# Compute the conditional expectation when brownian motion is involved

Consider the following conditional expectation

$$E[(x_t + \mu(s-t) + \sigma(\beta_s - \beta_t) )^2 | F_t]$$

where $$\beta_s$$ and $$\beta_t$$ are brownian motion.

Then,

= $$E[x^2_t + 2x_t\mu(s-t) + 2x_t\sigma(\beta_s - \beta_t) + 2\mu(s-t)\sigma(\beta_s - \beta_t) + \mu^2(s-t)^2 + \sigma^2(\beta_s - \beta_t)^2 |F_t]$$

Since $$(\beta_s - \beta_t)$$ is independent of $$F_t$$ and using the fact that $$\beta_s - \beta_t ∼N(0,s−t)$$ then the final answer would be

$$= x^2_t + 2x_t \mu(s-t) + \mu^2(s-t)^2$$

Is this correct?

If $$x_t$$ is $$F_t$$ measurable then you are almost right, but you dropped the last term. $$E\sigma^{2}(\beta_s-\beta_t)^{2}=\sigma^{2} (s-t)$$ since variance of $$\beta_s-\beta_t$$ is $$s-t$$.