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?