2
$\begingroup$

I am trying to solve the following exercise in Oksendal's book:

Let $B_t$ be Brownian motion and fix $t_0\ge 0$. Prove that $$\bar{B_t}:=B_{t_0+t}-B_{t_0};\quad t\ge 0$$ is a Brownian motion.


I try to prove it by definition. In the book, Brownian motion is defined as

$$P^x(B_{t_1}\in F_1, \cdots, B_{t_k}\in F_k) = \\ \int\limits_{F_1 \times \cdots \times F_k}p(t_1, x, x_1)\cdots p(t_k-t_{k-1}, x_{k-1}, x_k)dx_1 \ldots dx_k, $$ where $$p(t,x,y) = (2\pi t)^{-n/2}\cdot \exp(-\frac{|x-y|^2}{2t})$$

Hence for simplicity, we only need to show $$P^0(\bar{B}_{t_1}\in F_1, \bar{B}_{t_2}\in F_2) = \int\limits_{F_1 \times F_2}p(t_1, 0, x_2) p(t_2-t_1, x_3, x_2)dx_2 dx_3$$ Suppse $B_t$ starts at $x_0$. By definition of $\bar{B_t}$

$P^0(\bar{B}_{t_1}\in F_1,\bar{B}_{t_2}\in F_2)=P^0(B_{t_0+t_1}-B_{t_0}\in F_1, B_{t_0+t_2}-B_{t_0}\in F_2)\tag{1}$

Then I don't know how to handle it. I know I should use $B_t$'s finite dimensional distribution. So if I can write $(1)$ as

$$P^{x_0}[\bigcup_{x_1\in\mathbb{R}^n} \left(B_{t_0}=x_1,B_{t_0+t_1}\in F_1+x_1, B_{t_0+t_2}\in F_2+x_1\right)]\tag{2}\\=\int\limits_{\mathbb{R}^n\times (F_1+x_1) \times (F_2+x_1)}p(t_0, x_0, x_1)p(t_1, x_1, x_2) p(t_2-t_1, x_2, x_3)dx_1 dx_2 dx_3 $$

Then apply change of variable formula and Fubini's theorem, I can get the result.

But I don't know how to justify $(2)$ and the last equality. I think it should be true because it's the conditional probability. But the book didn't introduce this concept and since $\mathbb{R}^n$ is uncountable, I don't know how this is valid in terms of measure theory.

$\endgroup$
2
$\begingroup$

Why aren't you applying the characterization: "Any continuous real-valued process $(X_t)$ that is zero-mean Gaussian process with covariance $\mathrm{cov} (X_t,X_s)=t\wedge s$ is a Brownian motion" ? (Rogers and Williams, page 4).

I write $B'_t$ for your new process. It is a Gaussian process.

Then $B'_t$ has continuous sample paths (since $B_t$ does). Also

$$E(B'_t) = E(B_{t_o+t}) - E(B_{t_0}) = 0. $$

Finally for $s < t$

\begin{align} \mathrm{cov}(B'_t,B'_s) &= E((B_{t_o+t}-B_{t_o})(B_{t_o+s}-B_{t_o}))\\ &= \min(t_0+t, t_o+s) + \min(t_0,t_0) - \min(t_0+s,t_0) + \min(t_0+t,t_0)\\ &= (t_0 +s) + t_0 - t_0 -t_0 = s = \min(s,t). \end{align}

In the second line I have expanded the brackets, used linearity of expectation, and the characterization applied to $(B_t)$.

$\endgroup$
  • 3
    $\begingroup$ In my oppinion this does not answer the question. It is rather clear from the OP that JohnZhang wants to use the definition given in Oksendal's book. $\endgroup$ – saz Mar 26 '15 at 17:50
  • $\begingroup$ fair enough, but why make a mountain out of a mole hill $\endgroup$ – Frank Mar 26 '15 at 18:56
  • $\begingroup$ @Frank Thanks for sharing your answer. I didn't know this definition before. Since this is the first stochastic calculus book I have read, I will really appreciate it if you can prove the statement using the definition in Oksendal's book. By the way, in your characterization, does $(X_t)$ have to have zero-mean? Because in the construction given in Oksendal's book, Brownian motion can have mean $x$ if it starts at $x$. $\endgroup$ – John Mar 26 '15 at 19:30
  • $\begingroup$ @JohnZHANG I don't have his book to hand, but does he not proof this at some point in the text? If not, you can find it in any book on Brownian motion, including the reference I gave. And you are right about BM starting from $x$. The characterization I gave was for BM starting from $0$, i.e. a standard BM. $\endgroup$ – Frank Mar 26 '15 at 19:35
  • 1
    $\begingroup$ I have to say, in order to usefully set up Brownian motion you have to do the conditional probability calculation OP wants (or equivalent) at some point, so trying to "define away" the problem is just shifting work around. $\endgroup$ – aes Mar 27 '15 at 4:22

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.