What is the covariance of $\int\limits_0^1 e^{B(s)} dB_s$ and $\int\limits_0^2 e^{-3B(s)} dB_s$?

Question

Let $B(s)$ the standard Brownian motion. I know that \begin{equation} \operatorname{Cov} \Big( \int\limits_0^T X(s) \, dB_s, \int\limits_0^T Y(s) \, dB_s \Big) = \mathbb{E} \Big( \int\limits_0^T X(s) \cdot Y(s) \, ds \Big). \tag{1} \end{equation} So, by extending the first integrand on $[1,2]$, as identically zero, I have \begin{align*} \operatorname{Cov} \Big( \int\limits_0^1 e^{B(s)} \, dB_s + \int\limits_1^2 0 \, dB_s, \int\limits_0^1 e^{-3B(s)}\, dB_s + \int\limits_1^2 e^{-3B(s)} \, dB_s \Big) \end{align*} and, by bilinearity of covariance, I get four terms, of which two are zero (by $(1)$): \begin{align*} \operatorname{Cov} \Big( \int\limits_0^1 e^{B(s)} \, dB_s, \int\limits_0^1 e^{-3B(s)} \, dB_s \Big) + \operatorname{Cov} \Big( \int\limits_0^1 e^{B(s)} \, dB_s, \int\limits_1^2 e^{-3B(s)}\, dB_s \Big) + 0 + 0. \end{align*} Now, I know how to calculate the first covariance (again using $(1)$), but not the second, as the boundaries of integration don't match.

Answer 1

Note that

$$\int_0^1 e^{B_s} \, dB_s = \int_0^2 1_{[0,1]}(s) e^{B_s} \, dB_s$$

and $$\int_1^2 e^{-3B_s} \, dB_s = \int_0^2 1_{[1,2]}(s) e^{-3B_s} \, dB_s.$$

Consequently, by (1),

$$\text{cov} \left( \int_0^1 e^{B_s} \, dB_s, \int_1^2 e^{-3B_s} \, dB_s \right) = \mathbb{E} \left( \int_0^2 1_{[0,1]}(s) e^{B_s} 1_{[1,2]}(s) e^{-3B_s} \, ds \right)=0.$$

Remark: Actually, there is no need to split up the integrals into two parts (i.e. to write $\int_0^2 = \int_0^1 + \int_1^2$). Using the above reasoning, it follows directly from (1) that

$$\text{cov} \left( \int_0^1 e^{B_s} \, dB_s , \int_0^2 e^{-3B_s} \, dB_s \right) = \mathbb{E} \left( \int_0^1 e^{B_s} e^{-3B_s} \, ds \right).$$