X_sds$ a <Brownian motion> under $Q$.
At exit, $|X_T|=1$ and $\int_0^TX_s^2ds\leq T$, so $e^{1/2-T}\leq Z_T\leq e^{1/2}$. On $\{T\leq t\}$, integrate these likelihood bounds in $Q(T\leq t)=\mathbb E_P[Z_T\mathbf1_{\{T\leq t\}}]$ to obtain
$$
e^{-1/2}Q(T\leq t)\leq P(T\leq t)\leq e^{t-1/2}Q(T\leq t).
$$
A lower bound on $Z_T=dQ/dP$ gives an upper bound on $P$ in terms of $Q$; reversing that direction gives an invalid comparison.
Back to article page
