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Let $\Omega=C([0,1],\mathbb{R})$, $(\Pi_t)_{t\in [0,1]}$ the canonical process with $\Pi_t(\omega)=\omega_t$, $\mathcal{F}=\sigma(\Pi)$ and $\mathbb{F}$ the filtration generated by $\Pi$. Let $F:[0,1] \rightarrow \mathbb{R}$ be a continuous nondecreasing function.

I am currently reading a paper in which the following assertion is used without any reference.

There exists a unique probability measure $\mu$ on $(\Omega,\mathcal{F})$ such that $\Pi$ is a centered Gaussian process on $(\Omega,\mathcal{F},\mathbb{F},\mu)$ with $\mathrm{Cov}[\Pi_s,\Pi_t]=F(\min(s,t))$.

I can not find this claim anywhere. I would be very grateful for a reference to this assertion.

Learner
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    Hint . For $F(x)=x$ we know that the Gaussian process $\Pi_t$ is a Brownian motion . For general $F$ the Gaussian process $\int_0^t \Pi_s,dF(s)$ has covariance function $F(\min(s,t))$. So the measure $\mu$ should be the image of the Wiener measure under the map $\Pi\mapsto \int_0^. \Pi_s,dF(s)$. – Kurt G. Apr 21 '22 at 14:01
  • Thank you for the hint! But if I take $F(x)=x$ then $\int_0^t \Pi_s dF(s)$ has variance $s$ according to your hint. But here https://math.stackexchange.com/questions/243925/integral-of-brownian-motion-is-gaussian there is a different variance, namely $\int_0^t (s-t)^2 , ds$, calculated. Do I miss something? – Learner Apr 22 '22 at 07:25
  • Good find. I made a mistake which got hopefully erased in the posted answer. – Kurt G. Apr 22 '22 at 08:17

1 Answers1

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For $F(x)=x$ we know that the Gaussian process $\Pi_t$ is a Brownian motion. A general nondecreasing $F$ is differentiable almost everywhere on $[0,1]$. Then the process

$$ \int_0^t \sqrt{F'(s)}\,d\Pi_s=G(t)\Pi_t-G(0)\Pi_0-\int_0^t\Pi_s\,dG(s)\,,\quad\text{ where }G(s):=\sqrt{F'(s)} $$ has variance $\int_0^t F'(s)\,ds=F(t)$ and covariance function $F(\min(s,t))$.

The measure $\mu$ you are looking for is then the image (push forward) of the Wiener measure under the map $$ \Pi_{\,\textstyle.}\mapsto G\,\Pi_{\,\textstyle.}-G(0)\Pi_0-\int_0^{\,\textstyle\cdot}\Pi_s\,dG(s) $$ from $C([0,1],\mathbb R)$ to itself.

Kurt G.
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  • Hmm, why does $\int_0^t \sqrt{F'(s)},d\Pi_s=\int_0^t\Pi_s,dG(s)$ hold? We have a continuous local martingale on the left hand side and an adapted continuous process of locally bounded variation on the right hand side. If they were equal, they must both be 0, right? – Learner Apr 22 '22 at 16:37
  • Correct ! Fixed one more mistake . – Kurt G. Apr 22 '22 at 18:32
  • Great, thank you! – Learner Apr 22 '22 at 19:30