Question

Let \(\left\\{\xi_{n}\right\\}\) be independent identically distributed random variables having zero means and unit variances. Show that every moving average $$ X_{n}=\sum_{k=0}^{m} a_{k} \xi_{n-k+} \quad n=0, \pm 1, \ldots $$ is ergodic. Suppose \(\sum a_{k}^{2}<\infty .\) Is the same true of $$ Y_{n}=\sum_{k=0}^{\infty} a_{k} \xi_{n-k} ? $$

Short answer

In summary, for the moving average \(X_n\), we have shown that it is stationary and its autocorrelation converges to zero as the time difference goes to infinity, which proves that it is ergodic. However, for the infinite moving average \(Y_n\), we can't provide a general conclusion on its ergodicity. The ergodicity of \(Y_n\) will depend on the specific coefficients \(a_k\), and it's possible for \(Y_n\) to be ergodic under certain conditions, where the sum of squares of coefficients \(\sum a_k^2\) is finite.

Step-by-step solution

Prove \(X_n\) is stationary

Stationarity is characterized by having constant mean value and autocovariance that only depends on the time difference. Given that the random variables \(\xi_n\) have zero means and unit variances (given), we can show that \(X_n\) is stationary. Mean of \(X_n\): \[E[X_n] = E\left[\sum_{k=0}^m a_k \xi_{n-k}\right] = \sum_{k=0}^m a_k E[\xi_{n-k}] = 0\] Now calculate the autocovariance function \(R(n, n + \tau) = E[X_n X_{n + \tau}]\). \[R(n, n + \tau) = E\left[\left(\sum_{k=0}^m a_k \xi_{n-k}\right)\left(\sum_{j=0}^m a_j \xi_{n + \tau - j}\right)\right]\] Since \(\xi_n\) are uncorrelated random variables, the covariance of distinct random variables is zero, so only terms with matching indices will contribute to the sum. \[R(n, n+\tau) = \sum_{k=0}^m \sum_{j=0}^m a_k a_j E[\xi_{n-k}\xi_{n+\tau-j}]\] Thus, the autocovariance only depends on the time difference \(\tau\). This shows that \(X_n\) is stationary, as both its mean and autocovariance are constant or only depend on \(\tau\).

Prove the limit of autocorrelation converges to zero

Now, we need to show that the limit of the autocorrelation converges to zero as \(\tau \rightarrow \infty\). The autocorrelation function can be defined as: \[r(n, n+\tau) = \frac{R(n, n+\tau)}{\sqrt{R(n, n)R(n+\tau, n+\tau)}}\] Utilizing the autocovariance function \(R(n, n+\tau)\) derived earlier and the fact that \(R(n, n) = \sum_{k=0}^m a_k^2\), we can complete the calculations. We know that: \[r(n, n+\tau) \rightarrow 0\quad\textrm{if}\quad \tau \rightarrow \infty\] Since we have a finite summation in \(X_n\), as the time difference \(\tau\) goes to infinity, terms with matching indices will tend to zero, meaning the numerator of the autocorrelation function will go to zero. The denominator will remain constant for a finite sum \(R(n, n) = \sum_{k=0}^m a_k^2\). So, the limit of the autocorrelation converges to zero, proving that \(X_n\) is ergodic.

Analyze the ergodicity of \(Y_n\)

For the autocovariance function of \(Y_n\), the calculation is similar to the one in step 1, but this time, the summation is infinite. In this case, the condition for ergodicity states that \(\sum a_k^2 < \infty\). Mean of \(Y_n\): \[E[Y_n] = E\left[\sum_{k=0}^{\infty} a_k \xi_{n-k}\right] = \sum_{k=0}^{\infty} a_k E[\xi_{n-k}] = 0\] Additionally, the autocovariance of \(Y_n\): \[R(n, n + \tau) = E\left[\left(\sum_{k=0}^{\infty} a_k \xi_{n-k}\right)\left(\sum_{j=0}^{\infty} a_j \xi_{n + \tau - j}\right)\right]\] \[R(n, n+\tau) = \sum_{k=0}^{\infty} \sum_{j=0}^{\infty} a_k a_j E[\xi_{n-k}\xi_{n+\tau-j}]\] As with step 2, we need to show that the limit of the autocorrelation converges to zero. However, since the summation is now infinite, it's not guaranteed that the limit converges to zero for all cases. The given condition, \(\sum a_k^2 < \infty\), could suffice to determine the ergodicity of the series for a specific set of \(a_k\) coefficients. Summing up, for \(Y_n\), we can't provide a general conclusion about ergodicity. The ergodicity of \(Y_n\) will depend on the specific \(a_k\) coefficients, and given the conditions, it's possible for \(Y_n\) to be ergodic.