Calculate conditional expectation of integral

Question

I have to calculate

$$\mathbb{E}\left(\int_1^2 (t^2W_t+t^3 )\,dt\mid F_1\right)$$

My attempt:

$$\int_1^2 (t^2W_t+t^3 )\,dt=\int_1^2t^2W_t\,dt+\frac{15}{4}$$

Now I will focus on: $\int_1^2t^2W_t\,dt$

Let's apply Ito formula to a process $X_t(W_t,t)=t^3W_t$

We have

$$dX_t=3t^2W_t\,dt+t^3\,dW_t$$

So we have $$\left.t^3W_t\, \vphantom{\frac11} \right|_1^2=3\int_1^2t^2W_t\,dt+\int_1^2t^3\,dW_t$$

From the above we have that

$$\int_1^2 t^2W_t \, dt=\frac{1}{3}\left(8W_2-W_1-\int_1^2t^3 \, dW_t\right)$$

Going back to expectation I have:

$$\mathbb{E}\left(\int_1^2 (t^2W_t+t^3 )\,dt\mid F_1\right)=\mathbb{E}\left(\int_1^2 t^2W_t \, dt\mid F_1\right)+\frac{15}{4}=\frac{16}{3}-\frac{1}{3}W_1+\frac{15}{4}$$

Is this correct?

Answer 1

The last step is not correct. Note that

$$\mathbb{E}(W_2 \mid F_1) = W_1$$

since $(W_t)_{t \geq 0}$ is a martingale. Therefore

$$ \frac{1}{3} \mathbb{E}(8 W_2-W_1 \mid F_1) = \frac{7}{3} W_1.$$

The correct result is

$$\frac{7}{3} W_1 + \frac{15}{4}.$$

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Instead of applying Itô's formula (which is rather overkill, I would say), one can use the fact that the conditional expectation can be interchanged with integration (see this question), i.e.

$$\mathbb{E} \left( \int_1^2 t^2 W_t \, dt \mid F_1 \right) = \int_1^2 t^2 \mathbb{E}(W_2 \mid F_1) \, dt.$$

Using again $\mathbb{E}(W_2 \mid F_1) = W_1$, the claim follows.

Answer 2

By linearity of the conditional expectation, $$\mathbb{E}\left(\left.\int_1^2 (t^2W_t+t^3 )\,\mathrm dt\,\right|F_1\right)=\int_1^2 (t^2\mathbb{E}(W_t\mid F_1)+t^3 )\,\mathrm dt,$$ hence $$\mathbb{E}\left(\left.\int_1^2 (t^2W_t+t^3 )\,\mathrm dt\,\right|F_1\right)=\int_1^2 (t^2W_1+t^3 )\,\mathrm dt=\int_1^2 t^2\,\mathrm dt\cdot W_1+\int_1^2t^3 \,\mathrm dt=\cdots$$