Where did I go wrong? Show $P\{S_n≥x\}=P\{S_n≤-x\}$ if $X_i$ and $-X_i$ have the same distribution The question is

Let $\{X_i,1≤i≤n\}$ be independent discrete RVs. Suppose $X_i$ and $-X_i$ have the same distributions, show that for all $x$ we have $P\{S_n≥x\}=P\{S_n≤-x\}$ where $S_n=∑_{i=1}^nX_i$.

My attempt:
$Let X=(X_1,X_2,…,X_n)$ and $f(x)$ be the joint mass function of $X$. When ${X_i}$ are independent, then $f(x)=f(x_1 )f(x_2 )…f(x_n )$ where $f(x_i )$ is the mass function of $X_i$. Let $Ω_i=\{x∈\Bbb{R}:f(x_i )≠0\}$, and we have

$\begin{gathered}
P\left\{ {{S_n} \geqslant k} \right\} = \mathop \sum \limits_{{x_n} \in {{{\Omega }}_n}} \mathop \sum \limits_{{x_{n - 1}} \in {{{\Omega }}_{n - 1}}}  \ldots \mathop \sum \limits_{{x_2} \in {{{\Omega }}_2}} \mathop \sum \limits_{{x_1} \in \left\{ {x \in {{{\Omega }}_1}:x \geqslant k - \mathop \sum \limits_{i = 2}^n {x_i}} \right\}} f\left( x \right) \hfill \\
 = \mathop \sum \limits_{{x_n} \in {{{\Omega }}_n}} \mathop \sum \limits_{{x_{n - 1}} \in {{{\Omega }}_{n - 1}}}  \ldots \mathop \sum \limits_{{x_2} \in {{{\Omega }}_2}} \mathop \sum \limits_{{x_1} \in \left\{ {x \in {{{\Omega }}_1}:x \geqslant k - \mathop \sum \limits_{i = 2}^n {x_i}} \right\}} f\left( {{x_1}} \right)f\left( {{x_2}} \right) \ldots f\left( {{x_n}} \right) \hfill \\
   = \mathop \sum \limits_{{x_n} \in {{{\Omega }}_n}} f\left( {{x_n}} \right)\mathop \sum \limits_{{x_{n - 1}} \in {{{\Omega }}_{n - 1}}} f\left( {{x_{n - 1}}} \right) \ldots \mathop \sum \limits_{{x_2} \in {{{\Omega }}_2}} f\left( {{x_2}} \right)\mathop \sum \limits_{{x_1} \in \left\{ {x \in {{{\Omega }}_1}:x \geqslant k - \mathop \sum \limits_{i = 2}^n {x_i}} \right\}} f\left( {{x_1}} \right) \hfill \\ 
\end{gathered} $

Similarly,

$P\left\{ {{S_n} \leqslant  - k} \right\} = \mathop \sum \limits_{{x_n} \in {{{\Omega }}_n}} f\left( {{x_n}} \right)\mathop \sum \limits_{{x_{n - 1}} \in {{{\Omega }}_{n - 1}}} f\left( {{x_{n - 1}}} \right) \ldots \mathop \sum \limits_{{x_2} \in {{{\Omega }}_2}} f\left( {{x_2}} \right)\mathop \sum \limits_{{x_1} \in \left\{ {x \in {{{\Omega }}_1}:x \leqslant  - k - \mathop \sum \limits_{i = 2}^n {x_i}} \right\}} f\left( {{x_1}} \right)$

Then I got stuck, because in order for the two series to be equal, we need to show 

$\mathop \sum \limits_{{x_1} \in \left\{ {x \in {{{\Omega }}_1}:x \leqslant  - k - \mathop \sum \limits_{i = 2}^n {x_i}} \right\}} f\left( {{x_1}} \right) = \mathop \sum \limits_{{x_1} \in \left\{ {x \in {{{\Omega }}_1}:x \geqslant k - \mathop \sum \limits_{i = 2}^n {x_i}} \right\}} f\left( {{x_1}} \right)$ ......(*)

The conditions we have is "$X_i$" and $-X_i$ has the same distribution" and thus $P\left\{ {{X_i} \leqslant x} \right\} = P\left\{ {{X_i} \geqslant  - x} \right\}$, from which we can only get $\mathop \sum \limits_{{x_1} \in \left\{ {x \in {{{\Omega }}_1}:x \leqslant  - k - \mathop \sum \limits_{i = 2}^n {x_i}} \right\}} f\left( {{x_1}} \right) = \mathop \sum \limits_{{x_1} \in \left\{ {x \in {{{\Omega }}_1}:x \geqslant k + \mathop \sum \limits_{i = 2}^n {x_i}} \right\}} f\left( {{x_1}} \right)$, which is different from (*).
I think the above approach should work, but where did I go wrong? Or is there any other way to solve the problem? Thank you! 
 A: Here's a potentially helpful shift in perspective: define $S_n'=\sum (-X_i)$. Both $S_n$ and $S_n'$ can be completely described in terms of identical joint distributions of independent variables, so they must have the same distribution. That is, for any $k$:
$$
P(S_n \geq x)=P(S_n' \geq x)
$$

The point here is that given the distributions of the variables $X_i$, we can completely describe the distribution of $S_n$.  So, in particular, let $\Omega_k \subset \Omega$ be the set of outcomes $(x_1,\dots,x_n)$ in which $\sum_{i}x_i \geq k$.  Then, we have
$$
P(S_n \geq k) = 
\sum_{(x_1, \dots, x_n) \in \Omega_k} P(X_1 = x_1)P(X_2 = x_2)\cdots P(X_n = x_n)
$$
Similarly, we have
$$
P(S_n' \geq k) = 
\sum_{(x_1, \dots, x_n) \in \Omega_k} P(-X_1 = x_1)P(-X_2 = x_2)\cdots P(-X_n = x_n)
$$
Because the $X_i$ and $-X_i$ are identically distributed, the outcome of these sums must be identical.  The desired conclusion follows.
It is noteworthy that independence is an essential assumption here; if the $X_i$ are not independent, then we can't deduce the  distribution of $S_n$.
