Symmetrization as conditional expectation ($\mathbb{E}[Z\mid \mathcal{I}]=\frac{1}{|\Gamma |}{\sum }_{g\in \Gamma }Z\circ T_g$)

If a finite group $\Gamma$ acts by measure-preserving maps on a probability space, let $\mathcal{I}$ be its invariant sigma-algebra. For an integrable random variable $Z$, the displayed group average is $\mathcal{I}$-measurable. For $A\in \mathcal{I}$, changing variables under each group element gives ${\int }_{A}Z\circ T_{g}d\mathbb{P}={\int }_{A}Zd\mathbb{P}$. The average therefore satisfies the defining integral identity for conditional expectation. For permutations of $N$ exchangeable coordinates, averaging a product attached to $d$ distinct indices yields the sum over all ordered distinct $d$-tuples divided by the falling factorial $(N{)}_d=N!/(N-d)!$.

For arrays of values in $[0,1]$, let $U_N$ average a product of $d$ entries over all ordered index tuples from $N$ coordinates, and let $V_N$ average it over the tuples with distinct indices. Decompose the first average as $U_N=\alpha V_N+(1-\alpha )W_N$, where $\alpha =(N{)}_d/N^d$ and the repeated-index average $W_N$ lies in $[0,1]$. This gives $|U_N-V_N|\le 1-\alpha$. For uniformly sampled indices, a union bound over the $\left(\genfrac{}{}{0ex}{}{d}{2}\right)$ pairs bounds the collision probability by $\left(\genfrac{}{}{0ex}{}{d}{2}\right)/N$. This deterministic estimate is the bridge between empirical-product limits and the De Finetti theorem.