# Ito isometry for different time interval

Without throwing too much technical details, The Ito isometry says

$$\mathbb{E}\bigg[\Big(\int^t_0 X_s d W_s\Big) \Big(\int^t_0 X_s d W_s\Big)^T \bigg] = \int^t_0 X_s\,X_s^T ds,$$

for some suitable Brownian motion $$W_t$$.

How about in the above two integrals $$t$$ are not the same? That is,

$$\mathbb{E}\bigg[\Big(\int^{t_1}_0 X_s d W_s\Big) \Big(\int^{t_2}_0 X_s d W_s\Big)^T \bigg] = ?$$

Is it equal to $$\int^{\mathrm{min}(t_1, t_2)}_0 X_s\,X_s^T ds$$? If yes how to prove?

Well your intuition is right here is why, supposing $$t_1 :
$$\mathbb{E}\bigg[\Big(\int^{t_1}_0 X_s d W_s\Big) \Big(\int^{t_2}_0 X_s d W_s\Big)^T \bigg] =\mathbb{E}\bigg[\Big(\int^{t_1}_0 X_s d W_s\Big) \Big(\int^{t_1}_0 X_s d W_s+\int^{t_2}_{t_1} X_s d W_s\Big)^T \bigg]=$$ $$\mathbb{E}\bigg[\Big(\int^{t_1}_0 X_s d W_s\Big) \Big(\int^{t_1}_0 X_s d W_s\Big)^T \bigg]+\mathbb{E}\bigg[\Big(\int^{t_1}_0 X_s d W_s\Big) \Big(\int^{t_2}_{t_1} X_s d W_s\Big)^T \bigg]$$
Now observe that the second term is null as the two random variables are independent as the first integral belongs to $$\mathcal{F}_{t_1}$$ and the second to $$\mathcal{F}_{t_2, t_1}$$ that are independent. You are left with the term for which you know the result already.