I am trying to derive the differential of the product of two processes, but I got stuck. This is what I have until now:

We have the following two stochastic processes: $dX_t= \mu_t dt +\sigma_t dW_t$ and $dY_t = \eta_t dt + \vartheta_t d \bar{W}_t$, with the correlation between the two Brownian motions equal to $\rho$, that is

E[($W_t - W_s$)($\bar{W}_t - \bar{W}_s$)|$F_s$] = $\rho(t-s)$ for s $\leq$ t.

Then I can get an expression for $d(X_t Y_t)$ with the following trick:

I start with decomposition: $(X_t+Y_t)^2=X_t^2+Y_t^2+ 2X_t Y_t$,

Which leads by differentiation to $d(X_t Y_t)= \frac{1}{2}[d(\{X_t +Y_t\}^2) - d(X_t^2) - d(Y_t^2)]$

Next I applied Ito-lemma to all three parts separately as follows: $d(X_t^2)=(2\mu_tXt+\sigma_t^2)dt+ 2\sigma_t X_tdW_t= 2X_t dX_t + \sigma_t^2dt$ $d(Y_t^2)=(2\eta_t Y_t+\vartheta_t^2)dt+ 2\vartheta_t Y_t d\bar{W}_t= 2Y_t dY_t+\vartheta_t^2 dt$

Now I don't know how to apply Ito lemma to the last part, i.e., $d(\{X_t +Y_t\}^2)$. Particularly, I don't know how to account for the correlation between the two Brownian Motions. Can someone help me with this last step?

  • $\begingroup$ I don’t understand why are you proceeding that way. Why don’t you directly apply Ito’s lemma to $X_tY_t$? $\endgroup$ Oct 24, 2018 at 21:16
  • $\begingroup$ I'm following Baxter and Rennies (1996) procedure for proving the formula for the product of two correlated brownian motions. I'm able to do it in in case they're independent or perfectly correlated, but not for the imperfect correlation term. If I apply the rule directly then I'm not really proving it I think. Do you perhaps know how to apply Ito lemma to $d(\{X_t +Y_t\}^2)$? $\endgroup$
    – Gary
    Oct 24, 2018 at 21:22
  • $\begingroup$ I’m voting to close this question because what was done in the answer was exactly what OP's professor wanted OP to do. $\endgroup$
    – Kurt G.
    Sep 30, 2023 at 7:29

1 Answer 1


It can be shown that:

$$dW_td\bar{W}_t=\rho dt$$

Applying Ito’s lemma to $X_tY_t$ directly yields:

$$\begin{align} d(X_tY_t)&=X_tdY_t+Y_tdX_t+dX_tdY_t \\[4pt] &=(X_t\mu_t+Y_t\eta_t+\rho\sigma_t\vartheta_t)dt \\ &\qquad+X_t\vartheta_td\bar{W}_t+Y_t\sigma_tdW_t \end{align}$$

Edit based on your comment:

Applying Ito's lemma:


where: $$dX_tdY_t=\sigma_t\vartheta_tdW_td\bar{W}_t=\sigma_t\vartheta_t\rho dt$$

  • $\begingroup$ My professor wants us to find $d(X_tY_t)=X_t dY_t+Y_t dX_ t+dX_t dY_t$ by applying the Ito lemma to the three parts $\endgroup$
    – Gary
    Oct 25, 2018 at 5:38

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