Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 30 / 6 / c / Solution

Under Wiener measure, the endpoint has normal distribution $N(x,t{I}_d)$. For $t>0$ and $1+\lambda t>0$, the Gaussian integral isComplete the square:Evaluating the centered Gaussian gives $(1+\lambda t{)}^{-d/2}\exp [-\lambda |x{|}^2/(2(1+\lambda t))]$. With the preceding prefactor, the critical quadratic potential for the Ornstein-Uhlenbeck generator solution isThe initial value is one. As a direct PDE check, write $u=a(t)e^{b(t)|x{|}^2/2}$, whereSince $\nabla u=bxu$ and $\Delta u=(db+b^2|x{|}^2)u$, these identities give precisely the required time derivative and spatial operator.

For $\lambda \ge 0$, this is finite for every $t\ge 0$; at $\lambda =0$ it reduces to $u=1$. If the parameter is allowed to be negative, the expectation is finite only for $t<-1/\lambda$. At and beyond that time the quadratic coefficient in the endpoint Gaussian integral is nonnegative, so the integral diverges. A nonnegative classical solution on the entire half-line cannot then exist: stopping the Feynman-Kac martingale in balls and discarding its nonnegative boundary term bounds any such solution below by the truncated path expectations; their limit is infinite. Thus the printed global positive-solution premise implicitly requires the nonnegative parameter convention.