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I want to find the best predictor of $(B_3-B_2)(B_4-B_{\pi})$ given an observation of $B_1$

Where $B_t$ is brownian motion for time $t \geq 0$.

I am not sure how to approach this.

I know it will be the result of the conditional expecation: $\mathbb{E}[(B_3-B_2)(B_4-B_{\pi})|B_1]$ But have no idea how to compute this result.

Would $(B_3-B_2)$ be independent of $(B_4-B_\pi)$ ?

I know the property of an independence of increments in brownian motion exists, but was not sure if that could be applied here.

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  • $\begingroup$ The last line answers the question on the penultimate line. $\endgroup$ – whuber Apr 21 '15 at 17:02
  • $\begingroup$ So you are saying it is okay to say: $$\mathbb{E}[(B_3-B_2)(B_4-B_{\pi})|B_1] =$$ $$\mathbb{E}[(B_3-B_2)|B_1] + \mathbb{E}[(B_4-B_{\pi}) | B_1] = 0$$ ? $\endgroup$ – piman314 Apr 21 '15 at 17:18
  • $\begingroup$ No, I haven't said anything like that! Your last statement asserts non-overlapping increments are independent. The penultimate one involves non-overlapping increments, whence they must be independent. But the expectation of a product of (conditionally) independent variables is not the sum of the expectations. Examples of the correct formula can be found by searching our site. $\endgroup$ – whuber Apr 21 '15 at 17:32
  • $\begingroup$ Hmm. I searched the first page, and did not find anything relevant. I tried another attempt, at the problem, however: $$\mathbb{E}[(B_3-B_2)(B_4-B_\pi)|B_1] = \mathbb{E}(B_3-B_2)(B_4-B_\pi)$$ since independence implies $Cov((B_3-B_2)(B_4-B_\pi),B_1) = 0$ Is this a step in the right direction? Thanks for your help so far @whuber $\endgroup$ – piman314 Apr 21 '15 at 18:10
  • $\begingroup$ From the first page of hits: stats.stackexchange.com/questions/129372/…, stats.stackexchange.com/a/52699/919, stats.stackexchange.com/questions/134242, stats.stackexchange.com/a/28231/919, etc., etc. all cite and use a formula for expectations of products of independent random variables. $\endgroup$ – whuber Apr 21 '15 at 18:13

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