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Suppose: $u_t \sim N(0,1) \ iid.$, $X_t = g(X_{t-1}) \cdot u_t$ whereas $g(X)$ can be any deterministic function. Is this sufficient to define a martingale?

So does it hold: $E(X_t|X_{t-1}, \ldots , X_1) = E(g(X_{t-1}) \cdot u_t|X_{t-1}, \ldots , X_1) =^{\ pi.} E(g(X_{t-1})|X_{t-1}, \ldots , X_1) \cdot E(u_t|X_{t-1}, \ldots , X_1) = E(g(X_{t-1})|X_{t-1}, \ldots , X_1) \cdot 0 = 0$

It seems pretty obvious but I think I missed some assumptions for $g(X)$ or am I totally wrong?

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    $\begingroup$ No you didn't. It would be a martingale even if $g(x)$ was random but the $u$'s were independent from past $X$'s. $\endgroup$ – Alecos Papadopoulos Jan 10 '14 at 21:57
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Indeed, $(X_t)_{t\geqslant 0}$ is a martingale for the natural filtration if we assume that $g\colon\mathbb R\to\mathbb R$ is measurable. Since $g(X_{t-1})$ is measurable with respect to the $\sigma$-algebra generated by $X_{t-1},\dots,X_1$, we obtain $$\mathbb E[X_t\mid\sigma(X_i,1\leqslant i\leqslant t-1)]=g(X_{t-1})\cdot\mathbb E[u_t\mid \sigma(X_i,1\leqslant i\leqslant t-1)].$$

If we can show that $u_t$ is independent of $\sigma(X_i,1\leqslant i\leqslant t-1)$ we will be done. But this is a consequence of the fact that $X_s$ is measurable with respect to $\sigma(u_k,1\leqslant k\leqslant s)$ (and this can be shown by induction on $s$).

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