This problem is question number 25 in John A. Rice's Mathematical Statistics and Data Analysis (3rd ed.).
Let $X$ have the density $f$, and let $Y=X$ with probability $\frac{1}{2}$ and $Y=-X$ with probability $\frac{1}{2}$. Show that the density of $Y$ is symmetric about zero $-$ that is, $f_Y(y) = f_Y(-y)$.
What I've done is the following:
- We are given that $Y=X$ with probability $\frac{1}{2}$ and $Y=-X$ with probability $\frac{1}{2}$ so this means $f_{Y|X}(x|x) = \frac{1}{2}$ and $f_{Y|X}(-x|x) = \frac{1}{2}$; that is, conditional under $X$, $Y$ is discrete uniform.
- By the multiplication law, $f_{X,Y}(x,y) = f_{Y|X}(y|x)f_{X}(x)$. Hence, $f_{X,Y}(x,x) = \frac{1}{2}f_{X}(x)$ and $f_{X,Y}(x,-x) = \frac{1}{2}f_{X}(x)$. Similarly, $f_{X,Y}(-x,-x) = \frac{1}{2}f_{X}(-x)$ and $f_{X,Y}(-x,x) = \frac{1}{2}f_{X}(-x)$. This shows that the density of $Y$ is symmetric about 0.
...But I'm not too sure. First, is the proof correct? Second, if it is, does this mean that the joint density function is over the lines $y=x$ and $y=-x$?
