Past exam of the continuous martingales and stochastic calculus course of the University of Oxford / 2023 / 1 / b

Let $S_N={\sum }_{n=0}^N{\sum }_{k=0}^{2^n-1}{a}_{n,k}{H}_{n,k}$. The assumed orthonormality of the Haar wavelets givesEnumerate the coefficients level by level as $A_1,A_2,\dots$. They are independent and identically distributed standard normal random variables, so the squares have mean one. The strong law of large numbers impliesConsequently,A convergent sequence in a Hilbert space must be bounded in norm. Hence this random Haar wavelet series diverges in $\mathcal{H}$ almost surely.

Enumerate the orthonormal basis consisting of $H_0$ and the Haar wavelets $H_{n,k}$ as $(e_j{)}_{j\ge 1}$, and enumerate the corresponding independent standard normal random variables as $(A_j{)}_{j\ge 1}$. Set $c_j=⟨h,e_j{⟩}_{\mathcal{H}}$. The Parseval identity for a Hilbertian basis givesThe partial sums $T_m={\sum }_{j=1}^m{c}_j{A}_j$ form a martingale for $\sigma (A_1,\dots ,A_m)$, since the next summand is independent of the past and has mean zero. Moreover,The martingale convergence theorem for a martingale bounded in L2 space proves convergence almost surely and convergence in L2. Equivalently one may apply the Kolmogorov convergence theorem to the summable variances $c_j^2$.

A finite linear combination of independent normal random variables is normal. Its characteristic function is $\mathbb{E}{e}^{iu{T}_m}=\exp (-u^2{\sum }_{j\le m}{c}_j^2/2)$. Passing to the limit identifiesFor $h=0$ this is the constant zero random variable. Each fixed $h$ has its own permissible null set; simultaneous pathwise convergence for all $h\in \mathcal{H}$ is unnecessary.

For any $h_1,\dots ,h_m\in \mathcal{H}$ and real $u_1,\dots ,u_m$, linearity of the finite partial sums and convergence in L2 giveBy part (ii), this random variable has the normal distributionA finite family is Jointly Gaussian precisely when every real linear combination has a possibly degenerate normal distribution. Thus every finite subfamily of $(Wh{)}_{h\in \mathcal{H}}$ is Jointly Gaussian, which is the meaning of joint Gaussianity for this collection.

Use the same finite truncation $T_m(h)={\sum }_{j\le m}⟨h,e_j⟩A_j$ for $f$ and $g$. Since $\mathbb{E}[A_i{A}_j]={\delta }_{ij}$ by independence, zero means and unit variances,Here ${\delta }_{ij}$ is the Kronecker delta. The Cauchy-Schwarz inequality and convergence in L2 justify passing to the limit in the product's expectation. The Parseval identity for a Hilbertian basis now givesBecause each $Wh$ has mean zero, this is also its covariance identity. Thus $W$ is an isonormal Gaussian process over $\mathcal{H}$.

Write $e_t=1_{[0,t]}$ as an element of L2 space. Parts (iii) and (iv) show that $B_t=W{e}_t$ is a centered Gaussian process with covariance functionIn particular $B_t-B_s\sim N(0,|t-s|)$. The fourth moment of a centered normal random variable of variance $v$ is $3v^2$, soThe Kolmogorov continuity theorem, with moment exponent $4$ and time exponent $2>1$, provides a continuous modification $\hat{B}$. At each finite set of times this modification of a stochastic process has the same joint law as $B$.

For $0\le t_0<t_1<\cdots <t_m\le 1$, the successive increments are Jointly Gaussian. The covariance of two increments on disjoint intervals equals the inner product of their disjoint interval indicators and is zero. Zero cross-covariances of Jointly Gaussian variables imply independence, and each increment has variance $t_j-t_{j-1}$. Finally $e_0=0$ in L2 space, so ${\hat{B}}_0=0$ almost surely. ThereforeIndependence from the whole past in its natural Brownian filtration follows by extending finite-dimensional independence to the generated sigma-algebra.