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Covariance of Correlated Ornstein–Uhlenbeck Processes

Article Quant Q&A · Author: LenaH
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Summary

The document derives the covariance between two Ornstein–Uhlenbeck processes whose Brownian drivers have instantaneous correlation. Starting from each process’s centered stochastic-integral solution, it takes the expectation of their product and uses the cross-variation of the Brownian motions. The resulting covariance depends on both volatility parameters, the driver correlation, the sum of the two mean-reversion speeds, and elapsed time.

The derivation distinguishes cross-process covariance from simply multiplying the marginal standard deviations by a correlation coefficient: the two processes can have different mean-reversion rates, so their covariance accumulates over time at a rate set by their combined decay. The answer also sketches a proof route using a representation of correlated Brownian motions and a martingale identity. The result assumes constant parameters and a fixed instantaneous correlation in the stated OU setup; the document does not discuss estimation from data or extensions to time-varying coefficients.

Key ideas

  • The covariance is obtained from the product of the centered stochastic-integral solutions.
  • Correlated Brownian drivers contribute through their instantaneous correlation in the cross-variation.
  • The covariance depends on the sum of the processes’ mean-reversion speeds.
  • The derivation assumes constant volatilities, mean-reversion speeds, and driver correlation.

Tags

StatisticsMean reversion
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# What is the covariance of two correlated Ornstein-Uhlenbeck processes?


# What is the covariance of two correlated Ornstein-Uhlenbeck processes?












What is the covariance of two correlated Ornstein-Uhlenbeck processes? I was trying correlation(1,2)*Var1^(1/2)*Var2^(1/2), but I am not sure! I took Var1=(sigma1^2/(2*speedofmeanreversion1))*(1-exp(-2*speedofmeanreversion1*dt)) and Var2 accordingly. Thank you.

## Answer by M. Jeunesse (score 5, accepted)

https://quant.stackexchange.com/a/25536

Using https://en.wikipedia.org/wiki/Ornstein%E2%80%93Uhlenbeck_process#Solution

$$X^i_t = (X^i_0 + \int_0^t\sigma_i e^{a_i u} dB^i_u)e^{-a_it} $$

and

$$ X^i_t-\mathbb{E}[X^i_t] = e^{-a_it} \int_0^t\sigma_i e^{a_i u} dB^i_u $$

and thus :

$$\text{Cov}(X^1_t,X^2_t)=\mathbb{E}\left[e^{-a_1t} \int_0^t\sigma_1 e^{a_1 u} dB^1_u e^{-a_2t} \int_0^t\sigma_2 e^{a_2 u} dB^2_u\right] $$

and if $d\langle B^1_t,B^2_t \rangle=\rho_{12}dt$

$$\begin{split} \text{Cov}(X^1_t,X^2_t)=& \mathbb{E}\left[e^{-(a_1+a_2)t} \int_0^t \sigma_1\sigma_2 e^{(a_1+a_2) u} \rho_{12} du\right]\\ &=e^{-(a_1+a_2)t} \int_0^t \sigma_1\sigma_2 e^{(a_1+a_2) u} \rho_{12} du\\ & =\frac{\sigma_1\sigma_2\rho_{12}}{a_1+a_2}\left(1-e^{-(a_1+a_2)t}\right) \end{split} $$

If you want to prove the last formula, you will need :

- the fact $B^2_t = \rho_{12}B^1_t+\sqrt{1-\rho_{12}^2}B^\perp_t$

- https://en.wikipedia.org/wiki/Quadratic_variation#Martingales

- the fact that $$2\mathbb{E}[M_tN_t]=\mathbb{E}[(M+N)^2_t]-\mathbb{E}[M^2_t]-\mathbb{E}[N^2_t]$$ with $M_t=\int_0^t\sigma_1 e^{a_1 u} dB^1_u$ and $N_t=\int_0^t\sigma_2 e^{a_2 u} \rho_{12} dB^1_u$

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.