I want to compute the spectral measure of the 1-dimensional Ornstein-Uhlenbeck process with covariance function \begin{align} \rho(t) = e^{-|\beta|t}, \qquad t \in \mathbb{R}. \end{align} Since $\rho$ is a proper covariance function, Bochner's theorem says that $\rho$ is the covariance function of some stationary Gaussian random field if and only if \begin{align} \rho(t) = \int_\mathbb{R} e^{i\omega t}\ d\mu(\omega), \end{align} for some finite non-negative symmetric Borel measure on $\mathbb{R}$. Now, I know as well that if $\mu \ll \lambda$, when $\mu$ is absolutely continuous w.r.t. the Lebesgue-measure, $\rho$ could be written as \begin{align} \rho(t) = \int_\mathbb{R} e^{i\omega t}f(\omega)\ d \omega, \end{align} where $f$ is the spectral density, i.e. the Fourier transform of the covariance function $\rho$. Therefore, \begin{align} f(\omega)= \frac{1}{2\pi} \int_\mathbb{R} e^{-i\omega t} \rho(t)\ dt. \end{align} However, I do not know see how to compute the spectral measure explicitly. Any help is appreciated!


1 Answer 1


At first, the covariance function of OU process should be $\rho(t)=e^{-\beta|t|}$, $\beta>0$. If this is true, then the spectral density is the following: \begin{align} f(\omega)&=\frac1{2\pi}\int_{\mathbb{R}}e^{-i\omega t}\rho(t)\,dt =\frac1{2\pi}\biggl[\int_{-\infty}^0e^{-i\omega t}e^{-\beta|t|}\,dt+ \int_{0}^{\infty}e^{-i\omega t}e^{-\beta|t|}\,dt\biggr]\\ &=\frac1{2\pi}\Bigl[\frac{1}{\beta-i\omega}+\frac{1}{\beta+i\omega}\Bigr] =\frac{\beta}{\pi(\beta^2+\omega^2)}. \end{align}

  • $\begingroup$ Okido, thanks for your reaction. How to find the spectral measure from the spectral density? $\endgroup$
    – iJup
    Commented Apr 8, 2017 at 8:31

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