# How to prove/show that some distribution is symmetric at 0

I have to prove that the Cauchy distribution is symmetric at 0. However, I'm not entirely sure how to do this. I'm given the problem: Suppose that a particle is fired from the origin in the $(x,y)$ plane in a straight line in a direction at random angle $\Phi$ to the $x$ axis and let $Y$ be the $y$-coordinate of the place where the particle hits the line $x=1$. Show that if $\Phi$ has uniform $(-\frac\pi2 , \frac\pi2)$ distribution, then $$f_Y(y)=\frac{1}{\pi (1+y^2)}$$ Show that the Cauchy distribution is symmetric about $0$. Firstly I would like to know a general strategy at "showing" things like this. I have seen a variety of problems that ask to show or derive something. Secondly how would I show that a probability distribution is symmetric about 0? A great description on how to do these would be much much appreciated. I can show what I have gotten so far:

I started by identifying that $f_\Phi(\phi)$ equals $\frac1\pi$ if $-\frac\pi2 \lt \phi \lt \frac\pi2$ and $0$ otherwise. Then by change of variables I divided the PDF of $\Phi$ divided by the absolute value of the derivative of $Y$. I think $y=arctan(x)$, therefore making the derivative $\frac1{(1+x^2)}$ but then once I put it all together I only came up with $\frac{(1+x^2)}{\pi}$. So I feel I'm close but I'm obviously doing something wrong. So if you took the time to read all this it is greatly appreciated.

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You need to multiply by the absolute value of the derivative, not divide. So you get $$\frac1{\pi(1+y^2)}$$ as you desire.
To show that it's symmetric about zero, you need to show that for any $a$, $f_Y(a) = f_Y(-a)$. This should not be very difficult. You can restrict yourself to the case $a > 0$ if you want (why?).
You mean $f_Y(a)=f_Y(-a)$. – Did Nov 7 '12 at 22:45