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?
Brownian Motion(symmetry, time reversal and scaling)
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brownian-motion
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0for time reversal : W(T) - W(T-t) must hold good, scaling: 1/sqrt(k) * W(kt) also holds good. – 2012-11-09
1 Answers
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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:
- $X_0=0$
- $X_{t_n}-X_{t_{n-1}},X_{t_{n-1}}-X_{t_{n-2}},\ldots,X_{t_1}-X_{t_0}$ are indepenendent
- $X_t-X_s \sim N(0,t-s)$
- $X$ has continuous sample paths.
So let $(W_t)_{t \geq 0}$ be a Brownian motion and set $X_t :=-W_t$. Then
- $X_0 = -W_0=0$
- 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.
- $X_t-X_s = -(W_t-W_s) \sim - N(0,t-s) = N(0,t-s)$
- 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.