Hi all, Im struggling with this question, i understand this part P(e_n≥e_(n-2)∩e_(n-4)≤-e_(n-2) )=0.25 is required. However cannot figure out at all what happens next in the solutions. Please can someone explain? Thanks!
Hi Molly This is certainly a challenging question, and quite unusual so don't get bogged down in this one. Starting from where you got to, we have shown that to have the Markov property, the following equation must hold: \( P(X_n \ge 0 | X_{n-1} \le 0) = P(X_n \ge 0 | X_{n-1} \le 0, X_{n-2} \le 0) \) And we have shown this is equivalent to: \( P(e_n + e_{n-2} \ge 0 \cap e_{n-2} + e_{n-4} \le 0) = 0.25 \) Or alternatively: \( P(e_n \ge -e_{n-2} \cap e_{n-4} \le -e_{n-2}) = 0.25 \) \(e_{n-2}\) is an observation from a standard normal distribution so can take any value. Let's call this value \(\epsilon\). (We could alternatively call this value \( -\epsilon \) which would remove all the minus signs below, but let's stick with what the solutions say.) Then, since \(e_n\) and \(e_{n-4}\) are also observations from a standard normal, we can write (using the Phi function): \( P(e_n \ge -\epsilon) = 1 - \Phi(-\epsilon) \) and \( P(e_{n-4} \le -\epsilon) = \Phi(-\epsilon) \) Putting these probabilities together gives us: \( (1 - \Phi(-\epsilon))\Phi(-\epsilon) = 0.25 \) This is straightforward to solve to find that \( \Phi(-\epsilon) = 0.5 \). This means that \( e_{n-2} \) is at the median of the standard normal, or in other words \( e_{n-2} = 0 \). Therefore the equation \( P(X_n \ge 0 | X_{n-1} \le 0) = P(X_n \ge 0 | X_{n-1} \le 0, X_{n-2} \le 0) \) only holds when \(e_{n-2} = 0 \). Since it doesn't hold in general, this process cannot be Markov. Does that help? Dave
