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

Consider $$0\leq s where $$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.

1. Is this proof correct?

2. Is there an "easier" proof?

• By definition, a Brownian motion has independent increments, that is, $W_t-W_s$ and $W_s=W_s-W_0$ are independent. Commented Feb 10, 2020 at 17:41
• Could you please give us your defintion of a brownian motion as there are a lot of definition? Commented Feb 10, 2020 at 18:00
• See Section 1.3 of the book Levy Processes and Stochastic Calculus, where the Brownian motion is defined as a particular case. Commented Feb 10, 2020 at 20:03
• @Gordon The reason why I was puzzled is because when you say increments, I understood as one discrete time unit apart, and with $s<t$, we don't know how many time steps they are apart. Do you know what I mean? Commented Feb 18, 2020 at 21:49
• Increments refer to a partition. It can be in any length and doesn't have to be in unit length. Commented Feb 19, 2020 at 3:18

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.

1. 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.

2. 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*}