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$X = \frac{ B_1+ B_3 - B_2}{\sqrt{2}}$ and $Y = \frac{B_1 - B_3+ B_2}{\sqrt{2}}$ Where $B_t$ Is brownian motion at time $t\geq0$

I want to find $\mathbb{E} [Y + 3X | X]$

It is known to me that $X, Y$ are independent, and that $\mathbb{E}[X] = \mathbb{E}[Y] =0$

Is this the correct approach:

$\mathbb{E} [Y + 3X | X]$

$\mathbb{E}[Y|X] + \mathbb{E}[3X|X]$

$\mathbb{E}[Y] + 3\mathbb{E}[X]$

$ =0$

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  • $\begingroup$ @Mankind Why the detrimental edit of the title? $\endgroup$
    – Did
    Mar 15, 2016 at 11:24
  • $\begingroup$ @Did ah, I meant to remove only the part "Check my answer please", but deleted too much. I was on my phone. $\endgroup$
    – Mankind
    Mar 15, 2016 at 11:28
  • $\begingroup$ @Mankind Then stop editing while on your phone. $\endgroup$
    – Did
    Mar 15, 2016 at 11:29
  • $\begingroup$ @did Really? I've done it hundreds of times and have made one mistake. $\endgroup$
    – Mankind
    Mar 15, 2016 at 11:30

1 Answer 1

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Be careful about the step "$\mathbb{E}[3X|X] = 3\mathbb{E}[X]$" - it isn't true.

It looks like you are thinking that you can remove the conditioning on $X$, but you cannot do this: $\mathbb{E}[X|X]$ is not equal to $\mathbb{E}[X]$, because $X$ is not independent of $X$.

What does hold true is that $X$ is measurable with respect to (the sigma algebra generated by) $X$, and so $\mathbb{E}[X|X] = X\mathbb{E}[1|X] = X\mathbb{E}[1] = X$.

Thus $$\mathbb{E}[Y+3X|X] = E[Y]+3E[X|X] = 3X.$$

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  • $\begingroup$ Thanks. I should look into this further. But I recognise the property you have used when saying $X$ is $X-measurable$, and removing it out of the expectation operator. Is there any book on probability which you would recommend, by the way? $\endgroup$
    – plzprobably
    Apr 21, 2015 at 11:51
  • $\begingroup$ Good. :) I believe that I learned all these conditional expectation results from various notes, so I cannot recommend any book from personal experience. However, I know that Billingsley's book "Probability and Measure" is a standard reference for probability theory, and that this book is widely used everywhere. $\endgroup$
    – Mankind
    Apr 21, 2015 at 13:21

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