Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 30 / 6

Choose independent random variables ${\xi }_i\sim N(0,1)$ on a countable product probability space. For each fixed $t$, the diagonal assumption givesThe partial sums ${\sum }_{i\le n}{k}_i(t){\xi }_i$ are therefore Cauchy in $L^2$. DefineChoose a representative of this limit for each $t$. No path continuity or simultaneous series convergence over all uncountably many times is being asserted.

For any finite list $t_1,\dots ,t_m$ and real coefficients $a_j$, the linear combination ${\sum }_j{a}_j{X}_{t_j}$ is the $L^2$ limit of centered Gaussian variablesTheir variances converge, so their characteristic functions converge to that of a centered normal distribution. This proves that every finite-dimensional vector is Gaussian and hence that $X$ is a Gaussian process. Taking $L^2$ limits also givesThis is the Gaussian process construction from square-summable features.

In the real Hilbert space $L^2({\mathbb{R}}_{+})$, let $f_t=1_{[0,t]}$. Then $k_i(t)=⟨f_t,h_i⟩$. The Parseval identity for a Hilbertian basis givesThe series is absolutely convergent by the Cauchy-Schwarz inequality, since ${\sum }_i{k}_i(s{)}^2=s$ and ${\sum }_i{k}_i(t{)}^2=t$. Thusthe Brownian covariance kernel. This is the Brownian covariance from an integrated orthonormal basis.

The variance at zero is zero, so $X_0=0$ almost surely. For $s<t$, the increment is Gaussian with mean zero and varianceFor every $u\le s$,The increment and any finite vector of past values are jointly Gaussian. Zero cross-covariances imply their independence, by the factorization of the joint Gaussian characteristic function. The Monotone class theorem extends this independence to the sigma-field generated by all past values, ${\mathcal{F}}_s^X=\sigma (X_u:u\le s)$.

Together with the assumed continuity, these proveThis is the Gaussian-process characterization of Brownian motion. Completion of the natural filtration preserves the independence assertion.

For $t>0$, write $A_t=t^{-1}{\int }_0^t{W}_{u}du$. Deterministic integrals of the Gaussian process $W$ are Gaussian, as follows by taking $L^2$ limits of Riemann sums, so $(W,A)$ is jointly Gaussian. For $0<s\le t$, integrating the Brownian covariance kernel givesFor the last equality, integrate ${\int }_0^t\min (u,v)dv=ut-u^2/2$ over $0\le u\le s$ and divide by $st$. ConsequentlyThe only nonzero choice giving Brownian covariance is $\boxed{c=3}$; even the variance condition at $s=t$ requires $c(c-3)=0$.

Define ${\hat{W}}_0=0$. Since $|A_t|\le {\sup }_{u\le t}|W_u|\to 0$ almost surely as $t\downarrow 0$, the transformed process is continuous at zero, as well as at positive times. It is centered Gaussian and has covariance $s\wedge t$, so part (c) provesThe transformed process is Brownian in its own natural filtration. This is the Brownian motion transform by three times its running average. The filtration qualification matters: for $t>s>0$,which is not identically zero. The Gaussian covariance calculation identifies the Brownian filtration generated by $\hat{W}$; it does not make the transformed process a martingale in the larger original Brownian filtration.