The product of 2 independent markov chains Let $(X_n)_{n≥0}$ and $(Y_n)_{n≥0}$ be two independent Discrete-time Markov chains $\in Z$.
a) Define the process $Z_n = (X_n, Y_n)$ $\in Z^2$. Is $(Z_n)_{n≥0}$ a Markov chain?
b) Define the process $W_n = X_nY_n$ $\in Z$. Is $(W_n)_{n≥0}$ a Markov chain?
For a) I've got $$ P[Z_n=k | Z_{n-1}=z_{n-1}, Z_{n-2}=z_{n-2},...,Z_0=z_0] = P[Xn=k_x, Yn=k_y | X_{n-1}=z_{n-1_{x}}, Y_{n-1}=z_{n-1_{y}},..., X_{0}=z_{0_{x}}, Y_{0}=z_{0_{y}}] = \text{(due to the independence of {Xi} and {Yj} and Markov property for Xn and Yn}) = P[X_n=k_x | X_{n-1}=z_{n-1_{x}},Y_{n-1}=z_{n-1_{y}}]P[Y_n=k_y | Y_{n-1}=z_{n-1_{y}}, X_{n-1}=z_{n-1_{x}}] = P[Z_n=k|Z_{n-1}=z_{n-1}] $$ - so $(Z_n)$ is a Markov chain. Is my reasoning correct?
For b) I have no idea.

For a, your answer looks fine.
For b, you need to think about possible factorization from the primes of XY.
Let's say XY = 12, for all pairs of X and Y whose product is 12, namely, 1 and 12, 2 and 6, 3 and 4, the process is still memoryless. This is the result in part a, stating all possible pairing retains memorylessness.
