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Let $X$ have density function $f$ and let $Y = X$ with probability $0.5$ and $Y = -X$ with probability $0.5$. Show that $Y$ is symmetric about $0$, that is $f_Y(y) = f_Y(-y)$.

What I don't understand is how this is even possible considering that $Y=X$ and $Y=-X$ are two lines in the $x,y$ plane, the area of the joint distribution function over this domain should be relatively small ?

Lgate8
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  • Self study ;) ? Can you start writing the density of $f_Y$ – RUser4512 Oct 02 '15 at 14:07
  • Please add the `[self-study]` tag & read its [wiki](http://stats.stackexchange.com/tags/self-study/info). Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. – Silverfish Oct 02 '15 at 14:53
  • I never asked for the answer first of all, I specifically said I am having trouble understanding (from a visual perspective) how this could exist – Lgate8 Oct 02 '15 at 14:58
  • I see your point. The area is, in fact, zero. But, roughly speaking, it is not a "joint density per square unit" anyway. I think that [this question](http://stats.stackexchange.com/q/91045/22228) is getting at the same issue as yours. – Silverfish Oct 02 '15 at 15:25
  • What is the probability (density) that $Y=5$. – jlimahaverford Oct 02 '15 at 18:18
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    Compute the cumulative distribution function of $Y$ in terms of that of $X$. You will be able to show it is differentiable; you can compute its derivative $f_Y$ in terms of $f$; and it will then be obvious that $f_Y$ is symmetric. (Using the CDF avoids any technical difficulties surrounding the singular nature of the *joint* distribution of $(X,Y)$.) – whuber Oct 02 '15 at 21:03

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Hint: Given any real function $g(x)$ defined on all $\mathbb R$, $h(x) = \displaystyle \frac{g(x)+g(-x)}{2}$ is an even function, that is, $h(y) = h(-y)$ for all $y \in \mathbb R$.

Dilip Sarwate
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