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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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up vote 1 down vote accepted

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?).

As a general strategy to "show" a property, try to remember or look up the definition of that property and any theorems you have learned that imply that property. Choose one that looks promising and try to prove that it follows from what you know. If you can't make it work, try attacking it from a different angle. When I used to do math competitions in high-school this is what we called looking for the "penultimate step."

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You mean $f_Y(a)=f_Y(-a)$. – Did Nov 7 '12 at 22:45
Right, thanks. I'll fix it. – Jonathan Christensen Nov 7 '12 at 22:58

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