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.

  • $\begingroup$ do you mean the crochet between $X^1$ and $X^2$ i.e $\left\langle X^1,X^2\right\rangle_t$ or the true covariance i.e $\mathbb{E}(X^1_t X^2_t)-\mathbb{E}(X^1_t)\mathbb{E}(X^2_t)$ ? $\endgroup$ Apr 21, 2016 at 9:10
  • $\begingroup$ The true covariance, I would say. I need it for the maximum likelihood estimation of the parameters (long-term means, speeds of mean reversion, volatilities, correlation) of the two Ornstein-Uhlenbeck processes. Thank you in advance! $\endgroup$
    – LenaH
    Apr 21, 2016 at 11:20

1 Answer 1


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

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


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

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 $N_t=\int_0^t\sigma_2 e^{a_2 u} \rho_{12} dB^1_u$
  • $\begingroup$ Thank you!! Do you happen to have a reference for this? $\endgroup$
    – LenaH
    Apr 21, 2016 at 12:35
  • $\begingroup$ it is the basics of stochastic calculus, I add some references and hints in my answer, hope it is helpful $\endgroup$ Apr 21, 2016 at 15:17
  • $\begingroup$ Great, that helps! $\endgroup$
    – LenaH
    Apr 21, 2016 at 15:34

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