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

Choose an orthonormal basis $(e_j)$ of the separable Hilbert space and independent standard normal random variables $(Z_j)$ on a common probability space. In finite dimension the sums below are finite; in dimension zero take $X(0)=0$. In infinite dimension defineIndependence, centring and unit variance giveBy the Parseval identity for a Hilbertian basis, the coefficient sequence is square-summable. Completeness of $L^2(\Omega )$ therefore supplies a limit, and we defineEvery partial-sum map is linear, and passage to the $L^2$ limit preserves this identity. Thus for each fixed $a,b,g,h$, $X(ag+bh)=aX(g)+bX(h)$ as an $L^2$ identity, hence an almost sure equality. This constructs an isonormal Gaussian process.

The usual meaning is almost-sure linearity for each fixed choice of arguments. A version linear simultaneously for every argument can also be chosen: select a Hamel basis of $H$, choose a measurable representative of $X$ on each basis vector, and extend each sample algebraically by finite sums. For every fixed $h$, this extension equals the constructed $L^2$ variable almost surely, so all its required distributions are unchanged.

For the construction above, independence of the normal random variables makes $S_N(h)$ normal with mean zero and variance $v_N={\sum }_{j\le N}⟨h,e_j{⟩}^2$. Its characteristic function is $\exp (-{\theta }^2{v}_N/2)$. The $L^2$ convergence gives $L^1$ convergence, andSince $v_N\to \Vert h{\Vert }^2$ by the Parseval identity for a Hilbertian basis, the limiting characteristic function identifiesThis includes $h=0$, where the normal distribution is degenerate at zero.

For any $r_1,\dots ,r_n\in \mathbb{R}$, the linearity from part (a) givesalmost surely, and the right side has a normal distribution. This is the defining linear-combination criterion for a multivariate normal distribution; singular covariance matrices are allowed.

Passing to the limit in the inner products of the partial sums givesThe passage to the limit is justified by Cauchy-Schwarz inequality and $L^2$ convergence. Equivalently, the vector's characteristic function isThus both joint normality and the complete covariance matrix follow from the Hilbert-space inner product.

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.

The density $Z=\exp (X(k)-\Vert k{\Vert }^2/2)$ is strictly positive. The Gaussian moment-generating function gives $\mathbb{E}Z=1$, so it defines an equivalent probability measure.

The joint normal distribution of $(X(h),X(k))$, with covariance $⟨h,k⟩$, gives the mixed exponential formulaMultiplying by the normalizing and centring factors therefore yieldsThe characteristic function identifies the answer:This is exponential tilting of an isonormal Gaussian process: the mean shifts by the inner product while its covariance remains unchanged.