Given random variables $X_1$,$X_2$,...,$X_n$ i.i.d, $Z$ is random variable which is independent of $X_1$,$X_2$,...,$X_n$, Let $Y_i = X_i + Z, i = 1,2,...,n$, are $Y_1, Y_2, ... , Y_n$ exchangable(I guess so)?If so, how to show it?If not, what about $Z$ is also iid with $X_1$,$X_2$,...,$X_n$?


One way is to use characteristic functions. We have $Ee^{i\sum_j t_jY_j}=\prod_jEe^{it_jX_j} Ee^{i \sum_j t_jZ}$ which clearly doesn't change under any permutation of $t_j$'s. It is not necessary to assume that $Z$ has the same distribution as $X_i$'s.


Let $X,Z\colon (\Omega,\Sigma,\mathbb P)\to \mathbb R$ be two random variables and $X_1,\dots,X_n\stackrel{\rm i.i.d.}{\sim} X$ be i.i.d. random variables that are independent of $Z$. For $A\in\mathcal B(\mathbb R)$, let $A-z := \{a-z|a\in A\}$ denote the shift of the set $A$ by $z\in\mathbb R$.

It is easier to first think of $Z$ as being a discrete random variable. Then for $A_1,\dots,A_n\in\mathcal B(\mathbb R)$ we obtain (by the law of total probability and the independence of all random variables: \begin{align*} \mathbb P\big( Y_1\in A_1,\dots, Y_n\in A_n\big) &= \sum_z \mathbb P(Z=z) \mathbb P\big( X_1\in A_1-z,\dots, X_n\in A_n-z\big) \\ &= \sum_z \mathbb P(Z=z) \mathbb P\big( X\in (A_1-z)\big)\cdots \mathbb P\big( X\in (A_n-z)\big). \end{align*} Since $X_1,\dots,X_n$ are i.i.d., you obtain the same result for $\mathbb P\big( Y_{\sigma(1)}\in A_1,\dots, Y_{\sigma(n)}\in A_n\big)$, where $\sigma$ is some permutation of $\{1,\dots,n\}$.

If $Z$ is not discrete, we have to use integrals instead of sums: \begin{align*} \mathbb P\big( Y_1\in A_1,\dots, Y_n\in A_n\big) &= \int_\mathbb R \mathbb P\big( X_1\in A_1-z,\dots, X_n\in A_n-z\big) \, \mathrm d\mathbb P_Z(z) \\ &= \int_\mathbb R \mathbb P\big( X\in (A_1-z)\big)\cdots \mathbb P\big( X\in (A_n-z)\big) \, \mathrm d\mathbb P_Z(z), \end{align*} where $\mathbb P_Z$ is the distribution of $Z$.


  1. It is not necessary for $X$ and $Z$ to map into $\mathbb R$. It can be any space you like. In order to make your question well-posed, you have to be able to perform the addition $X+Z$, though, so probably they map into the same vector space $V$.
  2. We used the fact that the product $\sigma$-algebra is generated by the sets of the form $A_1\times\cdots\times A_n$.

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