# Wiener process - proof of independent increments

I have defined the Wiener process to be a stochastic process $X_t$ with values in $\mathbb{R}$ such that $X_0=0$, the paths $t \mapsto X_t$ are continuous, and for any times $0<t_1<\dots<t_n$ and Borel sets $A_1,\dots,A_n \subset \mathbb{R}$:

$$\mathbb{P}(X_{t_1} \in A_1, \dots, X_{t_n} \in A_n) = \int_{A_1}\dots\int_{A_n}p_{t_1}(0,x_1)\dots p_{t_n-t_{n-1}}(x_{n-1},x_n) \; \textrm{d}x_1\dots \textrm{d}x_n$$

where

$$p_t(x,y) = \frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}$$

is the transition density. From this definition, how do I prove that for any $0=t_0 \leq t_1 \leq \dots \leq t_n$, the increments

$$X_{t_1}-X_{t_0}, \dots, X_{t_n}-X_{t_{n-1}}$$

are independent? In general, the only way I know how to show that two RVs are independent is to show that their joint density function factorises into the product of the marginals, but I can't see how to do that here. Thanks for any help.

• Direct from the fact that, for every $t$, $p_t(x,y)=q_t(y-x)$ for some suitable function $q_t$. (But note that each $p_{t_k}$ in your displayed formula should read $p_{t_k-t_{k-1}}$.) – Did Apr 5 '15 at 17:12
• @Did Ah yes, silly me. I've edited it. Still, how does this show that the increments are independent? Could you elaborate a little? – Tom Offer Apr 5 '15 at 17:24
• Did you try proving it for only two points $X_{t_0}, X_{t_1}$ (i.e. $X_{t_0} \perp X_{t_1} - X_{t_0}$) and instead of general $(A_i)$, try well defined sets? – user3371583 Apr 5 '15 at 17:47
• @user3371583 Well, yes, I would tackle it by considering just two increments as you suggest. But I'm not sure where to start. And what do you mean by well defined? – Tom Offer Apr 5 '15 at 17:54

First consider the case $n=2$. By definition, the joint distribution of $(X_{t_1},X_{t_2})$ is given by

$$p_{t_1}(0,x_1) p_{t_2-t_1}(x_1,x_2) \, dx_1 \, dx_2 = q_{t_1}(x_1) q_{t_2-t_1}(x_2-x_1) \, dx_1 \, dx_2$$

where

$$q_t(x) := \frac{1}{\sqrt{2\pi t}} \exp \left(- \frac{x^2}{2t} \right), \qquad x \in \mathbb{R}.$$

Consequently,

\begin{align*} \mathbb{E}\exp(\imath \xi X_{t_1} + \imath \, \eta (X_{t_2}-X_{t_1})) = \iint \exp(\imath \, \xi x_1+\imath \, \eta (x_2-x_1)) q_{t_1}(x_1) q_{t_2-t_1}(x_2-x_1) \, dx_2 \, dx_1 \end{align*}

For fixed $x_1$ we substitute, $y_2 := x_2 - x_1$ and obtain \begin{align*} \mathbb{E}\exp(\imath \xi X_{t_1} + \imath \, \eta (X_{t_2}-X_{t_1}) &= \iint \exp(\imath \, \xi x_1+\imath \, \eta y_2) q_{t_1}(x_1) q_{t_2-t_1}(y_2) \, dy_2 \, dx_1. \\ &= \left( \int e^{\imath \, \xi x_1} q_{t_1}(x_1) dx_1 \right) \left( \int e^{\imath \, \eta y_2} q_{t_2-t_1}(y_2) \, dy_2 \right) \\ &= \mathbb{E}e^{\imath \, \eta X_{t_1}} \cdot \mathbb{E}e^{\imath \eta (X_{t_2}-X_{t_1})}. \end{align*}

This finishes the proof for $n=2$. The general case is treated by induction.

• Thank you. I did not think of taking the expectation of the exponential! Why do you need the $\imath$? – Tom Offer Apr 8 '15 at 8:44
• @TomOffer The proof shows that the characteristic function of the joint distribution equals the product of the characteristic functions of the marginal distributions. (And the characteristic function of a random variable $X$ equals $\mathbb{E}e^{\imath \, \xi X}$; that's why I need the $\imath$.) – saz Apr 8 '15 at 8:52
• Ooh. I'm with you now. Cheers! (I can't tag you for some reason so I hope you read this...) – Tom Offer Apr 8 '15 at 9:03
• @TomOffer You are welcome. – saz Apr 8 '15 at 9:07
• are you not assuming $(X_{t_1},X_{t_2})$ has the same distribution as $(X_{t_1},X_{t_2}-X_{t_1})$? – badatmath Sep 16 at 13:26