Why is $\sum_{n=1}^N\varepsilon_ka_k$ symmetric around the origin? I'm trying to understand a step in the proof of Khinchin's inequality in this set of lecture notes (page 12) by Tao: https://arxiv.org/pdf/math/0311181.pdf
Let $a_1,\cdots,a_n$ be $n$ real numbers and $\varepsilon_1,\cdots,\varepsilon_n$ be iid random variables with
$
P(\varepsilon_1= 1)=P(\varepsilon_1= -1)=\frac{1}{2}.
$
After showing that
$$
P(\sum_{n=1}^N\varepsilon_ka_k\geq\lambda)\leq C e^{-\lambda^2/2},
$$
Tao concludes that
$$
P(|\sum_{n=1}^N\varepsilon_ka_k|\geq\lambda)\le 2Ce^{-\lambda^2/2},\tag{1}
$$
by saying that (page 14)

since the random variable $\sum_{n=1}^N\varepsilon_ka_k$ is clearly symmetric around the origin.

I guess this means
$$
P(Y\leq -\lambda)=P(Y\geq\lambda),\quad Y:=\sum_{n=1}^N\varepsilon_ka_k,
$$
which implies immediately (1).
Why is this clearly true? (While the intuition seems "clear" to me since $\varepsilon_k$ can be viewed as random independent signs, I don't have a proof.)
 A: Note that $-\epsilon_{i}\stackrel{d}{=}\epsilon_{i}$ (equal in distribution) and $\epsilon_i$ are identically distributed whence
$$
P\left(
\sum_{n=1}^N \epsilon_{k}a_k\leq -\lambda
\right)=P\left(\sum_{n=1}^N(-\epsilon_k)a_k\geq \lambda\right)=P\left(
\sum_{n=1}^N(\epsilon_k)a_k\geq \lambda
\right)
$$
A: I just realized after posting the question that I have an answer. (Alternatives are welcome.)
It suffices to show that the random variable 
$$
Z=-Y=\sum_{k=1}^N(-\varepsilon_k)a_k
$$
has the same distribution as that of $Y$. But clearly, $\varepsilon_k$ is symmetric: 
$$
P(\varepsilon_k=1)=P(-\varepsilon_k=1),\quad P(\varepsilon_k=-1)=P(-\varepsilon_k=-1).
$$ Hence $Z_k=(-\varepsilon_k)a_k$ and $Y_k=\varepsilon_ka_k$ are identically distributed. Since $Y_1,\cdots,Y_N$ are independent, so are $Z_1,\cdots,Z_N$. It follows that $Z=\sum Z_k$ and $Y=\sum Y_k$ have the same Fourier transforms (characteristic functions):
$$
E(e^{itZ})=\prod_k E(e^{itZ_k})=\prod_k E(e^{itY_k})=E(e^{itY})
$$
and hence the same distribution. 
