I am struggling with trying to show that a martingale with bounded differences converges. The issue lies in the fact that there is double randomness involved and I am not sure whether or not I can apply Azuma/McDiarmid's to the problem. Let me state it formally: assume we have two sequences of random variables $X_1,\ldots,X_n, Z_1,\ldots,Z_n$ each of which is a sequence of iid random variables and that are independent among themselves. ($(X_1,\ldots,X_n)$ is independent of $(Z_1,\ldots,Z_n)$). Assume also I have a function $f(U,X^n,Z^n)$ that has bounded differences with respect to $X^n $and $Z^n$. Meaning that, for every $u$ and every $(x_1,\ldots,x_i,\,x_n,z_1,\ldots,z_n)$ and $(x_1,\ldots,x'_i,\,x_n,z_1,\ldots,z_n)$, $$ |f(u,x_1,\ldots,x_i,\,x_n,z_1,\ldots,z_n) - f(u,x_1,\ldots,x'_i,\,x_n,z_1,\ldots,z_n)|\leq 1/n.$$
The object that I would like to bound (exponentially in n) is the following: $$ \mathbb{E}_{Z^n}[ \max_u P_{X^n}(|f(u,X^n,Z^n) - \mathbb{E}_{\hat{Z}^n}\mathbb{E}_{\hat{X}^n}[f(u,\hat{X}^n,\hat{Z}^n)]| \geq \epsilon) ].$$
My idea was the following. If I consider the following filtration: $\mathcal{F}_0=\sigma(\emptyset) \subseteq \mathcal{F}_1 =\sigma(X_1,Z^n) \subseteq \ldots \subseteq \sigma(X^n,Z^n)=\mathcal{F}_n,$ then $M_0 = \mathbb{E}[f(X^n,Z^n)|\mathcal{F}_0] = \mathbb{E}_{\hat{Z}^n}\mathbb{E}_{\hat{X}^n} [f(\hat{X}^n,\hat{Z}^n)]$ and
$ M_n = \mathbb{E}[f(X^n,Z^n)|\mathcal{F}_0] = f(X^n,Z^n).$
Now I have that for each $i$ $M_i - M_{i-1} = D_i = f_i(X_1,\ldots,X_i, Z^n)$ a.s. where $f_i(x_1,\ldots,x_i, z^n)= \mathbb{E}[f(x_1,\ldots,x_i,X_{i+1},\ldots,X_n, z^n) - f(x_1,\ldots,X_i,X_{i+1},\ldots,X_n, z^n)] \leq 1/n$ because of my assumption on $f$.
Can I then conclude, fixing $Z^n$ and $u$ that the inner terms decays exponentially with $n$? Or does this argument only work if I consider $P_{Z^nX^n}(|f(u,X^n,Z^n) - \mathbb{E}_{\hat{Z}^n}\mathbb{E}_{\hat{X}^n}[f(u,\hat{X}^n,\hat{Z}^n)]| \geq \epsilon)$? In some sense, the Martingale is constructed in such a way that I only gain knowledge about the process $X$ (and the increments are indeed due to $X$)and $Z^n$ is considered to be random for the whole reasoning. I apologise if the question comes across as a bit confused, I am confused myself by this problem! Many thanks for any input you might have on this.