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When revising for exams, I recently came across the following question:

Suppose that $X$ is Cauchy distributed, ie has a density function $$f_X(x) = \frac{1}{\pi(1+x^2)}$$ Show that $1/X$ is also Cauchy distributed.

Wanting to do the question properly, not just applying a formula, I approached it as follows:

First let $Z = 1/X$ and consider $z<0$ then we have:

$F_Z(z) = Pr(\frac{1}{X} \leq z) = Pr( X \leq \frac{1}{z}) = F_X(\frac{1}{z}) $

then consider $z>0$. In this case I believe we have to separate it into two probabilities after the second inequality:

$F_Z(z) = Pr(\frac{1}{X} \leq z) = Pr( X \geq \frac{1}{z}<| X > 0) + Pr( X < 0) = 1- F_X(\frac{1}{z}) + \frac{1}{2} $

We then find the distribution function by taking derivatives, and it will be $f_Z(z) = -\frac{1}{z^2} f_X(\frac{1}{z})$ for $z<0$ and $f_Z(z) = \frac{1}{z^2} f_X(\frac{1}{z})$ for $x>0$. Writing these out indeed produces Cauchy distributions, with the exception for that minus sign!

What am I doing wrong? Why isn't my separation into cases correct?

The book simply suggests the following solution:

$F_Z(z) = Pr(\frac{1}{X} \leq z) = Pr( X \geq \frac{1}{z}) = 1-F_X(\frac{1}{z}) $

Indeed this produces the required answer, but isn't it overly simplified and just "lucky" that it works out? Don't we have to consider negatives in the rearrangement of the inequality?

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  • $\begingroup$ I think you are dividing on the wrong cases. The support for Z is along the entire real line (including zero), so there is no point dividing the derivation of the PDF into <0 and >0. If you insist on splitting into two cases, you should split on X instead, both of which will lead to the book formula anyway. $\endgroup$ – B.Liu Apr 12 '19 at 20:19
  • $\begingroup$ Related: stats.stackexchange.com/q/450921/119261, stats.stackexchange.com/q/461328/119261. $\endgroup$ – StubbornAtom Apr 19 at 18:34
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When $z < 0$, instead of writing $$ F_Z(z) = Pr(\frac{1}{X} \leq z) = Pr( X \leq \frac{1}{z}) = F_X(\frac{1}{z}) $$ you must write $$ F_Z(z) = Pr(\frac{1}{X} \leq z) = Pr(1 \geq Xz) = Pr( X \geq \frac{1}{z}) = 1-F_X(\frac{1}{z}) $$ because every time you multiply or divide by a negative number (or in general any time you apply a nonincreasing function to both sides of an inequality) you must flip that inequality.

There is a similar problem with your work when you assume $z > 0$. To avoid any unnecessary conditioning, you might consider finding the survival function as an intermediate step when $z > 0$: $$ 1 - F_Z(z) = Pr(\frac{1}{X} \geq z) = Pr(1 \geq Xz) = Pr(\frac{1}{z} \geq X) = F_X(\frac{1}{z}). $$ Observe that there is no "flipping" in this case.

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  • $\begingroup$ Thank you for your reply, I do understand what I am doing wrong with the $z<0$ case now. Could you however lend me some more guidance with respect to what is wrong with the $z>0$ case? As far as I can see, we can have $1/X < z$ in two ways: either $X<0$ or $X>0$ and $X>1/z$. I cannot see how the survival function approach takes account of the former case. $\endgroup$ – Jhonny Apr 25 '19 at 16:28
  • $\begingroup$ When $z > 0$, I want to look at the event that $1/X \ge z$ because then I know $1/X > 0$ as well. $\endgroup$ – Taylor Apr 25 '19 at 16:45

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