Cross-Variation of Correlated Itô Diffusions

Summary

The document explains how the covariance of two continuous Itô processes follows from the covariance of their Brownian drivers. Drift terms do not contribute to their quadratic covariation; the stochastic components determine it. Applying the covariation rule for stochastic integrals gives the integral of the two volatility processes multiplied by the instantaneous covariance of the Brownian motions.

With instantaneous correlation denoted by ρ over time, the cross-variation accumulates as the time integral of ρ times both volatilities. If correlation and volatilities are constant, this reduces to their product multiplied by elapsed time. The answer supplies the general formula but does not expand on assumptions such as adaptedness and integrability of the volatility processes, or distinguish quadratic covariation from ordinary covariance of terminal values.

Key ideas

• The drift components do not affect the quadratic covariation of continuous Itô processes.
• The cross-variation of stochastic integrals weights Brownian cross-variation by both volatility processes.
• Instantaneous correlation produces an accumulated cross-variation through time integration.
• When correlation and volatilities are constant, cross-variation grows linearly with time.

Tags

# Covariation of Ito semimartingales


# Covariation of Ito semimartingales












If we have two Ito semimartingales over $[0,T]$: $$d X_t^i=a^i_tdt+\sigma_t^idW_t^i,\quad i=1,2$$ What is the relationship between $$\langle X^1,X^2 \rangle_t \quad \text{and} \quad \langle W^1,W^2 \rangle_t, $$ where $\langle \rangle_t$ denotes the quadratic variation? Assuming the correlation coefficient $\rho$ is constant, I think we should have $$\langle W^1,W^2 \rangle_t=\rho dt$$ while if it's not constant we would have $$\langle W^1,W^2 \rangle_t=\int_0^t\rho_s ds$$ How is this related to the the quadratic variation of $X^1$ and $X^2$, in the two cases? Is there any relationship with $\int_0^t\rho_s \sigma_s^1 \sigma_s^2 ds$?

## Answer by Gordon (score 3, accepted)

https://quant.stackexchange.com/a/55159

Note that \begin{align*} \left\langle \int_0^t \sigma_s^1 dW_s^1, \int_0^t \sigma_s^2 dW_s^2\right\rangle &= \int_0^t \sigma_s^1 \sigma_s^2 d\langle W_s^1, W_s^2 \rangle\\ &=\int_0^t \rho_s\sigma_s^1 \sigma_s^2 ds. \end{align*}

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