Mean: \(E[y_t] = E[w_{t-1} + w_t] = E[w_{t-1}] + E[w_t] = 0 + 0 = 0\)
Variance: \(Var[y_t] = Var[w_{t-1} + w_t] = Var[w_{t-1}] + Var[w_t] = 2\sigma^2_w\)
Autocovariance: \(Cov[y_t, y_{t+k}] = Cov[w_{t-1} + w_t, w_{t+k-1} + w_{t+k}]\) for \(k \geq 1\)
Since \(w_t\) and \(w_{t+k}\) are independent for \(k \geq 1\), the autocovariance function is \(0\) for \(k \geq 1\).
Therefore, this process is not stationary.
Mean: \(E[y_t] = \frac{3+7}{2} = 5\)
Variance: \(Var[y_t] = \frac{(7-3)^2}{12} = \frac{4}{3}\)
Autocovariance: Since \(y_t\) is constant, the autocovariance function is \(0\) for \(k \geq 1\).
Therefore, this process is weak stationary.
Mean: \(E[y_t] = tE[A] = t \frac{3+7}{2} = 5t\)
Variance: \(Var[y_t] = t^2 Var[A] = t^2 \frac{4}{3} = \frac{4}{3}t^2\)
Autocovariance: Since the mean and variance are functions of \(t\), the autocovariance is also a function of \(t\), hence not stationary.
Therefore, this process is not stationary.
Mean: \(E[y_t] = E[y_{t-1}] + E[w_t] = E[y_{t-1}] = \delta\) (since the increments have zero mean)
Variance: \(Var[y_t] = Var[y_{t-1}] + Var[w_t] = Var[y_{t-1}] + \sigma^2_w\)
Autocovariance: \(Cov[y_t, y_{t+k}] = Cov[y_{t-1} + w_t, y_{t+k-1} + w_{t+k}]\) for \(k \geq 1\)
\(= Cov[y_{t-1}, y_{t+k-1}] = Cov[y_{t}, y_{t+k}]\) (since \(w_t\) and \(w_{t+k}\) are independent)
\(= Cov[y_{t}, y_{t+k}]\) (since the increments are identically distributed)
This implies that the autocovariance is independent of time, hence stationary.
Therefore, this process is strong stationary.