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Assuming that $W(t)$ is a Brownian motion, and considering two integrals

$$ X :=\int_{0}^{T} W(t) d W,\quad\text{and}\quad Y :=\int_{0}^{T}(W(t)+t)^{2} d W $$

I'm looking for the covariance $Cov(X, Y)$.

Here is my try:

$$ \operatorname{Cov}(X, Y)=E[(X(t)-E[X(t)]) \cdot (Y(t)-E[Y(t)])] $$

$$ X(t)=\int_{0}^{t} W(s) d W=\frac{1}{2} W^{2}(t)-\frac{1}{2} t $$

$$ E\left[X_{t}\right]=E\left[\frac{1}{2} W_{t}^{2}-\frac{1}{2} t\right]=E\left[\frac{1}{2} W_{t}^{2} \right]-\frac{1}{2}t=\frac{1}{2}E\left[ W_{t}^{2} \right]-\frac{1}{2}t = 0 $$

$$ \operatorname{Cov}(X, Y)=E[(\int_{0}^{T} W(t) d W) \cdot (\int_{0}^{T}(W(t)+t)^{2}-E[Y(t)])] $$

Here is what things get messy. First, Im struggling with finding expectation of Y(t). Second, multiplication of integrals get messy:

$$=E\left[\int_{0}^{T} W^{3}(t) d W+\int_{0}^{T} t^{2} W(t) d W+\int_{0}^{T} 2 t W^{2}(t) d W-(\int_{0}^{T} W(t) d W)\cdot E[Y(t)]\right]$$

Due to linearity of expectation, I can further write that:

$$=E\left[\int_{0}^{T} W^{3}(t) d W\right] + E\left[\int_{0}^{T} t^{2} W(t) d W\right] + E\left[\int_{0}^{T} 2 t W^{2}(t) d W\right] - E\left[(\int_{0}^{T} W(t) d W)\cdot E[Y(t)]\right]$$

$$=\int_{0}^{T}E\left[ W^{3}(t) \right]d W + \int_{0}^{T} E\left[t^{2} W(t)\right] d W + \int_{0}^{T} E\left[2 t W^{2}(t)\right] d W - E\left[(\int_{0}^{T} W(t) d W)\cdot E[Y(t)]\right]$$

Now I'm wondering if I can continue as $$=\int_{0}^{T}E\left[ W^{3}(t) \right]d W + \int_{0}^{T} E\left[t^{2} W(t)\right] d W + \int_{0}^{T} E\left[2 t W^{2}(t)\right] d W - E\left[(\int_{0}^{T} W(t) d W)\cdot E[Y(t)]\right]$$

I would appreciate your comments on how far this is correct and how to get things straight. Thanks.

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    $\begingroup$ Are you allowed to use Itô-isometry? This tells that $$ \operatorname{Cov}\left(\int_{0}^{T} W(t) \, dW, \int_{0}^{T} (W(t) + t)^2 \, dW \right) = \mathbf{E}\left[ \int_{0}^{T} W(t) (W(t) + t)^2 \, \mathrm{d}t \right]. $$ The expectation in the right-hand side can be easily computed by switching the order of expectation and integral, yielding $$=\int_{0}^{T} 2t^2\,\mathrm{d}t=\frac{2}{3}T^3.$$ $\endgroup$ – Sangchul Lee Mar 24 at 22:11
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    $\begingroup$ You can't take the expectation inside the integral against Brownian motion as you do here. Notice that with the expectation outside you get just a number but with it inside you get a random variable. $\endgroup$ – Rhys Steele Mar 24 at 22:12
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    $\begingroup$ As $cov(X,Y) = E[XY] - E[X]E[Y]$, we need to justify why $E[X]$ and $E[Y]$ are equal to $0$ before applying the Ito isometry. Let's define the functions $\phi_t = W_t$ and $\psi_t = (W_t + t)^2$ which are adapted and continuous. Given the fact that $W_t$ is $L^p$ with $p \geq 1$, we have that : \begin{align*} t \in \mathbb{R}_+, \quad E\left[\int_{0}^{t}\phi_s^2ds\right] < +\infty \quad \text{and} \quad E\left[\int_{0}^{t}\psi_s^2ds\right] < +\infty \end{align*} Thus, $X$ and $Y$ are true martingales and $E[X]$ and $E[Y]$ are indeed equal $0$. The rest follows @SangchulLee comment. $\endgroup$ – Sesame Mar 25 at 11:45
  • $\begingroup$ @SangchulLee (and others) In order to get $\frac{2}{3} T^{3}$, how do you show $\mathbb{E} [W^3(t)] = 0$? $\endgroup$ – Blade Mar 27 at 20:10
  • $\begingroup$ Given that $W(t)\sim\mathcal{N}(0,t)$, where did you stuck in evaluating the expectation? $\endgroup$ – Sangchul Lee Mar 27 at 22:15

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