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 inmathbb 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_0alpha_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^kprod_{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[gleft(Z_t sqrt{sum_{k = 0}^{t-s-1} alpha_0 alpha_1^kprod_{i=1}^k Z_{t-i}^2 + alpha_1^{t-s}prod_{i=1}^{t-s} Z_{t-i}^2 X_{s}^2} ight) Big| mathcal{F}_s ight] \ = Bbb{E}left[gleft(Z_t sqrt{sum_{k = 0}^{t-s-1} alpha_0 alpha_1^kprod_{i=1}^k Z_{t-i}^2 + alpha_1^{t-s}prod_{i=1}^{t-s} Z_{t-i}^2 x^2} ight) ight]_{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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