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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.

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  • $\begingroup$ 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
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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

$$\epsilon=E\{e^2(t)\}=R_{ee}(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}S_{ee}(\omega)d\omega\tag{2}$$

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

$$\epsilon=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left[|H(\omega)|^2S_x(\omega)-2\text{Re}\left\{H(\omega)S_x(\omega)e^{-j\omega\lambda}\right\}+S_x(\omega)\right]d\omega\tag{3}$$

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:

$$S_x(\omega)=F(\omega)F^*(\omega)\tag{4}$$

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:

$$|H(\omega)|^2S_x(\omega)-2\text{Re}\left\{H(\omega)S_x(\omega)e^{-j\omega\lambda}\right\}+S_x(\omega)=\left|H(\omega)F(\omega)-F(\omega)e^{j\omega\lambda}\right|^2\tag{5}$$

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

$$S_{xx}(\omega)=\frac{2\alpha}{\alpha^2+\omega^2}=\frac{\sqrt{2\alpha}}{\alpha+j\omega}\cdot\frac{\sqrt{2\alpha}}{\alpha-j\omega}\tag{6}$$

We choose

$$F(\omega)=\frac{\sqrt{2\alpha}}{\alpha+j\omega}\tag{7}$$

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

$$\mathcal{F}^{-1}\left\{F(\omega)e^{j\omega\lambda}\right\}=\sqrt{2\alpha}e^{-\alpha(t+\lambda)}u(t+\lambda)\tag{8}$$

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

$$H(\omega)=e^{-\alpha\lambda}\tag{11}$$

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

$$h(t)=e^{-\alpha\lambda}\delta(t)\tag{12}$$

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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