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Conditional Brownian motion

  1. Dec 14, 2012 #1
    I computed the distribution of [itex]B_s[/itex] given [itex]B_t[/itex], where [itex]0\leq s <t[/itex] and [itex]\left\{B_t\right\}_{t\geq 0}[/itex] is a standard brownian motion. It's normal obviously..

    My question is, how do I phrase what I've done exactly? Is it that I computed the distribution of [itex]B_s[/itex] over [itex]\sigma(B_t)[/itex]?
     
  2. jcsd
  3. Dec 22, 2012 #2

    chiro

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    Hey IniquiTrance.

    If you partition the distributions so that they don't overlap then you can use the properties of a Wiener (or Brownian motion) process and that should be enough in terms of the justification used.
     
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