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For an infinite sequence of real exchangeable random variables, conditional on its invariant sigma-algebra of an exchangeable sequence, the coordinates are independent and identically distributed. In particular, for Borel sets $A_0,\dots ,A_p$,Here is the core proof. Let ${\mathcal{S}}_m$ denote invariance under permutations of coordinates $0,\dots ,m$. Symmetrization as conditional expectation identifies the empirical average of each indicator with its conditional expectation on ${\mathcal{S}}_m$. The reverse martingale convergence theorem gives its limit conditional on $\mathcal{S}$. Multiplying the averages sums over all ordered coordinate tuples, including repetitions. Symmetrization of the product itself sums over the distinct tuples. The fraction of repeated tuples is at most $\left(\genfrac{}{}{0ex}{}{p+1}{2}\right)/(m+1)$, so these bounded averages have the same limit. This proves the displayed factorization. Permutation invariance also makes the individual conditional laws identical. A regular conditional distribution on the real line packages that common law as a random probability measure, giving the equivalent mixture-of-iid formulation.

Let $T_{\pi }$ permute coordinates $0,\dots ,m$, fixing the later coordinates. For any integrable random variable $V$, its group averageis invariant under this group, hence ${\mathcal{S}}_m$-measurable. If $C\in {\mathcal{S}}_m$, invariance of $C$ and measure preservation of each $T_{\pi }$ give $\mathbb{E}[1_{C}V\circ T_{\pi }]=\mathbb{E}[1_{C}V]$. Thus $A_{m}V$ satisfies the integral definition of $\mathbb{E}[V\mid {\mathcal{S}}_m]$. This proves symmetrization as conditional expectation.

Apply it to $V=1_{A_k}(X_k)$ for $m\ge k$. Under a uniform permutation of $m+1$ coordinates, its selected coordinate is uniform among all of them, soThe reverse martingale convergence theorem says that, for decreasing sigma-algebras ${\mathcal{H}}_m$ and an integrable $V$, these conditional expectations converge almost surely and in $L^1$ to $\mathbb{E}[V\mid {\bigcap }_m{\mathcal{H}}_m]$. Here ${\mathcal{S}}_m\downarrow \mathcal{S}$, because every finitely supported permutation belongs to one of the finite groups. ThereforeIntersect these probability-one events for the finitely many $k=0,\dots ,p$ and multiply the limits. Expanding the product of finite sums givesThe sum here includes repeated indices. No independence of the original coordinates was assumed; exchangeability and reverse conditional-expectation convergence are precisely what replaced it.

Put $N=m+1$ and $d=p+1$. Average $V={\prod }_{k=0}^p{1}_{A_k}(X_k)$ over all permutations of the first $N$ coordinates, as justified in part (a). For any prescribed ordered tuple of $d$ distinct images $({\ell }_0,\dots ,{\ell }_p)$, exactly $(N-d)!=(m-p)!$ permutations produce that tuple. ThereforeThe denominator is the falling factorial $(N{)}_d=N!/(N-d)!$, with exactly $p+1$ factors. The sum must be over ordered distinct tuples: its $k$th coordinate is attached to $A_k$, and these sets need not be equal. This is the finite-sample version of symmetrization as conditional expectation.

The original PDF, as well as the TeX, prints the final factor as $m+1-p+1$. That is an off-by-one error: it must be $m+1-p$. With every $A_k=\mathbb{R}$, the conditional expectation is one and the numerator counts exactly $(N{)}_d$ ordered distinct tuples. This forces the normalization above and immediately detects the printed error. Interpreting the sum as unordered subsets would likewise give the wrong normalization and would lose the assignments to the different $A_k$.