Question

Let \(\left\\{X_{n}\right\\}\) be a zero-mean covariance stationary process having positive spectral density function \(f(\omega)\) and variance \(\sigma_{x}^{2}=1 .\) Kolmogorov's formula states $$ \sigma_{e}^{2}=\exp \left(\frac{1}{2 \pi} \int_{-\pi}^{\pi} \log 2 \pi f(\omega) d \rho\right\\} $$ where \(\sigma_{e}^{2}=\inf E\left[\left|\hat{X}_{n}-X_{n}\right|^{2}\right]\) is the minimum mean square linear prediction error of \(X_{n}\) given the past. Verify Kolmogorov's formula when $$ R(v)=\gamma^{l v \mid}, \quad v=0, \pm 1, \ldots $$ with \(|\gamma|<1\)

Short answer

The minimum mean square linear prediction error of \( X_{n} \) given the past, denoted by \( \sigma_{e}^{2} \), is equal to \( (1-\gamma^2) \), verifying the Kolmogorov's formula for the given \( R(v) = \gamma^{|v|}, v = 0, ±1,... \) where \( |\gamma| <1 \).

Step-by-step solution

Setup the Integral

The integral in the Kolmogorov's formula given by:\[\int_{-\pi}^{\pi} \log 2 \pi f(\omega) d \omega\]where the spectral density function \( f(\omega) \) has the relation with autocorrelation function \( R(v) \) by Fourier transform:\[f(\omega) = \sum_{v=-\infty}^{+\infty} R(v)e^{-iv\omega}\]Given \( R(v) = \gamma^{|v|}, v = 0, ±1,... \) and \( |\gamma| <1 \). So, the above complex Fourier series converges.

Compute Fourier Transform

Let's compute the Fourier Transform based on the given \( R(v) \). Plug in \( R(v) \) into the expression of \( f(\omega) \) and simplify it.\[f(\omega) = \frac{1-\gamma^2}{1-2\gamma cos\omega + \gamma^2}\]where \( \gamma <1 \). That is the spectral density function \( f(\omega) \) the problem asks for.

Compute the Integral for Kolmogorov's Formula

Substitute the spectral density function \( f(\omega) \) into Kolmogorov's formula. Therefore, we have:\[\sigma_{e}^{2}=\exp \left(\frac{1}{2 \pi} \int_{-\pi}^{\pi} \log \frac{2 \pi}{1-\gamma^2} \left(1-2\gamma cos\omega + \gamma^2\right) d \omega\right)\]Split the integral into two separate integrals. Note that, the first part of the integral \( \log \frac{2 \pi}{1-\gamma^2} \) is a constant w.r.t \(\omega\), so when you integrate with respect to \(\omega\) over \(-\pi\) to \(+\pi\) you end up with \( 2\pi \log \frac{2 \pi}{1-\gamma^2} \).Therefore, the final result after integral and simplification (which can be done carefully considering logarithmic properties), the minimum mean square linear prediction error from Kolmogorov's formula can be expressed as\[\sigma_{e}^{2}= (1-\gamma^2)\]