Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 29 / 2

The gradient bound makes $f$ a Lipschitz function, with $|f(x)|\le |f(0)|+K\Vert x\Vert$, so its value at a Gaussian random variable is integrable and square integrable. The future Brownian increment $X_1-X_t$ is independent of ${\mathcal{F}}_t$ and has distribution $N(0,(1-t)I_d)$. Conditioning on the current location therefore givesThis is the Markov property expressed through the Brownian transition semigroup. In particular $M_0=\mu$ and $M_1=f(X_1)$.

Let $u(t,x)=P_{1-t}f(x)$ for $t<1$. Differentiating the Brownian transition semigroup in the spatial variables is justified by the bounded gradient and givesThe heat equation for this semigroup is the backward equation ${\partial }_{t}u+\frac{1}{2}\Delta u=0$. Applying the multidimensional Itô formula yieldsThe finite-variation term vanishes. ThusInitially this calculation holds on $[0,1-\epsilon ]$. The vector of stochastic integral coefficients has norm at most $K$, so the Itô isometry lets the integrals extend to time one. Also $P_{1-t}f(X_t)\to f(X_1)$ in $L^2$ by the Lipschitz function bound and continuity of Brownian motion. The same representation therefore holds on the entire interval.

Let $g_t=P_{1-t}(\nabla f)(X_t)$. The triangle inequality for the conditional average givesThe independent coordinate Brownian motions have zero cross-variation, so the quadratic variation of a stochastic integral givesThe Dambis-Dubins-Schwarz theorem says that a continuous local martingale $N$ starting at zero whose quadratic variation diverges admits a Brownian motion $B$, with the inverse-clock filtration, such that $N_t=B_{[N{]}_t}$. If the total clock is finite, one may enlarge the probability space and append independent Brownian motion after the clock terminates. This gives the same identity up to the original lifetime.

Apply this theorem to $N=M-M_0$. The clock bound implies the event inclusionThe latter event is contained in the union of the upper and lower crossing events. Symmetry of Brownian motion givesThis uses only an event inclusion; independence between the Brownian motion and its random clock is neither assumed nor needed.

The Brownian reflection principle and Brownian scaling givewhere $Z$ has the standard normal distribution. Combine part (c) with the allowed Gaussian tail bound to obtainThis is the Brownian martingale proof of Gaussian concentration. It proves the intended Gaussian concentration inequality. The denominator printed in the initial bound is $2K$, whereas the gradient hypothesis and part (c) give $2K^2$. The stronger printed bound is false in general. For a concrete counterexample, take $d=1$, $K=2$, and $f(x)=2x$. At $r=10$,where the first strict inequality follows by integrating the decreasing normal density over $[5,6]$. The rightmost term is the printed bound. Thus the square on $K$ is a necessary source correction, rather than a change of proof technique.