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I am wondering whether an analytical expression of the maximum likelihood estimates of an Ornstein-Uhlenbeck process is available. The setup is the following: Consider a one-dimensional Ornstein-Uhlenbeck process $(X_t)_{t\geq 0}$ with $X_0=x$ for some $x\in\mathbb{R}$, i.e. $(X_t)_{t\geq 0}$ solves the SDE $ \mathrm{d} X_t=\theta(\mu-X_t)\,\mathrm{d} t + \eta\,\mathrm{d} W_t,\quad X_0=x $ where $(W_t)_{t\geq 0}$ is a standard Wiener process and $\eta,\theta>0$, $\mu\in\mathbb{R}$. If $\lambda=(\eta,\theta,\mu)$ is the vector of parameters, then the transition densities are known and if $p_{\lambda}(t,x,\cdot)$ denotes the density of $X_t$ (remember $X_0=x$) with respect to the Lebesgue-measure, then $ p_{\lambda}(t,x,y)=(2\pi\beta)^{-1/2}\exp\left(-\frac{(y-\alpha)^2}{2\beta}\right),\quad y\in\mathbb{R}, $ where $\alpha=\mu+(x-\mu)e^{-\theta t}$ and $\beta=\frac{\eta^2}{2\theta}(1-e^{-2\theta t})$.

Suppose we have observed an Ornstein-Uhlenbeck process in equidistant time-instances (where the parameter $\lambda$ is unknown), i.e. the vector of observations is given by $ \mathbf{x}=\{x_0,x_{\Delta},\ldots,x_{N\Delta}\}, $ where $x_0=x$ and $\Delta>0$ and $N+1$ is the number of observations. Then by the Markov property of $(X_t)_{t\geq 0}$ we have that the log-likelihood function is given by $ l(\lambda)=l(\theta,\eta,\mu;\mathbf{x})=\sum_{i=1}^N \log\left(p_{\lambda} (\Delta,x_{(i-1)\Delta},x_{i\Delta})\right). $ Now i am asking if it is possible to maximize this expression with respect to $\lambda=(\eta,\theta,\mu)$ simultaneously and if so, how would one go about doing this. If anyone can point me in the direction of a paper/book where this is shown, it would be much appreciated. Thanks in advance!

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    Aha, that's a good point! I will surely look into that. Thanks.2012-06-25

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In the paper "Parameter estimation and bias correction for diffusion processes" by Tang and Chen explicit formulas for the MLE are given. Their formulas ignore $X_0$, but this makes little difference if the number of observations is reasonably large. I am puzzled how they managed to come up with this formulas, though. Solving $l'(\lambda)=0$ seem to be a difficult task.

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    That was exactly what I was looking for. Thanks.2012-06-27