Question

Compute the covariance function and spectral density function for the moving average process $$ X_{n}=\sum_{k=0}^{\infty} a_{k} \xi_{n-3} $$ where \(\left\\{\xi_{n}\right\\}\) are zero-mean uncorrelated random variables having unit variance. and \(a_{0}, a_{1}, \ldots\) are real numbers satisfying \(\sum a_{k}^{2}<\infty\).

Short answer

The covariance function is \(C_X(n,n) = \sum_{k=0}^{\infty} a_{k}^2\) for \(m=n\) and 0 for \(m \neq n\). The spectral density function is \(f_X(\omega) = \frac{1}{2\pi}[\sum_{k=0}^{\infty} a_{k}^2 + 2\sum_{k=1}^{\infty} a_{k}^2 cos(k\omega)]\).

Step-by-step solution

Write down the covariance function

The covariance function for a moving average process \(X_n\) is defined as \(C_X(m,n) = E[X_m X_n]\). We need to compute this for our given process \(X_n = \sum_{k=0}^{\infty} a_{k} \xi_{n-k}\) where \(\xi_n\) are zero-mean uncorrelated random variables with unit variance and \(a_k\) are constants.

Compute the Expectation

Let's compute the expectation for \(m=n\). We get that \(C_X(n,n) = E[X_n X_n] = E[(\sum_{k=0}^{\infty} a_{k} \xi_{n-k})^2]\). Since \(\xi_n\) are uncorrelated, the crossproducts vanish when expanding the square. This simplifies to \(C_X(n,n) = \sum_{k=0}^{\infty} a_{k}^2\).

Evaluate for other cases

For \(m \neq n\), since \( \xi_n \) are uncorrelated, this implies that \( E[\xi_n \xi_m] = 0 \). Hence, the covariance function \( C_X(m,n) = 0 \).

Spectral Density Function

The spectral density function is the Fourier transform of the covariance function. As \( C_X(m,n) = 0 \) for \( m \neq n \), its Fourier transform gives us 0. Hence, \( f_X(\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty}C_X(m,n)e^{i\omega(m-n)} dm dn = 0 \). For \( m = n \), the Fourier transform is \( f_X(\omega) = a_0^2 + 2\sum_{k=1}^{\infty} a_{k}^2 cos(k\omega) \). So, \( f_X(\omega) = \frac{1}{2\pi}[\sum_{k=0}^{\infty} a_{k}^2 + 2\sum_{k=1}^{\infty} a_{k}^2 cos(k\omega)] \).