Question

Let \(\left\\{X_{n}\right\\}\) be a zero-mean covariance stationary process having covariance function \(R_{X}(v)\) and spectral density function \(f_{X}(\omega),-\pi \leq \omega \leq \pi\). Suppose \(\left\\{a_{n}\right\\}\) is a real sequence for which \(\sum_{b, 0}^{\infty}\left|a_{i} a_{j} R(i-j)\right|<\infty\), and define $$ Y_{n}=\sum_{k=0}^{\infty} a_{k} X_{n-k} $$ Show that the spectral density function \(f_{Y}(\omega)\) for \(\left\\{Y_{n}\right\\}\) is given by $$ \begin{aligned} f_{Y}(\omega) &=\frac{\sigma_{X}^{2}}{\sigma_{Y}^{2}}\left|\sum_{k=0}^{\infty} a_{k} e^{i k \omega}\right|^{2} f_{x}(\omega) \\ &=\frac{\sigma_{X}^{2}}{\sigma_{Y}^{2}}\left[\sum_{X=0}^{\infty} a_{j} a_{k} \cos (j-k) \omega\right] f_{X}(\omega), \quad-\pi \leq \omega \leq \pi . \end{aligned} $$

Short answer

The spectral density function \(f_{Y}(\omega)\) for \(\left\{Y_{n}\right\}\) is then given by \[ f_{Y}(\omega) =\frac{\sigma_{X}^{2}}{\sigma_{Y}^{2}}\left[\sum_{k=0}^{\infty} a_{k} a_{j}\cos(k-j)\omega\right] f_{X}(\omega), \quad-\pi \leq \omega \leq \pi. \]

Step-by-step solution

Transform \(Y_{n}\) into the Frequency Domain

Firstly, transform \(Y_{n}\) into the frequency domain using the Fourier Transform. The result is \(\hat{Y}(\omega) = \sum_{k=0}^{\infty} a_{k}X(e^{i\omega(n-k)})\).

Apply the Power Spectral Density (PSD) Function

Now, apply the power spectral density (PSD) function to \(\hat{Y}(\omega)\): \[ \hat{Y}(\omega) \hat{Y}^{*}(\omega) = \left[\sum_{k=0}^{\infty} a_{k}X(e^{i\omega(n-k)})\right] \left[\sum_{j=0}^{\infty} a_{j}X(e^{-i\omega(n-j)})\right] \]

Apply Expectation Operator and Separate the Terms

Apply the expectation operator to both sides and separate the terms inside the sum operators: \[ E\left[\hat{Y}(\omega) \hat{Y}^{*}(\omega)\right] = \sum_{k=0}^{\infty}\sum_{j=0}^{\infty} a_{k}a_{j}E\left[X(e^{i\omega(n-k)})X(e^{-i\omega(n-j)})\right]. \]

Expanding Covariance Function and Apply Stationarity Property

Expand the covariance function. \[ R_{X}(k-j) = E[X_kX_j] \=torch (\sigma_{X}^{2} f_{X}(\omega)\) Due to the stationary property, the expectation does not depend on the index itself \(k\) or \(j\), but the difference between them. Replace it in above equation. We get \[ E\left[\hat{Y}(\omega) \hat{Y}^{*}(\omega)\right]\ =\sigma_{X}^{2} \sum_{k=0}^{\infty}\sum_{j=0}^{\infty} a_{k}a_{j}e^{i\omega(k-j)} f_{X}(\omega) \]

Simplify the Power Spectral Density Of \(Y(\omega)\)

Combine cosine terms and trigonometric identities to simplify the power spectral density of \(Y(\omega)\) to the required form. The final Expression would look like \[ f_{Y}(\omega) =\frac{\sigma_{X}^{2}}{\sigma_{Y}^{2}}\left[\sum_{k=0}^{\infty} a_{j} a_{k}\cos(k-j)\omega\right] f_{X}(\omega), \quad-\pi \leq \omega \leq \pi . \]