Give an example of a process in the 3-period model that is a Martingale process but is not a Markov process. Prove both assertions.

To address your question, let's first provide a brief definition to understand our terms clearly.

• Martingale Process: A stochastic process $X_n$ is a martingale with respect to another stochastic process $Y_n$ if for all $n$, $E[X_{n+1} | Y_1, Y_2, \ldots, Y_n] = X_n$.
• Markov Process: A stochastic process $X_n$ is a Markov process if the future state is independent of the past states given the present state, meaning for all $n$: $P(X_{n+1} | X_n, X_{n-1}, \ldots) = P(X_{n+1} | X_n)$.

Example of a 3-Period Model

1. Define the Process: Let's consider a process $X_n$ where $n = 0, 1, 2, 3$. Define the outcomes as follows:

   • At time 0, $X_0 = 0$.
   • At time 1, $X_1$ can be either $1$ or $-1$ with equal probability, i.e., $P(X_1 = 1) = P(X_1 = -1) = \frac{1}{2}$.
   • At time 2, $X_2$ can be either $2$, $0$, or $-2$ with the following probabilities:
     • $P(X_2 = 2 | X_1 = 1) = 1$
     • $P(X_2 = 0 | X_1 = 1) = 0$
     • $P(X_2 = -2 | X_1 = -1) = 1$
   • At time 3, let $X_3$ always equal 1, irrespective of $X_2$.

2. Verify Martingale Property: To prove that $X_n$ is a martingale:

   • Calculate $E[X_1 | X_0] = X_0 = 0$: $E[X_1 | X_0] = 1 \cdot P(X_1=1 | X_0) + (-1) \cdot P(X_1=-1 | X_0) = 1 \cdot \frac{1}{2} + (-1) \cdot \frac{1}{2} = 0$

   • Next, calculate $E[X_2 | X_1]$:

     • If $X_1 = 1$, then $X_2 = 2$ with probability 1: $E[X_2 | X_1 = 1] = 2$
     • If $X_1 = -1$, then $X_2 = -2$ with probability 1: $E[X_2 | X_1 = -1] = -2$
     • However, notice $E[X_2] = 0$ because each outcome is equally probable.

   • Finally, calculate $E[X_3 | X_2] = 1$ which does not depend on the value of $X_2$.

3. Check Non-Markov Property: For it to be a Markov process, the future state must depend only on the present state.

   Consider $X_3$: $P(X_3 = 1 | X_2 = 2) = 1 \quad \text{and} \quad P(X_3 = 1 | X_2 = -2) = 1$

   However, the transition from $X_1$ to $X_2$ also depends on $X_{0}$. Thus, the transition from $X_1$ to $X_2$ depends on the entire history (not only on $X_n$), hence, $X_n$ is not a Markov process.

Final Answer

The process $X_n$ defined above is a martingale because it satisfies the definition $E[X_{n+1} | X_n] = X_n$, but it is not a Markov process due to its dependence on the past states in determining $X_3$.