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!

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    $\begingroup$ Since $Y$ and $-Y$ have the same distribution and $Y$ is independent of $X$, $XY$ and $-XY=X(-Y)$ have the same distribution, QED. You don't even need the symmetry of $X$ or the existence of any PDFs. $\endgroup$ – whuber Nov 5 '14 at 23:12
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    $\begingroup$ @whuber +1 for a very elegant observation. But please do expand a little on the fact that the result does not hold when $X$ and $Y$ are dependent. It is easy to miss where exactly independence was needed in your proof. $\endgroup$ – Dilip Sarwate Nov 6 '14 at 3:00
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    $\begingroup$ @Dilip I agree that this is potentially a subtle point. Consideration of the case of perfect dependence, where $X=Y$, should make it clear, since $XY=X^2$ cannot be negative. $\endgroup$ – whuber Nov 6 '14 at 14:51
  • $\begingroup$ @whuber I think the point is that $-Y$ is also independent of $X$ and so $(X,Y)$ and $(X,-Y)$ have the same joint distribution which allows for the conclusion that $XY$ and $-XY = X(-Y)$ have the same distribution. With dependent $X$ and $Y$, we cannot be sure that $(X,Y)$ and $(X,-Y)$ have the same distribution. Perhaps you can use your moderator superpowers to change "$Y$ is independent of $X$" to "$-Y$ is also independent of $X$" in your original comment? $\endgroup$ – Dilip Sarwate Nov 6 '14 at 15:14

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 $y$.


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:

$$f_Z(z) = \int_{-\infty}^{\infty} \frac{1}{|y|} f_X\left(\frac{z}{y}\right) f_Y(y)~\mathrm{d}y$$


$$f_Z(-z) = \int_{-\infty}^{\infty} \frac{1}{|y|} f_X\left(\frac{-z}{y}\right) f_Y(y)~\mathrm{d}y = \int_{-\infty}^{\infty} \frac{1}{|-y|} f_X\left(\frac{z}{-y}\right) f_Y(-y)~\mathrm{d}y$$

since we're integrating over $y$ in $(-\infty,\infty)$, changing $y$ into $-y$ shows $f_Z(-z) = f_Z(z)$

  • $\begingroup$ It would be nice to see an approach that works generally, because not all random variables have PDFs--and even though part of the question implicitly assumed these RVs do have PDFs, the question as actually stated does not make that assumption. The details of your calculation get a little messier, too, once you realize they have to cope with the possibility that $y$ can be negative or even zero. $\endgroup$ – whuber Nov 5 '14 at 23:30
  • $\begingroup$ yeah I got carried away on deriving $f_Z$ instead of a actually answering the question, I need sleep $\endgroup$ – Arthur B. Nov 5 '14 at 23:34
  • $\begingroup$ I almost got it, but how $P(Y \in [y-dy/2,y+dy/2]$ comes from $P(y \in [y-dy/2, y+dy/2])$ $\endgroup$ – user133140 Nov 5 '14 at 23:34
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    $\begingroup$ The answer is really helpful to me, thanks very much! $\endgroup$ – user133140 Nov 5 '14 at 23:39
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    $\begingroup$ keep in mind that whuber's answer is a much, much better answer to your original question, but in general being able to connect p.d.f with infinitesimal probabilities is a useful sanity check. $\endgroup$ – Arthur B. Nov 5 '14 at 23:41

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