ARCH(1) process is a Markov process

Question

I have a question about the ARCH(1) process. Let $(\Omega, \mathcal F, P)$ be a probability space, let $(Z_t)_{t \in \mathbb Z}$ be a sequence of i.i.d. real-valued random variables with mean zero and variance one. A stochastic process $(X_t)_{t \in \mathbb Z}$ is an ARCH(1)-process if it is strictly stationary and if, for all $t$ and some process $(\sigma_t)$ with $\sigma_t > 0$ for every $t$, one has $X_t = \sigma_t Z_t$ and $\sigma_t^2 = \alpha_0 + \alpha_1 X_{t-1}^2$, where $\alpha_0 > 0$ and $\alpha_1 \ge 0$. Let $\mathcal F_t$ be the natural filtration of the process $(X_t)$, i.e. $\mathcal F_t = \sigma(X_s; s \le t)$.

I want to prove that $(X_t)$ has the Markov property, i.e. for each $B \in \mathcal B(\mathbb R)$ and for all $s, t \in\mathbb Z$ with $s < t$, one has $P[X_t \in B \mid \mathcal F_s] = P[X_t \in B \mid X_s]$. Is that possible?

Answer 1

For s < t we can write:

$$X_t = Z_t \sqrt{\alpha_0 + \alpha_1X_{t-1}^2} = Z_t \sqrt{\alpha_0 + \alpha_0\alpha_1 Z_{t-1}^2 + \alpha_1^2Z_{t-1}^2X_{t-1}^2}\\ = \ldots = Z_t \sqrt{\sum_{k = 0}^{t-s-1} \alpha_0 \alpha_1^k\prod_{i=1}^k Z_{t-i}^2 + \alpha_1^{t-s}\prod_{i=1}^{t-s} Z_{t-i}^2 X_{s}^2}$$ By construction $Z_t,\ldots, Z_{s+1}$ is independent to $\mathcal{F}_s$, so for a measureable function $g \geq 0$ it holds:

$$\Bbb{E}[g(X_t) | \mathcal{F}_s ] = \Bbb{E}\left[g\left(Z_t \sqrt{\sum_{k = 0}^{t-s-1} \alpha_0 \alpha_1^k\prod_{i=1}^k Z_{t-i}^2 + \alpha_1^{t-s}\prod_{i=1}^{t-s} Z_{t-i}^2 X_{s}^2} \right) \Big| \mathcal{F}_s \right] \\ = \Bbb{E}\left[g\left(Z_t \sqrt{\sum_{k = 0}^{t-s-1} \alpha_0 \alpha_1^k\prod_{i=1}^k Z_{t-i}^2 + \alpha_1^{t-s}\prod_{i=1}^{t-s} Z_{t-i}^2 x^2} \right) \right]_{x = X_s} = \Bbb{E}[g(X_t) | X_s ]$$

Here the notation $x = X_s$ denotes that we don't integrate over $X_s$ anymore, but set the value of $x$ in the expection by evaluating $X_s$ first. In the last step we just go back the steps before but with $\sigma$-Algebra $X_s$. This shows $\Bbb{P}[X_t \in B | \mathcal{F}_s] = \Bbb{P}[X_t \in B | X_s]$, but it is not the full Markov property. For the Markov property you need additionally $\Bbb{P}[X_t \in B | X_s] = \Bbb{P}[X_{t-s} \in B | X_0 = x]_{x= X_s}$, but this follow from the calculations above, because the $Z_t$ are i.i.d.