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.)
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.