Why must a product of symmetric random variables be symmetric? I was reading about weight initialization in neural networks (He et. al, 2015) when I came across this statement:
"If we let $w_{l-1}$ have a symmetric distribution around zero and $b_{l} = 0$, then $y_{l-1}$ has zero mean and a symmetric distribution around zero."
Where $y_{l-1}$ is given by $w_{l-1}x_{l-1}$, the product of the layer's weights and its inputs. Here $w_{l-1}$ is generated by a normal distribution with mean zero, so it is symmetric. I know how to prove that the mean of this product is zero, but how would I go about showing it is also symmetric? In other words, given two random variables $X$ and $Y$ with $X$ being symmetric, how can I show $XY$ is also symmetric?

To say that a random variable $W$ "has a symmetric distribution around zero" is saying that $W$ and $-W$ have the same distribution.
Let $X$ be another random variable and set $Y=WX.$  By the rules of algebra, $-Y = -WX = (-W)X$ must have the same distribution as $WX$ provided $W$ and $X$ are independent.  Therefore $Y$ is symmetric around zero.
To see why independence is a necessary assumption, let $(W,X)$ take on the values $(-1,0),$ $(-1,1),$ $(1,0),$ and $(1,1)$ with probabilities $1/3,1/6,1/6,$ and $1/3,$ respectively.  Because $W$ has equal chance of being $\pm 1,$ it is symmetric around zero.  ($X$ is symmetric about $1/2.$) But $Y=WX$ takes on the value $0$ with probability $1/3+1/6,$ the value $-1$ with probability $1/6,$ and the value $1$ with probability $1/3,$ and therefore is not symmetric at all.
$$\begin{array}{crrr}
& \Pr & W & X & Y \\
\hline
& \frac{1}{3} & -1 & 0 & 0\\
& \frac{1}{6} & -1 & 1 & -1\\
& \frac{1}{6} & 1 & 0 & 0\\
& \frac{1}{3} & 1 & 1 & 1
\end{array}$$
