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Indeed it should be because I have to prove it but I have some stupid I guess trouble.

Let $V(t)=-W(t)$ where $W(t)$ is Wiener process.

  1. $V(0)=0$ without doubts.

  2. let $t>s$ then $V(t)-V(s)=W(s)-W(t) \sim \mathcal{N}(0,s-t)$ but we wanted $\mathcal{N}(0,t-s)$$

  3. let $t_0<t_1<...<t_n$ then $ V(t_1)-V(t_0),..., V(t_n)-V(t_{n-1}) $ are independent? We obtain

$W(t_0)-W(t_1),..., W(t_{n-1})-W(t_{n})$ and it is quite not what we wanted to get.

  1. Of course we have continuous paths.

Maybe point 2 and 3 are not a problem but I want to understend it very well. Thanks for help.

nodis6
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1 Answers1

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Point2: $V(t) - V(s) = W(s) - W(t) \sim N(0, {\color{red}{t -s}})$ since $W(t) - W(s) \sim N(0, t-s)$ where $t > s$.

Point3: $V(t_2) - V(t_1) = W(t_1) - W(t_2)$ is independent of $V(t_1) - V(t_0) = W(t_0) - W(t_1)$ since $W(t_2) - W(t_1)$ is independent of $W(t_1) - W(t_0)$.

Chia
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  • Ok i got it. Since $Var(W(t)-W(s))=Var((-1)(W(s)-W(t))=(-1)^2Var((W(s)-W(t))=Var((W(s)-W(t))$ – nodis6 Jan 21 '23 at 13:02