Assume $\{B_t:t\ge0\}$ be a brownian motion process. Is $B_s-\frac{s}{t}B_t$ and $B_t$ independent given that ($s\le t$)
Are these 2 random variable independent???
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$\begingroup$
probability
brownian-motion
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3@SimonHayward On MSE, **ANYTHING** can garner upvotes (and I agree with everything you wrote in your comment). – 2012-12-03
1 Answers
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Hint: The random vector $(B_s-(s/t)B_t,B_t)$ is centered normal. There is a simple way to check that some entries of a normal vector are independent, you could try it.
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0@robjohn Not sure I understand what you mean. Note that to *satisfy* misnamed user Mathematics is the least of my concerns. Note also that I already indicated in a comment *how one might show independence* and that I do not intend to modify my answer: believe it or not, some people here *think* about their answers before posting them. (To tell you the truth, the more I ponder your comment, the less I like it.) – 2012-12-05