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A (1-dim) Brownian motion $(B_t)_{t \geq 0}$ satisfies the following properties:

  • (B0): $B_0=0$ a.s.
  • (B1): $(B_t)_t$ has independent increments
  • (B2): $(B_t)_t$ has stationary increments, i.e. $B_{t-s} \sim B_t-B_s$
  • (B3): $B_t \sim N(0,t)$
  • (B4): $t \mapsto B_t(w)$ continuous for almost all $w \in \Omega$

Prove that $(B0)-(B2),(B4)$ imply that $B_t$ is (possibly degenerate) Gaussian.

I tried to prove this using the central limit theorem: We have $$B_t = \sum_{j=1}^n B_{t_j}-B_{t_{j-1}}$$ for $t_j := \frac{j}{n}$ where $B_{t_j}-B_{t_{j-1}}$ ($j=1,\ldots,n$) are independent and identically distributed by assumption. Moreover, $$ \begin{align} \mathbb{E}(B_t) &= n \cdot \mathbb{E}(B_{t_j-t_{j-1}}) \tag{1} \\ \mathbb{V}(B_t) &= n \cdot \mathbb{V}(B_{t_j-t_{j-1}}) \tag{2} \end{align}$$ Thus

$$\begin{align} B_t- \mathbb{E}B_t &\stackrel{(1)}{=} \sum_{j=1}^n (B_{t_j}-B_{t_{j-1}})- \mathbb{E}(B_{t_j-t_{j-1}}) \\ &= \sum_{j=1}^n \frac{\sqrt{\mathbb{V}(B_{t_j}-B_{t_{j-1}})}}{\sqrt{\mathbb{V}(B_{t_j}-B_{t_{j-1}})}} \cdot \bigg((B_{t_j}-B_{t_{j-1}})- \mathbb{E}(B_{t_j-t_{j-1}})\bigg) \\ &\stackrel{(2)}{=} \sqrt{\mathbb{V}(B_t)} \cdot \frac{1}{\sqrt{n}} \sum_{j=1}^n \underbrace{\frac{(B_{t_j}-B_{t_{j-1}})-\mathbb{E}(B_{t_j}-B_{t_{j-1}})}{\sqrt{\mathbb{V}(B_{t_j}-B_{t_{j-1}})}}}_{=:G_j^n} \end{align}$$

where $G_j^n$ ($j=1,\ldots,n$) are independent and $G_j^n \sim Z^n$ (i.e. identically distributed), satisfying $\mathbb{E}Z^n=0$, $\mathbb{E}((Z^n)^2) = 1$. I would like to prove that the triangular array $(G_j^n)_{j=1,\ldots,n;n \in \mathbb{N}}$ satisfies the Lindeberg-Feller condition, i.e. that $$\begin{align} \frac{1}{s_n^2} \sum_{j=1}^n \int_{|G_j^n| > \varepsilon \cdot s_n} (G_j^n)^2 \, d\mathbb{P} &\stackrel{\text{iid},(3)}{=} \frac{1}{n} \cdot n \cdot \int_{|Z^n| > \varepsilon \cdot \sqrt{n}} (Z^n)^2 \, d\mathbb{P} \\ &= \int_{|B_{\frac{t}{n}}-\mathbb{E}B_{\frac{t}{n}}|>\varepsilon \cdot \sqrt{n \cdot \mathbb{V}B_{\frac{t}{n}}}} \frac{(B_{\frac{t}{n}}-\mathbb{E}B_{\frac{t}{n}})^2}{\mathbb{V}B_{\frac{t}{n}}} \, d\mathbb{P} \end{align}$$ converges to $0$ as $n \to \infty$ where $$s_n^2 := \sum_{j=1}^n \mathbb{V}G_j^n = n \cdot \mathbb{E}((Z^n)^2) = n \tag{3}$$ since $\mathbb{E}((Z^n)^2)=1$.

Somehow, this convergence has to be a consequence of the continuity of the sample paths (since Lévy processes are normal distributed iff they have continuous sample paths). How to prove that the sequence converges indeed to $0$?

Actually, the distribution of $Z^n$ does not depend on $n$ (this follows easily from (B3)) and if one could prove this (without using (B3)), the convergence would be obvious.

share|cite|improve this question
Could you expand on how you arrive at the expression for $B_t-\mathbb{E}[B_t]$? And also how the expression in the Lindeberg-Feller condition simplies to $\mathbb{P}[Z^n>\varepsilon\cdot n]$? – Stefan Hansen Mar 20 '13 at 15:43
@StefanHansen I expanded the first one. For the second one, I recognized that there was indeed a mistake (I skipped the $(Z^n)^2$) - so, probably, the Lindeberg-Feller condition is only fulfilled since the Brownian motion has continuous paths. – saz Mar 20 '13 at 16:54
up vote 1 down vote accepted

It seems rather difficult to prove the Lindeberg-Feller condition directly. Instead, one can imitate the proof of the central limit theorem. The path-continuity implies some kind of asymptotically negligibility.

A proof is found for instance in K. Ito: Lectures on stochastic processes; p. 136ff.

share|cite|improve this answer
Page 136 in this? – Stefan Hansen Mar 26 '13 at 8:24
@StefanHansen The original page numbers are set bold, outline to the left/right of the text. So, in this edition, it's page 111. – saz Mar 26 '13 at 8:38
I see, thanks! :) – Stefan Hansen Mar 26 '13 at 8:39

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