Itō's formula in stochastic calculus

Question

Let $$dI_t=K_t^1dt+H_t^1dB_t\text{ and }dJ_t=K_t^2dt+H_t^2dB_t.$$ With Itō's formula how to show that $$I_tJ_t=I_0J_0+\int_0^tI_sdJ_s+\int_0^t J_sdI_s+\int_0^tH_s^1H_s^2ds?$$

My idea: let $f(x,y)=xy$. How to use Itō's formula here?

Answer 1

Yes it can be derived from 2-dimensional Ito's Lemma with $f(x,y)=xy$.

Applying Ito's Lemma we obtain $$(\star) \ \ \ \ d(I_tJ_t) =I_tdJ_t + I_tdJ_t + \frac{1}{2}dI_tdJ_t + \frac{1}{2}dJ_tdI_t=I_tdJ_t + I_tdJ_t + dI_tdJ_t$$

Knowing Ito's table, that is, $dtdt = dtdB_t = dB_t dt = 0$ and $dB_tdB_t = dt$

we obtain that $$dI_tdJ_t =H_t^1H_t^2 dB_tdB_t =H_t^1H_t^2 dt $$

thus by integrating both sides of $(\star)$ we obtain

$$I_tJ_t - I_0J_0 = \int_0^t I_sdJ_s + \int_0^t J_sdI_s + \int_0^tH_s^1H_s^2 ds$$

and so

$$I_tJ_t = I_0J_0 + \int_0^t I_sdJ_s + \int_0^t J_sdI_s + \int_0^tH_s^1H_s^2 ds.$$

After rearranging the above we also obtain integration by parts formula

$$\int_0^t I_sdJ_s = I_tJ_t - I_0J_0 - \int_0^t J_sdI_s - \int_0^tH_s^1H_s^2 ds.$$