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As is proved, $X_t = tW\left(\frac{1}{t}\right)$ is a Brownian motion. For example see Theorem 4.2 in this paper http://math.uchicago.edu/~may/REU2012/REUPapers/Leiner.pdf

I'm just confused because it seems to me that this inverted Brownian Motion is not a martingale. Take $s>t\ge 1$, then

$$\mathbb{E}_t[X_s] =\mathbb{E}_t\left[sW\left(\frac{1}{s}\right)\right] $$

But isn't $W\left(\frac{1}{s}\right)$ a piece of known information by time $t$? Therefore we can take it out from the expectation, but the past path of Brownian Motion could be anything, and it seems to me that $X_t$ is not a martingale. In my humble knowledge, Brownian Motion has to be a martingale, and it's a contradiction to me. Where was I wrong?

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    $\begingroup$ Brownian motion is a martingale w.r.t. its natural filtration, whereas using filtration $\sigma(W_t)$ for $t W(1/t)$ is unnatural. $\endgroup$ – Ilya Nov 12 '15 at 14:55
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    $\begingroup$ Awesome, I guess that's what I'm missing. Can you briefly explain what is the natural filtration for $tW(1/t)$? Do you mind writing an answer so that I could choose your answer as the best? $\endgroup$ – Kenneth Chen Nov 12 '15 at 14:57
  • $\begingroup$ $$\sigma(X_s;s\leqslant t)=\sigma(W(1/s);s\leqslant t)=\sigma(W(s);s\geqslant1/t)$$ $\endgroup$ – Did Nov 12 '15 at 15:06
  • $\begingroup$ @Did was fast enough to do that before me, so please expect answer from him. $\endgroup$ – Ilya Nov 12 '15 at 15:07
  • $\begingroup$ @Did Do you mind explaining the second equality from your comment? Is the natural filtration of $X_t$ starting from infinity and goes to zero indexed of time? $\endgroup$ – Kenneth Chen Nov 12 '15 at 15:13

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