Given a stochastic signal $x(t)$ with autocorrelation function $R_{xx}(\tau)=\mathrm{exp}(- \alpha|\tau|)$, $\alpha>0$.

I want to predict $x(t+\lambda)$,$\lambda>0$ by $x(t-\tau)$, $\tau\ge0$ through a linear etimation,

i.e. I want to find estimator $\hat{x}(t+\lambda)=\int_0^{\infty}h(x)x(t-x)\mathrm{d}x$.

By the principle of linear minimal mean squared error, I can get Wiener Hopf equation of the integral form: $$R_{xx}(\tau+\lambda)=\int^{\infty}_{0}h(x)R_{xx}(\tau-x)\mathrm{d}x $$ $$\implies \mathrm{exp}(-\alpha|\tau+\lambda|)=\int^{\infty}_{0}h(x)\mathrm{exp}(-\alpha|\tau-x|)\mathrm{d}x$$ I tried to use Fourier transform to solve this integral equation for $h(x)$

$$\frac{e^{j\omega\lambda}}{\\\alpha^2+\omega^2}=H(\omega)\times \frac{1}{\\\alpha+j\omega} $$

But I am not sure the right-hand side is correct or not. Is it legitimate to view right hand side as convoution of $h(\tau)$ and $\mathrm{exp}(-\alpha\tau)\mathrm{u}(\tau)$, where $\mathrm{u(\tau)}$ is a unit step function? The textbook by Papoulis and Garcia use the factorization teqnique (it's on p594 of Papoulis 4th ed and p616 of Garcia 3rd ed) but I cannot see how to factor the transfer function to causal and non-causal part. I guess the answer of $h(x)$ is $c\times\delta(x)$ for some constant. Moreover, I want to find the mean sqaure error of this predictor.

  • $\begingroup$ Would you share with us why you think that the optimal predictor's impulse response is just a scaled impulse? Not that I don't agree, it's just that this does not appear to result from your calculations. $\endgroup$ – Matt L. Dec 25 '17 at 14:03
  • $\begingroup$ @MattL. Thank you for comment. It is based on the discrete case that the linear esitmator of signal with autocorrealtion of the form $r^{|t|}$. The predictor is only compsed of the nearst available signal. On Papoulis 4th ed,page 594, the best predictive estimator is $ \hat{x}(t+\lambda)=e^{-\alpha \lambda}x(t)$. Surely it is uncompatible with my calculation through Wiener Hopf integral equation. That is my question. $\endgroup$ – Rikeijin Dec 25 '17 at 15:08

Your Wiener-Hopf equation is correct, but you can't solve it by simply taking the Fourier transform of both sides. The reason for this is that the equation is only valid for $\tau\ge 0$ due to the causality requirement. Taking the Fourier transform of both sides implies that the equation is valid for all values of $\tau$. You have to take another route and proceed as follows.

The error signal is given by

$$e(t)=x(t+\lambda)-(h\star x)(t)\tag{1}$$

We want to minimize the MSE given by


where $R_{ee}(\tau)$ is the autocorrelation function of $e(t)$, and $S_{ee}(\omega)$ is its power spectral density, which is the Fourier transform of the autocorrelation function.

Using $(1)$ it is a rather straightforward exercise to show that $(2)$ can be written as


The trick is now to rewrite the integrand of $(3)$ by completing the square, which will explicitly show how $H(\omega)$ must be chosen. But first we need to factor $S_x(\omega)$. Since $S_x(\omega)$ is non-negative and real-valued we can factor it as follows:


where $F(\omega)$ corresponds to a real-valued causal and stable system with all its poles and zeros in the left half-plane (i.e., $F(\omega)$ is a minimum-phase system). Note that $|F(\omega)|=|F^*(\omega)|=\sqrt{S_x(\omega)}$.

Using $(4)$, we can write the integrand of $(3)$ as follows:


Clearly, we can minimize $(5)$ and $(3)$ by choosing $H(\omega)=e^{j\omega\lambda}$, which just means that we shift the input by $\lambda$ towards the future to get $x(t+\lambda)$, but that's unfortunately not how prediction works. We need to restrict our system to be causal. Note that if $H(\omega)$ is causal then also $F(\omega)H(\omega)$ is causal (because $F(\omega)$ is also causal). This means that the best we can do to minimze $(5)$ and $(3)$ is to use the term $H(\omega)F(\omega)$ to cancel the causal part of $F(\omega)e^{j\omega\lambda}$. So we need to compute the causal part of $F(\omega)e^{j\omega\lambda}$.

From the given autocorrelation function $R_{xx}(\tau)$ we get


We choose


corresponding to a causal and stable system. The inverse Fourier transform of $F(\omega)e^{j\omega\lambda}$ is given by


where $u(t)$ is the unit step function. The causal part of $(8)$ is simply obtained by multiplication with a step function:

$$\sqrt{2\alpha}e^{-\alpha\lambda}e^{-\alpha t}u(t)\tag{9}$$

The Fourier transform of $(9)$ is

$$\mathcal{F}\left\{\sqrt{2\alpha}e^{-\alpha\lambda}e^{-\alpha t}u(t)\right\}=e^{-\alpha\lambda}\frac{\sqrt{2\alpha}}{\alpha+j\omega}=e^{-\alpha\lambda}F(\omega)\tag{10}$$

From $(5)$ we have to choose $H(\omega)F(\omega)$ to equal $(10)$, the causal part of $F(\omega)e^{j\omega\lambda}$. So we finally obtain


as the optimal prediction filter frequency response, which corresponds to the impulse response


So for the given statistics of the input signal, the optimal predictor only uses the current input value to predict the future value $x(t+\lambda)$. It does not use any past values of $x(t)$.


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