Itô Rule for Correlated Processes

Article Quant Q&A · Author: Gary

Summary

The document derives the differential of the product of two stochastic processes driven by Brownian motions with correlation ρ. Its central method is Itô’s product rule: combine each process’s drift and diffusion terms, then include the cross-variation term. Since the Brownian increments have covariance ρ dt, the product of the processes’ diffusion increments contributes ρσ_tϑ_t dt to the drift of their product.

The answer also shows how the same covariance enters when applying Itô’s lemma to the square of the sum of the processes. This supplies the missing term in the question’s decomposition. The result assumes the stated diffusion forms and correlation structure; it is a calculus explanation rather than an empirical trading method, and it does not discuss estimation or practical modeling of time-varying correlation.

Key ideas

The product differential includes both processes’ drift and diffusion contributions.
The cross-variation of correlated Brownian motions is ρ dt.
The product’s drift therefore includes a covariance adjustment of ρσ_tϑ_t.
The same cross term appears when applying Itô’s lemma to the square of a sum.

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# Differential product Correlated processes


# Differential product Correlated processes












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?

## Answer by Daneel Olivaw (score 1)

https://quant.stackexchange.com/a/42366

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:

$$d\left((X_t+Y_t)^2\right)=2(X_t+Y_t)(dX_t+dY_t)+(dX_t)^2+(dY_t)^2+2dX_tdY_t$$

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

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.

Itô’s Product Rule for Correlated Diffusions

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