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Suppose we have an increasing sequence of stopping times $\{\tau_n\}$ such that $\tau_n-\tau_{n-1}$ are iid. Furthermore let $B$ be a Brownian Motion and we define $S_n:=B(\tau_n)$ which gives a random walk. Moreover


for $\frac{k}{n}\le t\le \frac{k+1}{n}$ and $k=0,\dots,n-1$. So $S^n$ is the piecewise linear interpolation. Why do we have the following: For every $t\in[\frac{k}{n},\frac{k+1}{n}]$ there is $\nu\in[\tau_k,\tau_{k+1}]$ such that $S^n(t)=\frac{1}{\sqrt{n}}B_{\nu}$? Obviously at the end point of the interval, this is clear. But why is it true for the $t\in(\frac{k}{n},\frac{k+1}{n})$?

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up vote 2 down vote accepted

This is a direct consequence of the intermediate value theorem: Let $t \in \left[ \frac{k}{n}, \frac{k+1}{n} \right]$, then

$$\sqrt{n} \cdot S^n(t) = \left(t- \frac{k}{n} \right) B_{\tau_{k+1}} + \left(\frac{k+1}{n}- t \right) \cdot B_{\tau_k} \in [\min\{B_{\tau_k},B_{\tau_{k+1}}\},\max\{B_{\tau_k},B_{\tau_{k+1}}\}]$$

Since $t \mapsto B(t,w)$ is continuous for almost all $w$, there exists by the intermediate value theorem $\nu(w) \in [\tau_k(w),\tau_{k+1}(w)]$ such that $$\sqrt{n} \cdot S^n(t,w) = B_{\nu}(w)$$ for almost all $w$.

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very nice! thanks a lot. – math Mar 19 '13 at 10:58

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