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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?

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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 $s-t$.

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