Suppose $B_1(t)$ and $B_2(t)$ are two independent Brownian motions, and $X(t)$ is defined by:
$X(t) = B_1(t) + tB_2(t)$
Find the conditional expectation:
$\mathbb{E}[X(2) | (X(3)-X(1))^2]$
My approach is using the theory of normal correlation:
$\mathbb{E}[X(2) | (X(3)-X(1))^2]$ $= \mathbb{E}[X(2)] + \frac{\text{Cov}(X(2), (X(3)-X(1))^2)}{\text{Var}((X(3)-X(1))^2)}((X(3)-X(1))^2 - \mathbb{E}[(X(3)-X(1))^2])\\ $
I've worked out the covariance function of $X(t)$ to be:
$\text{Cov}(X(t), X(s))$ = $min(s,t) + s\cdot min(t,\frac{1}{s}) + t \cdot min(\frac{1}{t}, s) + st \cdot min(\frac{1}{t},\frac{1}{s})$
However I am unsure how to proceed with the $(X(3)-X(2))^2$ terms inside the covariance and variance terms, is there a property between $X(1)$, $X(2)$, and $X(3)$ I can use, to avoid expanding the $(X(3)-X(2))^2$ terms?