Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do I prove the symmetry of Brownian motion? ( -w(t) is a Brownian motion?)? Also i read in many places about time reversal and scaling of brownian motions as prepositions. I would like to learn how the proof for the same could be carried out that these properties exist?

share|cite|improve this question
Do you now what you have to show? –  Davide Giraudo Nov 9 '12 at 21:40
The proof that -w(t) is a brownian motion for proof for symmetry, –  Probabilityman Nov 9 '12 at 22:49
for time reversal : W(T) - W(T-t) must hold good, scaling: 1/sqrt(k) * W(kt) also holds good. –  Probabilityman Nov 9 '12 at 22:51

1 Answer 1

Let $(X_t)_{t \geq 0}$ be a stochastic process. If we want to prove that $(X_t)_{t \geq 0}$ is a Brownian Motion, we have have to check the following properties:

  1. $X_0=0$
  2. $X_{t_n}-X_{t_{n-1}},X_{t_{n-1}}-X_{t_{n-2}},\ldots,X_{t_1}-X_{t_0}$ are indepenendent
  3. $X_t-X_s \sim N(0,t-s)$
  4. $X$ has continuous sample paths.

So let $(W_t)_{t \geq 0}$ be a Brownian motion and set $X_t :=-W_t$. Then

  1. $X_0 = -W_0=0$
  2. Since $X_{t_j}-X_{t_{j-1}} = -(W_{t_{j}}-W_{t_{j-1}})$ and we know that $W_{t_n}-W_{t_{n-1}},W_{t_{n-1}}-W_{t_{n-2}},\ldots,W_{t_1}-W_{t_0}$ are indepenendent, we obtain that the random variables $X_{t_n}-X_{t_{n-1}},X_{t_{n-1}}-X_{t_{n-2}},\ldots,X_{t_1}-X_{t_0}$ are indepenendent.
  3. $X_t-X_s = -(W_t-W_s) \sim - N(0,t-s) = N(0,t-s)$
  4. Clearly $t \mapsto X_t(w)=-W_t(w)$ is continuous, since $t \mapsto W_t(w)$ is continuous.

A very similar argumentation proves that

$$Y_t := \frac{1}{\sqrt{c}} \cdot W_{c \cdot t} \quad \text{and}\quad Z_t := W_T-W_{T-t}$$

are Brownian motions where $c>0$, $T>0$ are fixed constants.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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