Question

Suppose $$ W_{n}=\sum_{j=1}^{4} \sigma_{j} \sqrt{2} \cos \left(\lambda_{j} n-V_{j}\right) $$ where \(\sigma_{j}, \lambda_{j}\) are positive constants, \(j=1, \ldots, q\), and \(V_{1}, \ldots, V_{g}\) are independent, uniformly distributed in the interval \((0,2 \pi)\). Show that \(\left\\{W_{n}\right\\}\) is covariance stationary and compute the covariance function.

Short answer

In summary, to prove that the process \(W_n\) is covariance stationary, we first showed that its mean is constant and equal to zero. Then, we computed the covariance function and demonstrated that it depends only on the time difference \(k\) and not on the original time points \(n\) and \(n+k\). The covariance function is given by: \(Cov(W_{n}, W_{n+k}) = \sum_{j=1}^4 \sigma_j^2 E[\frac{1}{2}\cos(2\lambda_j n+(\lambda_j k-2V_j))+\frac{1}{2}\cos(\lambda_j k - 2 V_j)] \)

Step-by-step solution

Prove mean is constant

To show that the process \(W_n\) has constant mean, we first calculate its expected value given the process definition: \(E[W_n] = E[\sum_{j=1}^4 \sigma_j \sqrt{2} \cos(\lambda_j n - V_j)]\) Since the expectation is a linear operator, the expected value of the sum is equal to the sum of the expected values: \(E[W_n] = \sum_{j=1}^4 E[\sigma_j \sqrt{2} \cos(\lambda_j n - V_j)]\) Since \(\sigma_j\), \(\lambda_j\), and \(n\) are constants, we can take them out of the expectation: \(E[W_n] = \sum_{j=1}^4 \sigma_j \sqrt{2} E[\cos(\lambda_j n - V_j)]\) Now, we will analyze the expectation of each cosine term in the sum. Notice that \(V_j\) is uniformly distributed in the interval \((0, 2\pi)\), making \(\cos(\lambda_j n - V_j)\) also random. Since cosine is a symmetrical function around the y-axis, its expectation will be equal to zero due to the uniform distribution of \(V_j\). Hence: \(E[\cos(\lambda_j n - V_j)] = 0\) Therefore, the expected value of \(W_n\) is: \(E[W_n] = \sum_{j=1}^4 \sigma_j \sqrt{2} \cdot 0 = 0\) This proves that the mean of the process is constant and equal to zero.

Compute covariance function

To show that the covariance function depends only on the time difference, we need to compute the covariance between \(W_n\) and \(W_{n+k}\). The covariance is defined as: \(Cov(W_n, W_{n+k}) = E[W_n W_{n+k}] - E[W_n]E[W_{n+k}]\) Since the mean is zero, we have: \(Cov(W_n, W_{n+k}) = E[W_n W_{n+k}]\) We can now substitute the expressions for \(W_n\) and \(W_{n+k}\) and compute the covariance: \(Cov(W_n, W_{n+k}) = E[\sum_{j=1}^4 \sigma_j \sqrt{2} \cos(\lambda_j n - V_j) \sum_{m=1}^4 \sigma_m \sqrt{2} \cos(\lambda_m (n+k) - V_m)]\) \(Cov(W_n, W_{n+k}) = \sum_{j=1}^4 \sum_{m=1}^4 \sigma_j \sigma_m 2 E[\cos(\lambda_j n - V_j)\cos(\lambda_m (n+k) - V_m)]\) We have the expectation of a product of cosines, so we can use the trigonometric identity:\(\cos(A)\cos(B) = \frac{1}{2}[\cos(A+B) + \cos(A-B)]\) \(Cov(W_n, W_{n+k}) = \sum_{j=1}^4 \sum_{m=1}^4 \sigma_j \sigma_m E[\frac{1}{2}\cos((\lambda_j+\lambda_m)n+(\lambda_m k-2V_m))+\frac{1}{2}\cos((\lambda_j-\lambda_m)n+(\lambda_m k-2 V_j))] \) Now, we will analyze the expectation of each cosine term in the summation. Similar to Step 1, the expectation of each cosine term will be equal to zero when \(j \neq m\) because the phase is uniformly distributed. However, when \(j = m\), the expectation will not be zero. So, we should consider only the terms where \(j = m\): \(Cov(W_n, W_{n+k}) = \sum_{j=1}^4 \sigma_j^2 E[\frac{1}{2}\cos(2\lambda_j n+(\lambda_j k-2V_j))+\frac{1}{2}\cos(\lambda_j k - 2 V_j)] \) Finally, we find that the covariance function depends only on the time difference \(k\) and not on the original time points \(n\) and \(n+k\): \(Cov(W_{n}, W_{n+k}) = \sum_{j=1}^4 \sigma_j^2 E[\frac{1}{2}\cos(2\lambda_j n+(\lambda_j k-2V_j))+\frac{1}{2}\cos(\lambda_j k - 2 V_j)] \) Hence, the process \(W_n\) is covariance stationary, as its mean is constant and its covariance function depends only on the time difference.