0
$\begingroup$

Let $C[0,1]$ the space of continuous functions $x:[0,1]\to \mathbb{R}$ with $x(0)=0$. Equip $C[0,1]$ with the sigma algebra generated by the cylinder sets $$B_E=\{x\in C:(x(t_1),x(t_2),\dots,x(t_n))\in E\in B(\mathbb{R}^n)\}$$ which coincides with the Borel sigma algebra. Let $W$ denote the Wiener measure on $B(C[0,1])$ and consider the measure space $(C[0,1],B(C[0,1]),W)$. Then the functionals $$f_t:C[0,1]\to f_t(x)=x(t)$$ are normally distributed random variables $N(0,t)$. For $t>s$, $f_t-f_s$ is $N(0,t-s)$. And finally for $t_3>t_2>t_1$, $f_{t_3}-f_{t_2}$ is stochastically independent of $f_{t_2}-f_{t_1}$.

We see that $(f_t)_{t\in[0,1]}$ satisfies the properties of Brownian motion.

Is $f_t(\omega)=B(t,\omega)$ a Brownian motion for $\omega\in C[0,1]$? How do I show almost sure continuity?

$\endgroup$
0
$\begingroup$

Isn't it there in the very definition? For every fixed $x \in C[0,1]$ the sample path corresponding to it simply the function $x$ on $[0,1]$ which is continuous.

$\endgroup$
  • $\begingroup$ Is this the correct definition of the Wiener process? $\endgroup$ – badatmath Jul 18 at 9:22
  • $\begingroup$ Yes. The definition is $B(t,\omega)=\omega (t)$ for $\omega \in C[0,1]$ and $t \in [0,1]$. $\endgroup$ – Kabo Murphy Jul 18 at 9:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.