Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 25 / 1 / c

Apply the construction of part (a) to the real Hilbert space $H=L^2([0,\infty ),du)$, and setChoose $X(0)=0$. The interval of length zero represents the zero vector of $H$, so $W_0=0$. All variables are defined on the single probability space already used for the isonormal Gaussian process.

For $0\le s<t$, linearity of the isonormal Gaussian process givesThe squared $L^2$ norm of this indicator is $t-s$, so part (a) yieldsAlso, the covariance formula gives $\mathrm{Cov}(W_s,W_t)=s\wedge t$, the Brownian covariance kernel. The Gaussian statement concerns the signed increment.

Let ${\Delta }_j=W_{t_j}-W_{t_{j-1}}$ for $1\le j\le n$. These increments are jointly normal, because each is obtained by evaluating the isonormal Gaussian process on an interval indicator. Indicators of distinct intervals are orthogonal in $L^2([0,\infty ))$, soBy uncorrelated jointly Gaussian variables are independent, the increments on all these disjoint intervals are independent. Reversing the sign of the first increment, as in the printed list, preserves this independence.

The three requested properties have now been obtained without an existence theorem for Brownian motion. One can also obtain continuous paths: the Gaussian fourth moment gives $\mathbb{E}|W_t-W_s{|}^4=3|t-s{|}^2$. The Kolmogorov continuity theorem therefore supplies a continuous modification on every finite time interval, which can be chosen consistently on the half-line. Modification preserves every finite-dimensional distribution and hence the independent Gaussian increments. This yields Brownian motion itself.