Let $\mathcal G_n=\sigma(S_n,S_{n+1},\ldots)$. Since $X_{n+j}=S_{n+j}-S_{n+j-1}$,
$$
\mathcal G_n=\sigma(S_n,X_{n+1},X_{n+2},\ldots).
$$
Permuting $X_1,\ldots,X_n$ preserves their <joint probability distribution> and leaves every generator of $\mathcal G_n$ unchanged. Therefore, for each bounded $\mathcal G_n$-measurable $H$,
$$
\mathbb E[X_1H]=\mathbb E[X_jH]\qquad(1\leq j\leq n).
$$
This is the symmetry of <exchangeable random variables>. By the defining identity of <conditional expectation>, all $\mathbb E[X_j\mid\mathcal G_n]$ are equal <almost surely>. Summing these <conditional expectations> and using that $S_n$ is $\mathcal G_n$-measurable gives
$$
n\mathbb E[X_1\mid\mathcal G_n]=\mathbb E[S_n\mid\mathcal G_n]=S_n.
$$
Hence the <conditional expectation of a summand given future partial sums> is
$$
\boxed{\mathbb E[X_1\mid S_n,S_{n+1},\ldots]=\frac{S_n}{n}\quad\text{almost surely}.}
$$
The <integrable random variable>[integrability] of each $X_j$ justifies every <conditional expectation> above.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 201 / 3 / b / Solution

For <independent and identically distributed random variables> $X_j$ that are <integrable random variables>, with $S_n=\sum_{j=1}^nX_j$, symmetry gives $\mathbb E[X_1\mid S_n,S_{n+1},\ldots]=S_n/n$. The decreasing <sigma-algebras> generated by these future sums permit application of the <reverse martingale convergence theorem>.

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