Characteristic function of $\sum_{t=1}^N a_t X_t$ given certain independence conditions 
Let $S=\sum_{t=1}^N a_tX_t$ where each $a_t$ is Bernoulli with probability $\frac{1}{2}$ for $1$ and also $\frac{1}{2}$ for $0$. Moreover, it is also given that the vector $(a_1,\ldots,a_N)$ is independent with the vector $(X_1,\ldots,X_N)$; $a_1,\ldots, a_N$ are mutually independent; and $\{X_1,\ldots,X_N\}$ is a random permutation of $\{R_1,\ldots,R_N\}$ (each $R_t$ is a (nonrandom) constant). I'm trying to understand why it is that
  $$
E[\exp(i\tau S)]=\left(\frac{1}{2}\right)^N\prod_{t=1}^N(1+\exp(i\tau R_t))\tag{$*$}.
$$

I'm thinking that all the independence conditions allow us to write
$$
S=\sum_{t=1}^Nb_tR_t\tag{$**$}
$$
where $b_1,\ldots,b_N$ are i.i.d. Bernoulli with prob. $\frac{1}{2}$ for either $1$ and $0$. If ($**$) is true, then ($*$) follows immediately but I don't know how to justify ($**$) or am even sure that ($**$) is necessarily true.
($*$) is claimed in passing in my reading. Thank you for any help.
 A: For concreteness, let $\pi$ be the random permutation such that $X_t = R_{\pi(t)}$. Then we have
$$
S = \sum_{t=1}^N a_t X_t = \sum_{t=1}^N a_t R_{\pi(t)} = \sum_{t=1}^N a_{\pi^{-1}(t)} R_t.
$$
It remains to show that for each $\pi$, the vector $(a_{\pi^{-1}(1)},\ldots,a_{\pi^{-1}(n)})$ has the same distribution as $(a_1,\ldots,a_n)$. One way to show this is to compute, for each fixed $(b_1,\ldots,b_n) \in \{0,1\}^n$,
$$
\Pr[(\forall i) a_{\pi^{-1}(i)} = b_i] = 
\Pr[(\forall i) a_i = b_{\pi(i)}] = 
\frac{1}{2^n} =
\Pr[(\forall i) a_i = b_i].
$$
Let now $\Pi$ be the law of $\pi$ (which doesn't have to be the uniform distribution over $S_N$). Then for each $s$,
$$
\begin{align*}
\Pr[S = s] &= \sum_{\pi \sim \Pi} \Pr[\pi] \Pr[\sum_{t=1}^n a_{\pi^{-1}(t)} R_t = s] \\ &=
\sum_{\pi \sim \Pi} \sum_{b_1,\ldots,b_n} \Pr[\pi] \Pr[(\forall t)a_{\pi^{-1}(t)} = b_t | \pi] \Pr[\sum_{t=1}^n b_t R_t = s] \\ &=
\sum_{\pi \sim \Pi} \sum_{b_1,\ldots,b_n} \Pr[\pi] \Pr[(\forall t)a_t = b_t] \Pr[\sum_{t=1}^n b_t R_t = s] \\ &=
\sum_{b_1,\ldots,b_n} \Pr[(\forall t)a_t = b_t] \Pr[\sum_{t=1}^n b_t R_t = s] \\ &=
\Pr[\sum_{t=1}^n a_t R_t = s].
\end{align*}
$$
The third equality holds since the vector $(a_i)$ is independent of $\pi$.
