Let $w_0,w_1,…$ be independent Wiener processes on $[0,1]$ and define $$W(t)=w_0(1)+w_1(1)+…+w_{\left \lfloor{t}\right \rfloor-1}(1)+w_{\left \lfloor{t}\right \rfloor}(t-\left \lfloor{t}\right \rfloor).$$
How do I prove that
- with probability $1$ the map $t\mapsto W(t)$ is continuous,
- for $t\geq0,s>0$ that $W(t+s)-W(t)$ is $N(0,s)$-distributed?
What I know:
1. I have shown that for a countable number of events $E_n$ with $P(E_n)=1$ for all $n\in\mathbb{N}$ that $P(\bigcap_{n\in\mathbb{N}} E_n)=1$. I know that all maps $t\mapsto w_i(t)$ are continuous with probability $1$. How do I use the above to prove the continuity of $t\mapsto W(t)$?
- I see that $W(t+s)-W(t)=w_0(1)+w_1(1)+…+w_{\left \lfloor{t+s}\right \rfloor-1}(1)+w_{\left \lfloor{t+s}\right \rfloor}(t+s-\left \lfloor{t+s}\right \rfloor)-w_0(1)-w_1(1)-…-w_{\left \lfloor{t}\right \rfloor-1}(1)-w_{\left \lfloor{t}\right \rfloor}(t-\left \lfloor{t}\right \rfloor).$
If $\left \lfloor{t+s}\right \rfloor=\left \lfloor{t}\right \rfloor$, then $W(t+s)-W(t)=w_{\left \lfloor{t}\right \rfloor}(t+s-\left \lfloor{t}\right \rfloor)-w_{\left \lfloor{t}\right \rfloor}(t-\left \lfloor{t}\right \rfloor)$. This is $N(0,s)$ distributed, because $w_{\left \lfloor{t}\right \rfloor}$ is Wiener.
What do I do if $\left \lfloor{t+s}\right \rfloor>\left \lfloor{t}\right \rfloor$? I know that $W(t+s)-W(t)=w_{\left \lfloor{t}\right \rfloor+1}(1)+…+w_{\left \lfloor{t+s}\right \rfloor}(t+s-\left \lfloor{t+s}\right \rfloor)$. How do I show that this is $N(0,s)$-distributed?
Best Answer
Your process is $W(t)=\sum_{j=0}^{\lfloor t\rfloor-1} w_j(1)+w_{\lfloor t\rfloor}(t-\lfloor t\rfloor)$. By the continuity of the individual Wiener processes $w_j$, this process has continuous sample paths for $t$ in $[0,1)$, $(1,2)$, etc. The only possible trouble spots are when $t$ is a positive integer. At such $t$ the left limit is $\sum_{j=0}^{t-2} w_j(1)+w_{t-1}(1)$ while the right limit is $\sum_{j=0}^{t-1} w_j(1)+w_t(0),$ and these are equal. So $W(t)$ has continuous sample paths.
You left off a term in your variance calculation. In fact \begin{eqnarray*}W(t+s)-W(t)&=&w_{\lfloor t+s\rfloor}(t+s-\lfloor t+s\rfloor)+ \sum_{j=\lfloor t\rfloor}^{\lfloor t+s\rfloor-1} w_j(1)- w_{\lfloor t\rfloor}(t-\lfloor t\rfloor)\\ &=&w_{\lfloor t+s\rfloor}(t+s-\lfloor t+s\rfloor)+ \sum_{j=\lfloor t\rfloor+1}^{\lfloor t+s\rfloor-1} w_j(1)+ [w_{\lfloor t\rfloor}(1)-w_{\lfloor t\rfloor}(t-\lfloor t\rfloor)]. \end{eqnarray*} This is the sum of independent, mean zero normal random variables, whose variances add up to $(t+s-\lfloor t+s\rfloor)+(\lfloor t+s\rfloor-\lfloor t\rfloor-1)+(1-(t-\lfloor t\rfloor))=s.$