ARCH(1) process is a Markov process

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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

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.

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