Past exam of the stochastic differential equations and Brownian motion in complex analysis course of the University of Oxford / 2007 / A / 4

Put $a(t)=e^{{\int }_0^t\mu (s)ds}$ and $q(t)=a(t{)}^{-1}$. Since $q^{\prime }=-\mu q$ and $q$ has finite variation, the product rule has no quadratic-covariation contribution:Integrating and multiplying by $a(t)$ yieldsThe deterministic integrand is square integrable on every compact interval. This constructs a strong solution. Two solutions driven by the same Brownian motion have difference solving $dD=\mu Ddt$, $D_0=0$, hence their difference vanishes, proving pathwise uniqueness.

The deterministic stochastic integral in the solution is centered Gaussian with varianceTherefore $X(t,x)$ has mean $a(t)x$ and varianceFor $t>0$ this is strictly positive, since $\sigma >0$. Its density at $y$ isThe boundedness assumptions on $\sigma$ and its reciprocal in particular ensure nondegeneracy; compact-interval continuity is already sufficient for the displayed finite-time construction.

With $G$ a standard normal random variable,For $f\in C_b^1$, in the usual sense that $f$ and its continuous derivative are bounded, the mean value theorem bounds the difference quotient by $a(t)\Vert f^{\prime }{\Vert }_{\infty }$. Dominated convergence theorem therefore allows differentiation inside the integral:This also follows pathwise from $X(t,x+h)-X(t,x)=a(t)h$. If the textual description were read as assuming only bounded differentiability, some control of $f^{\prime }$ is needed for the displayed right side; the stated $C_b^1$ hypothesis supplies it. Gaussian smoothing itself still differentiates a bounded measurable $f$, using the derivative of the Normal distribution instead.