Find conditional expectation of $\int_1^2W_t^2dt$ with respect to $F_1$

Question

$$\mathbb{E}(\int_1^2W_t^2dt|F_1)=\int_1^2\mathbb{E}((W_t-W_1+W_1)^2|F_1)dt=\int_1^2\mathbb{E}((W_t-W_1)^2|F_1)dt+\int_1^2 2\mathbb{E}((W_t-W_1)W_1|F_1)dt+\int_1^2 \mathbb{E}(W_1^2|F_1)dt=\int_1^2(t-1)dt+2 \cdot\int_1^2 W_1\cdot 0 dt+\int_1^2 W_1^2dt=\frac{1}{2}+\int_1^2 W_1^2dt$$

Is this the correct anwser?

Answer 1

Given $\mathscr{F}_1$, $\beta_ t = W_{t + 1 } - W_1$ is a standard Brownian Motion independent of $W_1$. So, for $t \geq 1$, $$ \mathbf{E}[ W_t^2 | \mathscr{F}_1 ] = \mathbf{E}[ (\beta_{t-1} + W_1 )^2 | \mathscr{F}_1 ] = t-1 + W_1^2. $$ So that, $$ \mathbf{E}\left[ \int_1^2 W_r^2 dr \big | \mathscr{F}_1 \right] = 1/2 + W_1^2. $$