# If $Z_1,Z_2,…$ are i.i.d., $X_0$ is independent of $Z_1,Z_2,…$ and $X_n=φ(X_{n-1},Z_n)$, then $X_0,…,X_{n-1}$ is independent of $φ(x,Z_n)$

Let

• $(\Omega,\mathcal A),(D,\mathcal D)$ and $(E,\mathcal E)$ be measurable spaces
• $Z_1,Z_2,\ldots:\Omega\to D$ be independent and identically distributed random variables
• $X_0:\Omega\to E$ be a random variable independent of $Z_1,Z_2,\ldots$
• $\varphi:E\times D\to E$ be $(\mathcal E\otimes\mathcal D,\mathcal E)$-measurable and $$X_n:=\varphi(X_{n-1},Z_n)\;\;\;\text{for }n\in\mathbb N$$

Fix $x\in E$. How can we show that $X_0,\ldots,X_{n-1}$ is independent of $\varphi(x,Z_n)$ for all $n\in\mathbb N$?

Clearly, since $X_0$ is independent of $Z_1$, $X_0$ is independent of $\varphi(x,Z_1)$. This follows from the fat that if $X,Y$ are independent and $f,g$ are measurable, then $f\circ X,g\circ Y$ are independent.

However, I don't see how I need to proceed for $n>1$. The problem is that $X$ being independent of $Y,Z$ doesn't imply independence of $X$ and $(Y,Z)$.

Each of the variables $X_0,X_1,...,X_{n-1}$ is measurable w.r..t $\sigma \{X_0,Z_1,...,Z_{n-1}\}$ and $Z_n$ is independent of this sigma algebra. Hence $\phi (x,Z_n)$ is independent of this sigma algebra which makes it independent of $X_0,X_1,...,X_{n-1}$.