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I got a definition of BM as

Definition 2.3. The Brownian motion is a continuous time stochastic process $\{W(t), t \geq 0\}$ that satisfies the following conditions:

  • (i) $W(0)=0$ a.s.;
  • (ii) the paths $t \longmapsto W(t)$ are continuous a.s.;
  • (iii) for $0 \leq s<t<\infty$, the increment $W(t)-W(s)$ is independent of $W(s)$;
  • (iv) for $0 \leq s<t<\infty$, the increment $W(t)-W(s)$ has the normal distribution with mean 0 and variance $t-s$.

and then a proposition

Proposition 2.2. For any $0=t_{0} \leq t_{1} \leq \ldots \leq t_{n}$ the increments $$W\left(t_{1}\right)-W\left(t_{0}\right), \ldots, W\left(t_{n}\right)-W\left(t_{n-1}\right)$$ are independent random variables.

I could not prove this proposition from part (iii) of the definition. From other sources, the part (iii) of this definition is quite non-standard. Could you confirm if this definition is correct?

Update 1: It seems I found how to prove Prop 2.2. Could you confirm if my proof is correct?

It suffices to show $W(t_4) - W(t_3) \perp W(t_2) - W(t_1)$. We have $W(t_4) - W(t_2) \perp W(t_2)$ and $W(t_3) - W(t_2) \perp W(t_2)$ by definition. Then $W(t_4) - W(t_3) \perp W(t_2)$. Similarly, we get $W(t_4) - W(t_3) \perp W(t_1)$ and thus $W(t_4) - W(t_3) \perp -W(t_1)$. Finally, $W(t_4) - W(t_3) \perp W(t_2) - W(t_1)$.

Akira
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    You seem to be saying that if $X$ is independent of $Z$ and $Y$ is independent of $Z$ then $X+Y$ is independent of $Z$. That is not necessarily true – Henry Sep 27 '21 at 16:06
  • @Henry I thought so and it's my bad :( – Akira Sep 27 '21 at 16:09
  • @Henry Do you think the definition 2.3. is correct? – Akira Sep 27 '21 at 16:14
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    I have my doubts about $2.3(iii)$ and would prefer it to say something like "for $0 \le r \leq s<t<\infty$, the increment $W(t)-W(s)$ is independent of all $W(r)$" though I suspect there is a standard definition – Henry Sep 27 '21 at 16:33

1 Answers1

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EDIT: this proof is valid, if $W(t)$ is defined to be Gaussian stochastic process, i.e. having jointly Gaussian finite-dimensional distributions. There are some steps in proving this proposition.

  1. From (iv) we deduce that $\mathbb{E}W(t) = 0$: as $W(0) = 0$ a.s., $W(t) \sim N(\mu = 0, \sigma^2 = t)$.
  2. Let $C(s, t) = \mathbb{E}\left(W(s) - \mathbb{E}W(s)\right)\mathbb{E}\left(W(t) - \mathbb{E}W(t)\right)$ denote the covariance function. From 1, it is equal to $\mathbb{E} W(s)W(t)$. As $W(t) - W(s) \perp W(s)$ for $0 \leq s < t$, $$0 = \mathbb{E}\left(W(t) - W(s)\right)W(s) = \mathbb{E} W(s)W(t) - \mathbb{E} W(s)W(s) = C(s,t) - s \Rightarrow C(s,t) = s$$ since $\mathbb{E} W(s)W(s)$ is the variance of $W(s) \sim N(\mu=0,\sigma^2=s)$. Similarly, for $0 \leq t < s$ we have $W(s) - W(t) \perp W(t)$ and $C(s,t) = t$. So, it leads to $C(s, t) = \min\{s, t\}$.
  3. As $W(t_1) - W(t_0), \dots, W(t_n) - W(t_{n-1})$ are all Gaussian, their independence is equivalent to being uncorrelated, which can be proven the following way: let $t_{i-1} < t_i < t_{j-1} < t_j$, then $$ cov\left(W(t_i)-W(t_{i-1}),W(t_j) - W(t_{j-1})\right) = cov(W(t_i),W(t_j)) - cov(W(t_i),W(t_{j-1})) - cov(W(t_{i-1}),W(t_{j})) + cov(W(t_{i-1}),W(t_{j-1})) = C(t_i,t_j) - C(t_i,t_{j-1}) - C(t_{i-1},t_j) + C(t_{i-1},t_{j-1}) = \min\{t_i,t_j\} - \min\{t_i,t_{j-1}\} - \min\{t_{i-1},t_j\} + \min\{t_{i-1},t_{j-1}\} = t_i - t_i - t_{i-1} + t_{i-1} = 0 $$
Yalikesifulei
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  • I think that this proof is not correct. Indepence and being uncorrelated is equivalent, if the RV are jointly Gaussian. But the problem of the above definition is that this fact in not mentioned. – user940347 Sep 27 '21 at 20:28
  • Thanks, I’ve missed that step in the proof. Joint Gaussian criteria (via linear combinations) can be used, but there is no condition that $W(t)$ has Gaussian finite-dimensional distributions. I’ll add this to edit. – Yalikesifulei Sep 27 '21 at 20:44