Let X and Y be independent random variables and suppose Y is symmetric(around 0). Show that XY is symmetric Let X and Y be independent random variables and suppose Y is symmetric(around 0). Show that XY is symmetric.
What I thought is "Y is symmetric, so we have $f_{Y}(y)=f_{Y}(-y)$,then if we let Z=XY, we need to show that then $f_{Z}(z)=f_{Z}(-z)$". Am I right? And how could I do this?
Does anyone could help me? Thanks!
 A: To expand on whuber's comment on the OP's question
and the discussion thereafter, 
when $X$ and $Y$ are independent random
variables, so are $X$ and $-Y$ independent random variables. Since $Y$
has a symmetric distribution meaning that the (marginal)
distribution of $-Y$ is
the same as the (marginal) distribution of $Y$, it is also true that
the joint distribution of $(X,Y)$ (which, because of independence,
is the product of the marginal
distributions of $X$ and $Y$) is the same as the joint distribution of
$(X,-Y)$ (which is the product of the marginal distributions of $X$ and
$-Y$ since $X$ and $-Y$ are also independent). Consequently, the 
distribution of $XY$ is the same as the
distribution of $X(-Y) = -XY$, that is, $XY$ has a symmetric distribution.
This result cannot be shown to hold when $X$ and $Y$ are dependent
random variables: that the marginal distribution of $Y$ is symmetric
does not guarantee that the joint distribution of $(X,Y)$ is the
same as the joint distribution of $(X,-Y)$. As whuber points out,
in the extreme case of $X = Y$, $XY = X^2$ cannot take on negative values and
so cannot have the same distribution as $-XY=-X^2$ which cannot take
on positive values.

For the special case when $X$ and $Y$ are jointly continuous and
thus $XY = Z$ is a continuous random variable (as in Arthur's answer), 
note that for $z > 0$,
$$\begin{align}
P\{Z > z\} &= \int_{x=0}^\infty \int_{y=\frac zx}^\infty f_{X,Y}(x,y)
\,\mathrm dy\, \mathrm dx
+ \int_{x=-\infty}^0\int_{y=-\infty}^{\frac zx} f_{X,Y}(x,y)
\,\mathrm dy\, \mathrm dy\\
P\{Z < -z\} &= \int_{x=0}^\infty \int_{y=-\infty}^{\frac{-z}{x}} f_{X,Y}(x,y)
\,\mathrm dy\, \mathrm dx
+ \int_{x=-\infty}^0\int_{y=\frac{-z}{x}}^{\infty} f_{X,Y}(x,y)
\,\mathrm dy\, \mathrm dx
\end{align}$$
which upon differentiating with respect to $z$ leads us to 
$$f_Z(z) = \int_{x=-\infty}^\infty \frac{1}{|x|}f_{X,Y}\left(x,\frac zx\right)
 \, \mathrm dx ~ -\infty < z < \infty.$$
From this, we get that $f_Z(z) = f_Z(-z)$ holds whenever $f_{X,Y}(x,y)$
enjoys the property that $f_{X,Y}(x,y) = f_{X,Y}(x,-y)$ for all 
$x,y \in (-\infty, \infty)$; $X$ and $Y$ need not be independent
e.g., this property holds if $(X,Y)$ is uniformly distributed on the 
interior of the triangle with vertices at $(0,1), (0,-1), (1,0)$.
Note that $f_{X,Y}(x,y) = f_{X,Y}(x,-y)$ implies that $f_Y(y)$ is an
even function of $y$, that is, the distribution of $Y$ is symmetric.
For the special case when $X$ and $Y$ are independent random
variables, we have that 
$f_{X,Y}(x,y) = f_X(x)f_Y(y)$ equals $f_{X,Y}(x,-y)=f_X(x)f_Y(-y)$
whenever $f_Y(y) = f_Y(-y)$ for all $y$.
A: edit: the answer given by whuber in the comments is by all means the right one, it's concise and doesn't make any unnecessary assumptions. However, if you were interested in writing $f_Z$ explicitly, you can do it and use that to prove the property.

$f_Z(z)$ is the probability density of random variable $z$. Or if you prefer, $f_Z(z)~dz$ is the probability that $Z$ falls within an infinitesimal interval of size $dz$ centered around $z$.
Let's be a little sloppy to get a good intuition
$$f_Z(z) dz = P(X Y \in [z-dz/2, z+dz/2]$$
We won't go very far without knowing $Y$, so we condition on $Y$ and sum over every possible value
$$f_Z(z) dz = \sum_y P( X y \in [z-dz/2, z+dz/2] ) P(Y \in [y-dy/2, y+dy/2])$$
that is
$$\sum_y P\left(X \in \left[\min\left(\frac{z}{y} -\frac{dz}{2y}, \frac{z}{y} +\frac{dz}{2y}\right),\max\left(\frac{z}{y} -\frac{dz}{2y}, \frac{z}{y} +\frac{dz}{2y}\right), \right]  \right)P(Y \in [y-dy/2,y+dy/2]$$ 
We can identify
$$P\left(X \in \left[\min\left(\frac{z}{y} -\frac{dz}{2y}, \frac{z}{y} +\frac{dz}{2y}\right),\max\left(\frac{z}{y} -\frac{dz}{2y}, \frac{z}{y} +\frac{dz}{2y}\right), \right] \right) = f_X(z/y,z/y) \frac{dz}{|y|}$$
(note the division by $|y|$! The interval has shrunk and it matters!)
and also
$$P(Y \in [y-dy/2, y+dy/2]) = f_Y(y)dy$$
The way to sum over $y$ is with an integral since y takes continuous values. This finally gives us:
$$f_Z(z) = \int_{-\infty}^{\infty} \frac{1}{|y|} f_X\left(\frac{z}{y}\right) f_Y(y)~\mathrm{d}y$$
now
$$f_Z(-z) = \int_{-\infty}^{\infty} \frac{1}{|y|} f_X\left(\frac{-z}{y}\right) f_Y(y)~\mathrm{d}y = \int_{-\infty}^{\infty} \frac{1}{|-y|} f_X\left(\frac{z}{-y}\right) f_Y(-y)~\mathrm{d}y$$
since we're integrating over $y$ in $(-\infty,\infty)$, changing $y$ into $-y$ shows $f_Z(-z) = f_Z(z)$
