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 $A_1,\dots,A_n \subset \mathbb{R}$:

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


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

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.

  • 1
    $\begingroup$ 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 $p_{t_k-t_{k-1}}$.) $\endgroup$ – Did Apr 5 '15 at 17:12
  • $\begingroup$ @Did Ah yes, silly me. I've edited it. Still, how does this show that the increments are independent? Could you elaborate a little? $\endgroup$ – Tom Offer Apr 5 '15 at 17:24
  • $\begingroup$ 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? $\endgroup$ – user3371583 Apr 5 '15 at 17:47
  • $\begingroup$ @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? $\endgroup$ – 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

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


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


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

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.

  • $\begingroup$ Thank you. I did not think of taking the expectation of the exponential! Why do you need the $\imath$? $\endgroup$ – Tom Offer Apr 8 '15 at 8:44
  • 1
    $\begingroup$ @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 $\imath$.) $\endgroup$ – saz Apr 8 '15 at 8:52
  • $\begingroup$ Ooh. I'm with you now. Cheers! (I can't tag you for some reason so I hope you read this...) $\endgroup$ – Tom Offer Apr 8 '15 at 9:03
  • $\begingroup$ @TomOffer You are welcome. $\endgroup$ – saz Apr 8 '15 at 9:07
  • $\begingroup$ are you not assuming $(X_{t_1},X_{t_2})$ has the same distribution as $(X_{t_1},X_{t_2}-X_{t_1})$? $\endgroup$ – badatmath Sep 16 at 13:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.