Hello I'm unable to recall the formula used here for calculating the covariance . This is the solution of Question 1.2(i) from chapter 1 practice questions ( on page 29)
Thanks Calm and Aisha. A related question: The solution for Q1.7 uses the fact that (0,t) and (t,t+s) are non-overlapping time periods. But why does this justify why Cov((X(t), X(t+s)-X(t)) =0? Surely this should be broken down to Cov(X(t),-X(t)) +Cov(X(t),X(t+s)). And the first expression Cov(X(t),-X(t)) = var (X(t)) so the solution to the question should be 2var (X(t)) Tia
Hello The process has independent increments, so the increment \( X_{t+s} - X_{t} \) is independent of \( { X_u, 0 \le u \le t } \). This means that \( X_{t+s} - X_{t} \) is independent of \( X_t \) and so \(cov(X_{t+s}- X_{t}, X_t ) \) is 0. Hope this helps Andy