the infinite sum of symmetric random variables is also symmetric Definition. Let $(\Omega, {\mathcal F}, \mathbb{P})$ be a probability space and $X$ a random variable in $\Omega$. $X$ is said to be ${\mathbf symmetric}$ (about $0$) if $X$ and $-X$ are equal in law.That is, $P(X\in B)=P(-X\in B) $ for any Borel set $B$
My question is, if $X_1,X_2,\dots X_n,\dots $ are symmetric random variables, then the finite sum $$S_n=\sum_1^n X_i$$ is also symmetric. Furthermore, if $S_n$ converges to a random variable $S$, then $S$, as an infinite sum of symmetric random variables, is again symmetric.
I know nothing other than definition about symmetric random variables and have no idea how to deal with such a problem, especially the infinite sum case. Do we need $X_1,X_2,\dots X_n,\dots $ to be independent symmetric random variables?
 A: Assume the $X_n$ are independent and symmetric.
Let $\Phi=\mathbb{R}^{\mathbb{N}}$. This is the space of all sequences $(a_n)_{n \in \mathbb{N}}$ of real numbers (in other words, the space of all functions $\mathbb{N} \to \mathbb{R}$). Let $\mathcal{H}$ denote the $\sigma$-algebra on $\Phi$ which is generated by all of the natural projection maps $\pi_n:\Phi \to \mathbb{R}$ which are defined by $\pi_n(a)=a_n$.
Let $X:\Omega \to \Phi$ be the random variable defined by $X(\omega)=(X_n(\omega))_{n\in \mathbb{N}}=(X_1(\omega),X_2(\omega),\dots).$ Note that $X$ is $(\mathcal{F},\mathcal{H})$-measurable (since $\pi_n \circ X=X_n$ is Borel measurable for all $n$).
Let $-X$ denote the random variable defined by $(-X)(\omega)=(-X_n(\omega))_{n \in \mathbb{N}}$.
I claim that $X$ and $-X$ have the same law. In other words, $P(X \in C)=P(-X \in C)$ for all $C \in \mathcal{H}$. To prove this, note that $\mathcal{H}$ is generated by the $\pi$-system consisting of events of the form $\bigcap_{i=1}^n \{ \pi_i \in B_i \}$ where the $B_i$ are Borel sets and $n \in \mathbb{N}$, so it suffices to prove the claim when $C$ has this form. But this is true precisely because the $X_i$ are independent and symmetric (you can check this; this is where independence is needed).
Now suppose that $f: \Phi \to \mathbb{R}$ is any mapping which satisfies the following two properties: $f$ is $(\mathcal{H},\mathcal{B}_{\mathbb{R}})$-measurable, and $f(-X)=-f(X)$. Then I claim that $f(X)$ is a symmetric random variable. This is because if $B \subset \mathbb{R}$ is any Borel set, then $$P(f(X) \in B) = P(X \in f^{-1}(B)) = P(-X \in f^{-1}(B)) = P(f(-X) \in B) = P(-f(X) \in B)$$In particular, define $f_n=\sum_{i=1}^n \pi_i$. Then $f_n$ clearly satisfies the two properties stated above, and so $f_n(X)$ is a symmetric random variable for all $n$. But $f_n(X)=\sum_1^n X_i=S_n$, which proves that $S_n$ is symmetric. This proves the finite case.
For the infinite case, define $f=\limsup_n f_n$. If the $S_n$ converge to a random variable $S$, then $f$ satisfies the two properties from above, and $S=f(X)$. Thus $S$ is symmetric.
Note #1: Instead of assuming that $X_n$ is symmetric for all $n$, suppose we assumed the stronger hypothesis that the random vectors $(X_1, ...,X_n)$ are symmetric for all $n$. In other words: $$ P(X_1\in B_1, ...,X_n \in B_n) = P(-X_1 \in B_1, ... , -X_n \in B_n) $$ for all Borel sets $B_i$ and all $n \in \mathbb{N}$. Then independence is not needed in the above proof.
Note #2: I feel as though it is somewhat undesirable to work in the infinite product space $\mathbb{R}^{\mathbb{N}}$. I feel like there should be a more elementary proof of this which uses simpler methods. Does anyone know of one?
