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The autocovariance of an Ornstein–Uhlenbeck process

$$ dX(t) = \theta (\mu - X(t))dt + \sigma dW(t) $$

is given on Wikipedia as

$$ Cov(X(s),X(t)) = \frac{\sigma^2}{2\theta}(e^{-\theta|t-s|} - e^{-\theta(t+s)}) \quad \quad (1).$$

which seems to depend on the time of origin since it has $t+s$ term.

On the other hand, the discreet-time analogue of the O-U process is the AR(1) process $$ X_{i+1} = \theta' (\mu' - X_i) + \sigma' Z_{i+1} $$

which has autocovariance (again according to Wikipedia)

$$Cov(X_{i+n},X_i) = \frac{(\sigma')^2}{1-(\theta')^2}(\theta')^{|n|} \quad \quad (2)$$

which only depend on the lag $n$. How does one reconcile the two? I can see that in the limit of $t,s \to \infty$ in such a way that $t-s = n$, $(1)$ becomes

$$ Cov(X(s),X(t)) = \frac{\sigma^2}{2\theta}e^{-\theta|n|} \quad \quad (3)$$

but it is not clear how this is related to $(2)$.

Also, if we have a time series of O-U realisation (for which we do not know the origin of time), what do we actually get when we compute sample autocovarince: $(1)$ or $(2)$?

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If I discretise the O-U process, then I get

$$ X_{t+1} - X_t = \theta (\mu - X_t) \delta t + \sigma \sqrt{\delta t} Z_{t+1} $$

or after re-arranging $$ X_{t+1} = \theta \mu \delta t + (1- \theta \delta t) X_t + \sigma \sqrt{\delta t} Z_{t+1} .$$

If I compare this now to $(2)$, I see that $\theta'= \theta \delta t - 1$ and $\sigma' = \sigma \sqrt{\delta t}$ so that on substitution into $(3)$ I get

$$ Cov(X(s),X(t)) = \frac{(\sigma')^2 /\delta t}{2(1+\theta')/\delta t}e^{-\theta|n|} = \frac{(\sigma')^2}{2(1+\theta')}e^{-\theta|n|}\quad \quad (4)$$

but it still does not have the form of $(2)$.

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Instead of using the discretization, you can use the continuous time solution. Following the substitutions from this answer, we have that

$$\begin{align} \sigma' &= \frac1{2\theta}\sigma^2(1-e^{-2\theta\delta t})\\ \theta' &= e^{-\theta\delta t} \end{align}$$

which, when applied to your formula for the AR(1) covariance, yields

$$ \frac{{\sigma'}^2}{1-{\theta'}^2}{\theta'}^{|n|} = \frac{\sigma^2}{2\theta}\frac{1-e^{-2\theta\delta t}}{1-e^{-2\theta\delta t}}{e^{-\theta\delta t}}^{|n|} = \frac{\sigma^2}{2\theta}e^{-\theta\delta t|n|} $$

Now if you instead let $t-s \to n/\delta t$, you should have your answer.

I think that because your derivation would depends on an approximation (i.e. discretization), your answer would be an approximation.


There are two different covariances, one conditional (usually on $0$), and one unconditional. The one at Wikipedia calculates the conditional one

$$\begin{align} cov(x_s, x_t) &= E((x_s - E(x_s))(x_t-E(x_t))) \\&= E\left(\int_{\color{red}0}^s\sigma e^{\theta(u-s)}\,\mathrm d W_u \int_{\color{red}0}^t\sigma e^{\theta(v-t)}\,\mathrm d W_v\right) \\&= \sigma^2 e^{-\theta(s+t)}E\left(\int_{\color{red}0}^s e^{\theta(u)}\,\mathrm d W_u \int_{\color{red}0}^t e^{\theta(v)}\,\mathrm d W_v\right) \\&= \frac{\sigma^2}{2\theta} e^{-\theta(s+t)}(e^{2\theta \min(s,t)} \color{red}{- 1}) \\&= \frac{\sigma^2}{2\theta}(e^{-\theta|s-t|} \color{red}{- e^{-\theta(s+t)}}) \end{align}$$

Changing the lower limit to $-\infty$, causes the red term to vanish.

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    $\begingroup$ Thank you for your reply. I guess what I find peculiar is that the continuous process has a dependency on $|t+s|$ while the discreet only on the lag $|t-s|$. OK, the $|t+s|$ component is exponentially decreasing, but still... I suppose this is because the discreet result assumes origin is infinitely back in the past... $\endgroup$
    – Confounded
    Commented Apr 29, 2020 at 17:40
  • $\begingroup$ In this paper this paper the unconditional variance only depends on $|t-s|$, as you expected. $\endgroup$
    – Frank Vel
    Commented Apr 29, 2020 at 20:03
  • $\begingroup$ Both continuous and discrete should have the $t+s$ dependence in the conditional case. Remember for conditional, it's three points $(x_0, x_s, x_t)$. For unconditional, I think either taking the origin to infinity or solving for the steady state solution of the SDE and integrating over that will given the same result for this particular process. Someone please jump in if you know for sure. $\endgroup$
    – safetyduck
    Commented Mar 13, 2023 at 15:48

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