91Ó°ÊÓ

The given equation to prove is: \[ \operatorname{Var}(X(t+s)-X(t))=2 R_{X}(0)-2 R_{X}(t) \] To begin, remember that the variance of a function is equal to the covariance of the function with itself: \[ \operatorname{Var}(X(t+s)-X(t)) = \operatorname{Cov}(X(t+s)-X(t), X(t+s)-X(t)) \] Now we can utilize the property of covariance which states that \(\operatorname{Cov}(a+b, c+d) = \operatorname{Cov}(a, c) + \operatorname{Cov}(a, d) + \operatorname{Cov}(b, c) + \operatorname{Cov}(b, d)\). Applying this to our current equation gives: \[ \operatorname{Cov}(X(t+s)-X(t), X(t+s)-X(t)) = \operatorname{Cov}(X(t+s), X(t+s)) - \operatorname{Cov}(X(t+s), X(t)) - \operatorname{Cov}(X(t), X(t+s)) + \operatorname{Cov}(X(t), X(t)) \] Next, notice that: 1. The first term is just the variance of \(X(t+s)\) (i.e., \(R_X(s)\)). 2. The second term is the covariance between \(X(t+s)\) and \(X(t)\) (i.e., \(R_X(t)\)). 3. The third term is the covariance between \(X(t)\) and \(X(t+s)\), which is also \(R_X(t)\) by definition of a weakly stationary process. 4. The fourth term is just the variance of \(X(t)\) (i.e., \(R_X(0)\)). Using this information, rewrite the equation as: \[ \operatorname{Var}(X(t+s)-X(t)) = R_{X}(s) - R_{X}(t) - R_{X}(t) + R_{X}(0) \] By combining like terms, we arrive at the desired equation: \[ \operatorname{Var}(X(t+s)-X(t))=2 R_{X}(0)-2 R_{X}(t) \]

To show that the process \(Y(t)=X(t+1)-X(t)\) is weakly stationary, we must prove that its covariance function, \(R_{Y}(s)\), is correctly defined by the given equation. We are given: \[ R_{Y}(s)=2 R_{X}(s)-R_{X}(s-1)-R_{X}(s+1) \] First, recall from the problem definition that \(Y(t) = X(t+1) - X(t)\). Then, by definition, the covariance function of \(Y(t)\) is: \[ R_{Y}(s) = \operatorname{Cov}(Y(t), Y(t+s)) \] We already know from part (a) that: \[ \operatorname{Var}(X(t+s)-X(t)) = 2R_{X}(0) - 2R_{X}(t) \] Applying this result to the current situation, we can rewrite \(R_{Y}(s)\) as: \[ R_{Y}(s) = \operatorname{Cov}(X(t+1)-X(t), X(t+s+1)-X(t+s)) \] Now, we can once again apply the property of covariance mentioned in part (a): \[ R_Y(s) = \operatorname{Cov}(X(t+1), X(t+s+1)) - \operatorname{Cov}(X(t+1), X(t+s)) - \operatorname{Cov}(X(t), X(t+s+1)) + \operatorname{Cov}(X(t), X(t+s)) \] From this equation, we can see that: 1. The first term is just \(R_X(s+1)\). 2. The second term is \(R_X(s)\). 3. The third term is \(R_X(s-1)\). 4. The fourth term is also \(R_X(s)\). Now, we can substitute these values back into the equation: \[ R_Y(s) = R_X(s+1) - R_X(s) - R_X(s-1) + R_X(s) \] Combining like terms, we get the desired equation: \[ R_{Y}(s)=2 R_{X}(s)-R_{X}(s-1)-R_{X}(s+1) \] So the process \(Y(t) = X(t+1) - X(t)\) is indeed weakly stationary having a covariance function of \(R_Y(s) = 2R_X(s) - R_X(s-1) - R_X(s+1)\).