# Probability integral transforms - Cauchy distribution of 1/x and X

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?

• 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. – B.Liu Apr 12 '19 at 20:19
• – StubbornAtom Apr 19 at 18:34

## 1 Answer

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.

• 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. – Jhonny Apr 25 '19 at 16:28
• When $z > 0$, I want to look at the event that $1/X \ge z$ because then I know $1/X > 0$ as well. – Taylor Apr 25 '19 at 16:45