Let $W$ be a Wiener process.
For each fixed $t>\frac12$, is it true that,$$\sup_{s\in[0,\frac12]}|W(t-s)|$$ has the same law as $$\sup_{s\in[0,\frac12]}|W(t)-W(s)| ?$$
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$\begingroup$ Yes. What do you know about the process $W$? $\endgroup$– DidSep 22, 2013 at 21:09
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$\begingroup$ $W$ is 1-dimensional Brownian motion starting at $0$. Could you please justify why it is true? $\endgroup$– galanSep 22, 2013 at 21:21
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$\begingroup$ What. Do. You. Know. About. W? $\endgroup$– DidSep 22, 2013 at 21:44
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$\begingroup$ @Did I'm not able to understand your question. What additional information is required? I know that for each fixed $s$ and $t$, $W(t-s)$ is of the same law as $W(t)-W(s)$. But the question I ask requires considering $W$ sample path values at all $s\in [0,\frac12]$. $\endgroup$– galanSep 22, 2013 at 23:25
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1$\begingroup$ Got it. Define $\hat{W}(v)=W(s+v)-W(v)$. Then $\hat{W}$ is a BM starting at $0$. And $W(t)-W(s)=\int_0^{t-s}dW(s+v)=\int_0^{t-s}d\hat{W}(v)=\hat{W}(t-s)$. $\endgroup$– galanSep 23, 2013 at 1:45
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