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!
Best Answer
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 $y$.