Covariance of stochastic integral

Question

I have a big problem with such a task:

Calculate $\text{Cov} \, (X_t,X_r)$ where $X_t=\int_0^ts^3W_s \, dW_s$, $t \ge 0$.

I've tried to do this in this way: setting up $t \le r$ $$\text{Cov} \, (X_t,X_r)=\text{Cov} \, \left(\int_0^ts^3W_s \, dW_s,\int_0^rs^3W_s \, dW_s \right)=\int_0^t s^6 W^2_s \, ds$$ And I've stacked.

Answer 1

$\text{Cov} \, (X_t,X_r)=\text{Cov} \, (\int_0^ts^3W_sdW_s,\int_0^rs^3W_sdW_s)=\int_0^t s^6 W^2_s ds$

This identity does not hold true. Note that the left-hand side is a fixed real number whereas the right-hand side is a random variable.

If you apply Itô's isometry correctly, you find

$$\text{cov} \, (X_t,X_r) = \color{red}{\mathbb{E} \bigg( }\int_0^t s^6 W_s^2 \, ds \color{red}{\bigg)}.$$

Applying Tonelli's theorem yields

$$\text{cov} \, (X_t,X_r) = \int_0^t s^6 \mathbb{E}(W_s^2) \, ds.$$

Now, since $W_s \sim N(0,s)$, we have $\mathbb{E}(W_s^2) = s$, and therefore the integral can be easily calculated.