Set $X_t= \int_{0}^{t}\sqrt{s}\sin(B_s)dB_s, t\geq 0$.

How can I compute de covariance between $X_t$ and $X_u$, $0\leq u \leq t$?

I started using Itô isometry but I can´t go any further.




  1. Fix $u \leq t$. Use Itô's isometry to show that $$\mathbb{E}(X_u X_t) = \mathbb{E} \left( \int_0^u s \sin^2(B_s) \, ds \right) = \int_0^u s \mathbb{E}(\sin^2(B_s)) \, ds. \tag{1}$$ In order to calculate the integral we need to compute $\mathbb{E}(\sin^2(B_s))$.
  2. Apply Itô's formula to show that $$\mathbb{E}(\sin^2(B_s)) = \int_0^s (1-2 \mathbb{E}\sin^2(B_u)) \, du.$$ This means that the function $f(s) := \mathbb{E}\sin^2(B_s)$ satisfies the ordinary differential equation $$\frac{d}{ds} f(s) = 1- 2 f(s) \qquad f(0)=0. \tag{2}$$
  3. Solve $(2)$ and plug the result into $(1)$ to compute $\mathbb{E}(X_u X_t)$.

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