Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 27 / 3

For a finite list of real coefficients $(a_j{)}_{j\le m}$,A deterministic Itô integral is centered Gaussian: first verify it for step functions as a linear combination of independent Gaussian Brownian increments, then pass to an L2 approximation using the Itô isometry and characteristic functions. Its variance is the squared L2 norm of the integrand, which is ${\sum }_j{a}_j^2$ by the supplied orthonormality. ThereforeThis factors as the joint characteristic function of independent standard normals. Since every finite subfamily has this law, the entire sequence is independent and each ${\xi }_n$ has law N(0,1). This is the Gaussian coordinates of deterministic orthonormal Wiener integrands principle.

Put ${\omega }_n=(n-\frac{1}{2})\pi$. The deterministic derivative is $g_n^{\prime }=-{\omega }_n{h}_n$, and $g_n(1)=0$, while $W_0=0$. The Itô product rule with a deterministic smooth function givesAlmost every Brownian path is continuous, hence belongs to $L^2[0,1]$. Its Fourier coefficient in the given orthonormal basis $(h_n)$ is therefore ${\xi }_n/{\omega }_n$. The Parseval identity for a Hilbertian basis gives, pathwise on a probability-one event,This is the squared-norm consequence of the Brownian half-integer sine expansion; completeness, rather than pointwise convergence of a Fourier series, is all that is needed.

The nonnegative series in part (b) has independent squared-standard-normal terms. If $G$ is standard normal, direct Gaussian integration gives $\mathbb{E}{e}^{-a{G}^2}=(1+2a{)}^{-1/2}$ for $a\ge 0$. Consequently, by independence and dominated convergence applied to the exponentials of increasing partial sums,Use the permitted product identity with $x=\sqrt{2\lambda }$. The answer isThis Laplace transform of the integrated square of Brownian motion equals $1$ at zero. Its first derivative there gives mean $1/2$, in agreement with ${\int }_0^1\mathbb{E}{W}_t^{2}dt$.