Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 217 / 2 / Solution

Construct a mean-square derivative of a Gaussian process on the original probability space before choosing its continuous modification. The real covariance function satisfies $K(h)=K(-h)$, because covariance is symmetric. Three-times differentiability implies that $K^{\prime \prime }$ is continuous everywhere and differentiable at zero. Since $K^{\prime \prime }$ is an even function,This uses differentiability of $K^{\prime \prime }$ at zero, not a bound on $K^{\prime \prime \prime }$ throughout a neighborhood.

Let $D_{h}X(t)=(X(t+h)-X(t))/h$ for $h\ne 0$. Twice applying the fundamental theorem of calculus to the covariance function givesOriented integrals make this identity valid for negative as well as positive $h,k$. For a fixed $t$, the right side tends to $-K^{\prime \prime }(0)$ as $h,k\to 0$. In particular,By completeness of L2 space, the mean-square derivative of a Gaussian processexists. Finite vectors consisting of values of $X$ and $Y$ have a centered multivariate normal distribution: approximate the $Y$ coordinates by difference quotients and use mean-square convergence and characteristic functions. Thus $Y$ is a centered Gaussian process jointly constructed with $X$, withIn particular $-K^{\prime \prime }(0)\ge 0$, and $-K^{\prime \prime }$ is a positive-semidefinite kernel because it is an actual covariance function. Merely constructing an unrelated Gaussian process with that covariance function would not yet construct a modification of a stochastic process of $X$.

The increment variance of $Y$ isFor small $|t-s|$, differentiability of $K^{\prime \prime }$ at zero bounds this by $C|t-s|$. For the remaining distances in $[0,1]$, continuity of $K^{\prime \prime }$ enlarges $C$ to give the same bound. The fourth moment of a centered normal distribution is three times its variance squared, soThe one-dimensional Kolmogorov continuity theorem states that $\mathbb{E}|U(t)-U(s){|}^p\le C|t-s{|}^{1+\eta }$ with $p,\eta >0$ gives a continuous modification, with sample-path Hölder continuity of every order less than $\eta /p$. Apply it here with $p=4,\eta =1$. We obtain a continuous modification $\tilde{Y}$ on $[0,1]$. On its null exceptional event, set the entire path equal to zero.

Define, on this same probability space,The path integral exists because its integrand is a continuous function. It agrees in L2 space with the integral of $Y$: Riemann sums converge in L2 space because $Y$ is mean-square continuous; sums of $\tilde{Y}$ have the same values almost surely at each finite collection of mesh points and converge pathwise to the integral. These two limits therefore agree almost surely for each fixed $t$. Equivalently, the integral is the Bochner integral of the L2 space-valued map $s\mapsto Y(s)$.

To verify the modification of a stochastic process property directly, put $A_t=X(t)-X(0)$ and $I_t={\int }_0^{t}Y(s)ds$. The preceding covariance identities and the fundamental theorem of calculus giveTherefore $\mathbb{E}(A_t-I_t{)}^2=0$ and $\tilde{X}(t)=X(t)$ almost surely for every fixed $t$. This is also the mean-square fundamental theorem of calculus in this stationary setting. If a version on all of $\mathbb{R}$ is desired, retain the original $X(t)$ outside $[0,1]$; equality at every fixed parameter still holds.

The ordinary fundamental theorem of calculus now gives the simultaneous pathwise conclusionThus the requested differentiable modification of a stationary Gaussian process exists. Integrability of $K$ is compatible with the hypotheses but is unnecessary for this direct argument. It also avoids any assumption that a third absolute moment of a spectral density follows from three-times differentiability.