Show equality of joint distribution Let $X_i$ be i.i.d. real random variables and consider $Z_n := \sum_{i=1}^n X_i$. I need to show
$$\mathbb{P}^{(X_1, Z_n)} = \mathbb{P}^{(X_i, Z_n)}$$
for all $i \in \mathbb{N}$. My current approach looks like this: Let $B_1, B_2 \in \mathcal{B}_\mathbb{R}$ be Borel sets. Then we get
$$\mathbb{P}^{(X_1, Z_n)}(B_1 \times B_2) = \mathbb{P}((X_1, Z_n)^{-1}(B_1 \times B_2)) = \mathbb{P}(X_1^{-1}(B_1) \cap Z_n^{-1}(B_2)) = \,\,\stackrel{?}{\dots}\,\, = \mathbb{P}^{(X_i, Z_i)}(B_1 \times B_2)$$
Then I could use measure completion to get the equality. Sadly I'm stuck at the $\stackrel{?}{\dots}$ part. What clue I missing here?
 A: Fact 1
$P^X=P^{X'}\Rightarrow P^{f(X)}=P^{f(X')}$ for any random vectors $X,X'$ and any measurable function $f$ (possibly vector-valued).
This is immediate from the fact that $P^X=P^{X'}\Leftrightarrow Eg(X)=Eg(X')$ for any measurable $g$
Fact 2
let $\sigma$ be any permutation of the indices $1,\dots,n$. Then $P^{(X_1,\dots, X_n)}=P^{(X_{\sigma(1)},\dots, X_{\sigma(n)})}$.
This is easy to see from the fact that $X_i$ are iid. For example, you can verify that the characteristic functions of the vectors $(X_1,\dots, X_n)$ and $(X_{\sigma(1)},\dots, X_{\sigma(n)})$ coincide.
Now, consider the measurable function $(x_1,\dots, x_n)\mapsto (x_1,\sum_i x_i)$. Putting together the two facts, we have $P^{(X_1,Z_n)}=P^{(X_{\sigma(1)},\sum_i X_{\sigma(i)})}$ for any permutation $\sigma$. But $Z_n=\sum_i X_{\sigma(i)}$, so $P^{(X_1,Z_n)}=P^{(X_{\sigma(1)},Z_n)}$, and since $\sigma$ is an arbitrary permutation, $\sigma(1)$ could be any index in $\{1,\dots, n\}$, so we are done.
A: Note that
$$
\mathbb P ^{(X_j,Z_n)}(B_1\times B_2)=\mathbb P^{(X_1,\ldots ,X_n)} \left(\left\{(x_1,\ldots ,x_n)\in \mathbb{R}^n:\sum_{k=1}^n x_k\in B_2\,\land\, x_j\in B_1\right\}\right)
$$
and $\mathbb P ^{(X_1,\ldots ,X_n)}=\prod_{k=1}^n \mathbb P ^{(X_k)}$ because the $X_k$ are independent, and $\mathbb P ^{(X_k)}=\mathbb P ^{(X_j)}$ for all $j,k\in\{1,\ldots,n\}$ (because they have the same distribution) then the measure of $B_1\times B_2$ doesn't depends of the choice of $j$.$\Box$
