Prove that the process $W_t=(1+t)U_{t/(1+t)}$ on $[0,\infty)$ is a Brownian motion.

$\text{(b)}$ Clearly $Y_0=U_0=0$, and inherits continuity of sample paths from $U_t$ (and hence from $W_t$).

Now, for $s,t\ge0$ the difference $$\begin{align} Y_{t+s}-Y_t&=(1+t+s)U_{(t+s)/(1+t+s)}-(1+t)U_{t/(1+t)}\\ &=(1+t+s)(W_{(t+s)/(1+t+s)}-(t+s)/(1+t+s)W_1)\\&\,\,\,\,\,\,-(1+t)(W_{t/(1+t)}-t/(1+t)W_1)\\ &=(1+t+s)W_{(t+s)/(1+t+s)}-(1+t)W_{t/1+t}-sW_1 \end{align}$$ Since $t/(1+t)\le (s+t)/(1+t+s)\le 1$, we can write this as $$(1+t)\left[W_{(t+s)/(1+t+s)}-W_{t/(1+t)}\right]+s\left[W_{s+t/1+s+t}-W_1\right]\,\,,$$ which is the sum of $(1+t)$ times a $N(0,s/(1+t)/(1+t+s))$ plus $s$ times an independent $N(0,1/(1+t+s)$, that is a $N(0,s(1+t)/(1+t+s)+s^2/(1+t+s))=N(0,s)$ as required.

This increment is independent of the history of $Y_t$ up to $t$, since that corresponds to $W_u$ for $u\lt t/(1+t)$.

I don't really understand anything about the structure of this proof. Could anyone please describe what steps they are taking and why?

  • 3
    $\begingroup$ It is not clear how the processes $(Y_t)_{t \geq 0}$, $(W_t)_{t \geq 0}$ and $(U_t)_{t \geq 0}$ are defined/related. (The proof shows that some process $(Y_t)_t$ is a Brownian motion, but the definition of $(Y_t)_t$ is not given. And is $(W_t)_t$ a Brownian motion? Or $(U_t)_t$?) $\endgroup$ – saz Nov 21 '13 at 19:06

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