Let $(B)_{t\in \mathbb R_+}$ be a real-valued Levy Process (i.e. independent and stationary increments) with almost surely continuous paths on the probability space $(\Omega, \mathcal A, P)$.

Let us fix some $t>0$ and let us define for $j,n \in \mathbb N$:

\begin{align} X_{n,j}:=\triangle\Big[B(tj/n)-B(t(j-1)/n)\Big] \: \qquad \text{with} \quad \triangle(x):= x \: \cdot \mathbf{1}_{(-1,1)}(x). \end{align}

How can I deduce now, that $\lim\limits_{n \rightarrow \infty}{\sum_{j=1}^{n}X_{n,j}=B(t)-B(0)}$ almost surely.

  • 1
    $\begingroup$ Well, if $(B_t)_{t \geq 0}$ has almost surely continuous sample paths, then $(B_t)_{t \geq 0}$ is just a Brownian motion, right? $\endgroup$ – saz May 30 '17 at 17:36
  • $\begingroup$ That is the aim of this exercise. To show that this $(B_t)_t$ has increments, which are normal distributed. This is only the First of six steps to show the existence of the brownian motion. So, unfortunately, I can not simply say it is a brownian motion here. $\endgroup$ – Frodo361 May 30 '17 at 19:47

Recall that a continuous function on a compact interval is uniformly continuous. Since the process $(B_t)_{t \geq 0}$ has almost surely continuous sample paths, this means that

$$[0,t] \ni s \mapsto B_s(\omega)$$

is uniformly continuous with probability $1$. In particular, we can choose $N=N(\omega) \in \mathbb{N}$ such that

$$|B_u(\omega)-B_v(\omega)| < 1 \qquad \text{for all} \, \, u,v \in [0,t], |u-v| \leq \frac{1}{N}.$$

By the very definition of the mapping $\Delta(x) := x 1_{(-1,1)}(x)$, we get

$$X_{n,j}(\omega) := \Delta \left[ B \left( \frac{tj}{n},\omega \right)- B \left( t \frac{j-1}{n},\omega \right) \right] = B \left( \frac{tj}{n},\omega \right)- B \left( t \frac{j-1}{n},\omega \right)$$

for all $j=1,\ldots,n$ and $n \geq N$. This implies that

$$\sum_{j=1}^n X_{n,j}(\omega) = \sum_{j=1}^n \left[ B \left( \frac{tj}{n},\omega \right)- B \left( t \frac{j-1}{n},\omega \right) \right] = B(t,\omega)-B(0,\omega)$$

for all $n \geq N$. This proves the assertion.


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.