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The question is as stated. I have thought about this for a while and can't really get anywhere. Here strictly positive means non-zero on non-empty open sets (in this case with a finite interval we are dealing with the topology of uniform convergence). I am pretty sure by considering drifts that it is enough to consider an open ball around a path which starts and ends at 0 but I can't see beyond this. Thanks in advance!

Mathmo
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  • Have a look at this question: http://math.stackexchange.com/q/1064700/ – saz Jan 07 '16 at 14:49
  • So I think that post answers the question without considering drifts since $B_t$ corresponds to the coordinate functions? Is there a way using drifts? @saz – Mathmo Jan 07 '16 at 15:27
  • That's how the Wiener measure is defined, isn't it? The Wiener measure assigns measure $0$ to any (msb.) set contained in $C[0,1] \backslash {f \in C[0,1]; f(0) = 0}$. – saz Jan 07 '16 at 15:51
  • Yes. Ah I have had a think about it and I think one could simplify the proof in the other post slightly. If one considers adding a drift it's easy to get rid of $f$ and $x$ in the formula (1) and hence one just gets a simple exit time from a symmetric interval around the origin. – Mathmo Jan 07 '16 at 15:56
  • Of course one has to introduce a change of measure but this doesn't affect the result. – Mathmo Jan 07 '16 at 15:57
  • You mean a shift and not a drift, right? Not sure whether this simplifies things a lot, but you are welcome to write an answer and present your idea. – saz Jan 07 '16 at 16:01
  • Just thought, in your answer you state that $M_t$ has the same distribution as $|B_t|$ via the reflection principle. I can do this by building up successive reflections around a point. Is there an easier way to do it with 1 or 2 reflections? – Mathmo Jan 07 '16 at 18:47

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