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Sep 17, 2020 at 13:52 comment added Appliqué @Iván If you plug in $\tau=0$ you do not get the expression for the variance in your post
Dec 14, 2016 at 9:53 comment added Iván Thank you very much. If anyone needs the results $Cov(Y_t, Y_{t+\tau})= \frac{\sigma^2}{2 \theta^3}( -e^{-\sigma (2t+\tau)} - 2 e^{-\theta t} + 2 e^{-\theta (t+\tau)} -2 - e^{-\theta \tau} + 2t\theta )$
Nov 28, 2016 at 13:13 vote accept Iván
Nov 18, 2016 at 2:45 history answered Nate Eldredge CC BY-SA 3.0