Conditional expectations one more time

Question

Please someone verifies my results:

1) $E \Big( \int_0^3W_t^2dt|F_1\Big)=$(editing in progress)

---

2) $E \Big( \int_0^2 (tW_t+t^2)dt|F_1\Big)=E \Big( \int_0^2 tW_tdt|F_1\Big)+E \Big( \int_0^2 t^2dt\Big)=E \Big( \int_0^1 tW_tdt|F_1\Big)+E \Big( \int_1^2 tW_tdt|F_1\Big)+\frac{8}{3}=\int_0^1 tW_tdt+\frac{8}{3} $

Is this correct?

Answer 1

1. No, the equality $$\mathbb{E} \left( \int_1^3 W_t^2 \, dt \mid \mathcal{F}_1 \right) = \mathbb{E} \left( \int_1^3 W_t^2 \, dt \right)$$ does not hold (mind that $W_t^2$, $t \geq 1$, is not independent of $\mathcal{F}_1$). Write instead $$\begin{align*} \int_1^3 W_t^2 \, dt &= \int_1^3 ((W_t-W_1)+W_1)^2 \, dt \\ &= \int_1^3 (W_t-W_1)^2 \, dt + 2 W_1 \int_1^3 (W_t-W_1) \, dt+ 2 W_1^2. \end{align*}$$

Solution: $$\int_0^1 W_t^2 \, dt +2 + 2 W_1^2.$$

2. No, this is not correct. Note that $\mathbb{E}(W_t \mid \mathcal{F}_1) \neq W_1$ for all $t \leq 1$.

Solution: $$\int_0^1 t W_t \, dt + \frac{3}{2} W_1 + \frac{8}{3}.$$