# Filtering Sum of Brownian Motions

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!

## 1 Answer

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 $$\frac{1}{2} Y_t = E[B_t | \mathcal{F}_t^Y] = E[W_t | \mathcal{F}_t^Y]$$.