Consider the N-period binomial model and processes Y0,Y1,...YN and C0,C1,...CN. Suppose that Y is Markov and C is deterministic (i.e., Cn does not depend on the coin flips ù).
a) Show: If Cn doesn't equal 0 for all n, then C0Y0,C1Y1,....,CNYN is Markov.
b) Give an example of processes C and Y such that at least one Cn is zero and C0Y0,C1Y1,....,CNYN is not Markov.
To answer these questions, let's start with the definition of a Markov process. A stochastic process is said to be Markov if the conditional distribution of its future states depends only on its current state and is independent of its past states.
a) To show that C0Y0, C1Y1, ..., CNYN is Markov when Cn ≠ 0 for all n, we need to show that the conditional distribution of its future states depends only on its current state and is independent of its past states.
Since C is deterministic and does not depend on the coin flips, the values of Cn are predetermined for each n. This means that the future values of Cn, namely Cn+1, Cn+2, ..., CN, are fixed and do not change regardless of the realization of Y0, Y1, ..., Yn.
On the other hand, the process Y is Markov, which means that the conditional distribution of Yn+1 given Yn depends only on Yn.
Now, when we consider the joint process C0Y0, C1Y1, ..., CNYN, we can observe that the future values of Cn+1, Cn+2, ..., CN are predetermined by C, and the future values of Yn+1, Yn+2, ..., YN are determined solely based on Yn.
Therefore, the joint process C0Y0, C1Y1, ..., CNYN follows the Markov property, where the conditional distribution of its future states depends only on its current state, which is CNYN.
b) To provide an example where at least one Cn is zero and the joint process C0Y0, C1Y1, ..., CNYN is not Markov, let's consider the following scenario:
Let N = 2, and the processes C and Y be defined as follows:
C0 = 1, C1 = 0, C2 = 2
Y0 = 1, Y1 = 2, Y2 = 3
In this example, C1 is zero, and Cn ≠ 0 for all other n.
Now let's examine the joint process C0Y0, C1Y1, C2Y2:
C0Y0 = 1 * 1 = 1
C1Y1 = 0 * 2 = 0
C2Y2 = 2 * 3 = 6
We can observe that the joint process C0Y0, C1Y1, C2Y2 is not a Markov process. This is because, in this example, the value of C1 is zero, and the conditional distribution of C2Y2 depends on the past value of C1Y1, which is zero. Hence, the joint process does not satisfy the Markov property.