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Let us assume that there exist two independent Brownian Motions $B_t$ and $W_t$, and consider their sum $Y_t=B_t + W_t$. Next, define the filtration generated by the sum, $\mathcal{F}_t^{Y}=\sigma(Y_u)_{0 \leq u \leq t}$.

How would one compute the filter $E[ B_t | \mathcal{F}_t^{Y}]$?

My intuition tells me that the solution to this filtering problem should just be $E[B_t | \mathcal{F}_t^{Y}] = \frac{1}{2} Y_t$, although I cannot prove it. As a secondary question, can we generalize to having independent continuous martingales instead of two Brownian Motions?

Any help is appreciated!

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Looks like I solved this one already, so I will share the solution. $$ Y_t = E[Y_t | \mathcal{F}_t^Y] = E[B_t | \mathcal{F}_t^Y] + E[W_t | \mathcal{F}_t^Y] $$ since $B_t$ and $W_t$ are identically distributed, we should have $$ E[B_t | \mathcal{F}_t^Y] = E[W_t | \mathcal{F}_t^Y] $$ and so we get that $\frac{1}{2} Y_t = E[B_t | \mathcal{F}_t^Y] = E[W_t | \mathcal{F}_t^Y]$.

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