Random variables having the same distribition We consider a symmetric function $f:\mathbb{R}^d \to \mathbb{R}$ and a sequence $(X_n)_n$ of i.i.d random variables.
For all $n \geq d,$ let $U_n:=\sum_{1\leq i_1<...<i_d\leq n}f(X_{i_1},...,X_{i_d}).$
Fix $n \geq d,k\in \mathbb{N}$ and $1\leq q_1<...<q_d \leq n.$
How to prove-explain that $(f(X_{q_1},...,X_{q_d}),U_n,...,U_{n+k})$ and $(f(X_{1},...,X_{d}),U_n,...,U_{n+k})$ have the same distribution?
 A: The short answer is that permuting the indices of $X$ according to $q$ doesn't change the distribution of the sequence and doesn't change the values of $U_n, \dotsc, U_{n+k}$ at all.
This means that each $f(X_{q_1}, \dotsc, X_{q_d})$ makes the same contribution to the distribution of $(U_n, \dotsc, U_{n+k})$, regardless of $q$.

To see this formally, let $s \colon \mathbf{Z}_{>0} \rightarrow \mathbf{Z}_{>0}$ be a bijection such that

*

*$s(i) = q_i$ for $1 \leq i \leq d$,

*$s$ is a permutation on $\{1, \dotsc, n\}$, and

*$s(i) = i$ for $i > n$.

For your sequence $X$, define a sequence $X'$ by $X'_n = X_{s(n)}$ so that $X'_i = X_{q_i}$ for $1\leq i \leq d$.
Since the terms of $X$ are iid, we have that $X' \stackrel{\mathrm{d}}{=} X$.
Now, notice that, for any $m \geq n$, we have that
$$
U_m(X')
= \sum_{1 \leq i_i < \dotsb < i_d \leq m} f(X'_{i_1}, \dotsc, X'_{i_d})
= \sum_{1 \leq i_i < \dotsb < i_d \leq m} f(X_{i_1}, \dotsc, X_{i_d})
= U_m(X),
$$
since the terms $f(X'_{i_1}, \dotsc, X'_{i_d})$ are just re-ordered versions of $f(X_{i_1}, \dotsc, X_{i_d})$, so each appears the same number of times in the two sums.
(Note that this isn't true if $m < n$, since if $s(i) = n$, then terms containing $X'_i = X_n$ would appear in the left-hand but not the right-hand sum.)
But now, we use the fact that $X' \stackrel{\mathrm{d}}{=} X$ to conclude
\begin{align*}
(f(X_{q_1}, \dotsc, X_{q_d}), U_n(X), \dotsc, U_{n+k}(X))
&= (f(X'_1, \dotsc, X'_d), U_n(X'), \dotsc, U_{n+k}(X')) \\
&\stackrel{\mathrm{d}}{=} (f(X_1, \dotsc, X_d), U_n(X), \dotsc, U_{n+k}(X)).
\end{align*}
