Is $Y_s=sB_{1\over s},s>0$ a brownian motion Suppose $\{B_s,s>0\}$ is a standard brownian motion process. Is $Y_s=sB_{1\over s},\ s>0$ a brownian motion or (stardard). I have found that $Y_0=0$ and $Y_s\sim N(0,1)$ as $B_s\sim N(0,s)$, so it remains to show that it is stationary increment and independent increment. But i am not sure how to do it.

Have you heard of Gaussian processes ? If you have, you only have to check that $(Y_s)$ has the same covariance function as the Brownian motion.
If you haven't, don't worry, it's very simple here: you are interested in the law of the couple $(sB_{1/s},tB_{1/t}-sB_{1/s})$ when $0 < s <t$. This is a 2 dimensional centered Gaussian vector, so its law is entirely determined by its covariance matrix. In the end, you have to compute $E(sB_{1/s} tB_{1/t})$.
