Suppose $f: [0,1]^{k+1} \to [0,1]$ and let $(a_0, \dots, a_k, b)$ be a tuple of $[0,1]$-valued random variables. Suppose for each $0 \leq i \leq k$ there is a collection of tuples $$ A_i = \{(a_0^n,\dots, a_{i-1}^n, a_i, a_{i+1}^n, \dots, a_{k}^n, b^n): n \in \omega\} $$ i.e. tuples which agree only on the $i$th coordinate, such that

- $\bigcup_{ i \leq k} A_i$ consists of i.i.d. random variables. In particular this implies $(a_0^n,\dots, a_{i-1}^n, a_i, a_{i+1}^n, \dots, a_{k}^n, b^n)$ and $(a_0, \dots, a_{k}, b)$ have the same distribution for all $0 \leq i \leq k$ and $n \in \omega$

Now I have two questions in decreasing level of generality

(1) Is it the case that $f(a_0, \dots, a_k, b)$ is independent from $(a_0, \dots, a_k)$ over $\{f(a^*_0, \dots, a^*_k, b^*) : (a^*_0, \dots, a^*_k,b^*)\in \bigcup_{i \leq k}A_i\}$?

Note we are not asking for independence over $b$ as we have not assumed we can find corresponding tuples which fix $b$.

(2) Is (1) the case if we assume $\{f(a^*_0, \dots, a^*_k, b^*) : (a^*_0, \dots, a^*_k,b^*)\in \bigcup_{i \leq k}A_i\}$ are independent? Note by Kolmogorov 0-1 law this reduces to showing $f(a_0, \dots, a_k, b)$ is independent from $(a_0, \dots, a_k, b)$.

It might be worth noting, by this question, Does independence of the sequence $f(A_i, B)$ imply the sequence is independent of $B$?, for any $0 \leq i \leq k$ we have $f(a_0, \dots, a_k, b)$ is independent from $a_i$.