$X = \frac{ B_1+ B_3 - B_2}{\sqrt{2}}$ 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$
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 $$\mathbb{E}[Y+3X|X] = E[Y]+3E[X|X] = 3X.$$