Sequence of a product of Bernoulli random variables is a Markov Chain?

Question

Suppose we have a sequence of random variables $X_n$ defined as $X_n = \prod_{i=0}^n Z_i $ where $Z_i$ for i=0,1,2,3... is a sequence of i.i.d variables with the Bernoulli distribution with parameter p. I wanted to ask if $X_n$ is a Markov chain? I had concerns over the fact that knowing if one of the $X_n = 0$ would break the required independence?

Answer 1

Assume $X_0=0$. For all $n\geqslant0$ we have $$X_{n+1} = Z_{n+1}\prod_{i=1}^{n+1} Z_i = Z_{n+1}X_n.$$ Since $X_n$ is $\sigma(X_n)$-measurable and $Z_{n+1}$ is indepenent of $\{X_i: 1\leqslant i\leqslant n\}$, it follows that for any continuous bounded $f:\{0,1\}\to\mathbb R$, $$ \mathbb E[f(X_{n+1})\mid\mathcal F_n] = \mathbb E[f(Z_{n+1}X_n)\mid\mathcal F_n] = \mathbb E[f(Z_{n+1}X_n)\mid X_n], $$ so that $\{X_n\}$ is a Markov process. Indeed, the $n$-step transition matrix is given by $$ P^n = \begin{pmatrix}p^n&1-p^n\\0&1\end{pmatrix}. $$