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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?

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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 $(X_t)_t$ be a stochastic process. If you want to proof that $(X_t)_t$ is a Brownian Motion, you'll have to proof 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 $(W_t)_t$ a Brownian motion and $X_t :=-W_t$. 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.

Similar argumentation proofs 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 $c>0$, $T>0$.

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