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.
Is $Y_s=sB_{1\over s},s>0$ a brownian motion
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probability
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2You have $Y_s \sim N(0,s)$ (as it should be) ;-) – 2012-12-05
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0I changed $B_s$~$N(0,s)$ to $B_s\sim N(0,s)$, coded as B_s\sim N(0,s). That is standard usage. – 2012-12-05
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0@martini why? $B_{1\over s}\sim N(0,{1\over s})$ isn't it? – 2012-12-05
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0It is, but $\lambda N(0,\sigma^2) \sim N(0,\lambda^2\sigma^2)$. – 2012-12-05
1 Answers
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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
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0i am checking about the gussian process. By the way, for the second method, i wonder if you are trying to show $cov(Y_t,Y_s)=s$? – 2012-12-05
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1First, the two methods are in fact the same. Second, I'm not trying to show anything, you are. Anyway trying to show that $cov(Y_t,Y_s)=\min(s,t)$ is a good intuition. – 2012-12-05