Let $X$ be a random variable such that $E[|X| < \infty$, and let $\{Z_t: :t = 0,1,\ldots\}$ be a random sequence. We define the random sequence $\{X_t: t = 0,1,\ldots\}$ by $X_t = E[X\mid Z_0, Z_1, \ldots, Z_t\}$.
We must show that $\{Xt : t = 0,1,\ldots\}$ is a martingale.
I have already shown that $E[|X_t|] < \infty$, and it is obvious that $X_t$ is determined by the value of $(Z_0, Z_1, \ldots, Z_t).$ So I already miss the 3rd condition for martingality : I need to show that $X_{t-1} = E[X_t \mid Z_0, Z_1, \ldots, Z_{t-1}]$.
But I can't see the link between $X$ and $Z$, so I can't see how does the sequence $\{Z_0, Z_1, \ldots, Z_t\}$ impact the value of $X$. But if it doesn't impact it, then $E[X\mid Z_0, Z_1, \ldots, Z_t]$ would be equal to $E[X]$ and $X_t$ wouldn't be a martingale with respect to $Z_t$. Does anyone have an idea ? Thanks.
The tower property of conditional expectation gives that if $\mathcal{F}$ and $\mathcal{G}$ are two sigma algebras such that $\mathcal{F}\subseteq\mathcal{G}$, then $E(X|\mathcal{F})=E(E(X|\mathcal{G})|\mathcal{F})$ (note that conditional expectation is a random variable and not necessarily a fixed real number like expectation).
Okay, if the above does not make total sense to you now, consider a following non-rigorous version. Say there are variates $Y,Z$. Then $E(E(X|Y=y,Z=z)|Y=y)=E(X|Y=y)$. Intuitively, conditional expectation is actually a projection (in a Hilbert space). So the heuristic explanation is that projecting a 3D vector to a plane and then to a line inside the plane, is the same as directly projecting directly to the line.
So here, $Y=y,Z=z$ contains more information than just $Y=y$. So if you are asked to compute $E(E(X|Y,Z)|Y)$ i.e. first conditioned on $\{Y=y,Z=z\}$(the plane contaning more information) then it will just be $E(X|Y)$ as $Y=y$(the line containing less information).
Armed with this,
$E(X_{t}|Z_{0},...,Z_{t-1})=E(E(X|Z_{0},...,Z_{t})|Z_{0},...,Z_{t-1})=E(X|Z_{0},...,Z_{t-1})=X_{t-1}$ as required due to the above heuristic that I gave.
Formally, you should set $\mathcal{F}=\sigma(Z_{0},...,Z_{t-1})$ and $\mathcal{G}=\sigma(Z_{0},...,Z_{t})$ to see that obviously $\mathcal{F}\subseteq\mathcal{G}$