Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

$$S^n(t):=\frac{1}{\sqrt{n}}\left[\left(t-\frac{k}{n}\right)S_{k+1}+\left(\frac{k+1}{n}-t\right)S_k\right]$$

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 $t\in(\frac{k}{n},\frac{k+1}{n})$?

share|improve this question
add comment

1 Answer

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 $w$.

share|improve this answer
    
very nice! thanks a lot. –  math Mar 19 '13 at 10:58
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.