Proving Independence of Brownian Motion Increments

# Showing BM $W(s)$ is independent of $W(t)-W(s)$


# Showing BM $W(s)$ is independent of $W(t)-W(s)$












Consider $0\leq s<t$ where $t,s$ represent time index.

I want to show a Brownian motion $W(s)$ is independent of $W(t)-W(s)$.

Specifically, show that $E[W(s)(W(t)-W(s))]=0$

Proof:

Writing $W(s)$ as a telescoping sum and using the definition $W(0)=0$,

$W(s)=W(s)-W(s-1)+W(s-1)-W(s-2)+...-W(1)+W(1)-W(0).$

You can do the same for $W(t)-W(s).$

Denote the telescoping series of $W(s)$ as A and $W(t)-W(s)$ as B.

Consider $E[W(s)(W(t)-W(s))]$.

This is $E[AB].$

But since $AB$ is simply a sum of cross product of indepdent increments and each increment is normally distributed with mean zero, $E[AB]=0$. QED.

Question.

- Is this proof correct?

- Is there an "easier" proof?

## Answer by user39119 (score 1, accepted)

https://quant.stackexchange.com/a/51081

In some books, what you want to prove is just part of the definition of the Brownian motion. In others, as part of the definition of the B.M., they give the following condition:

$$\text { for } 0 \leq s < t < \infty, W_t - W_s \ \text{is independent of } \mathscr{F}_s \tag*{(*)}$$ So, I'm assuming that given (*) you want to prove that the random variables $W_s$ and $W_t-W_s$ are independent.

Answer to your questions:

- No, your proof is not correct. If we have two random variables $X$ and $Y$, covariance$(X,Y)$=0 doesn't imply that $X$ and $Y$ are independent.

- A possible proof is the following. We will need the following theorem.

> Theorem. [Kac's theorem] Let $X_1$ and $X_2$ be $\mathbb{R}$- valued random variables. Then the following statements are equivalent: (i) $X_1$ and $X_2$ are independent. (ii) For all $\lambda_1, \lambda_2 \in \mathbb{R}$ $$Ee^{\lambda_1 X_1 > + \lambda_2 X_2} = E e^{\lambda_1 X_1} Ee^{\lambda_2 X_2}.$$

Proof ($W_t -W_s$ and $W_s$ are independent).

Let $a,b \in \mathbb R.$ Then \begin{align*} E[e^{aW_s+b(W_t-W_s)}]&=E[E[e^{aW_s+b(W_t-W_s)}|\mathscr{F}_s]] \tag*{tower property} \\ &=E[e^{aW_s} E[e^{b(W_t-W_s)}|\mathscr{F}_s]] \tag*{$W_s \in \mathscr{F}_s$} \\ &= E[e^{aW_s} E[e^{b(W_t-W_s)}]] \tag*{condition (*)} \\ &= E[e^{aW_s}] E[e^{b(W_t-W_s)}] \end{align*}