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?

  • $\begingroup$ 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, 2019 at 20:19
  • $\begingroup$ Related: stats.stackexchange.com/q/450921/119261, stats.stackexchange.com/q/461328/119261. $\endgroup$ Apr 19, 2020 at 18:34

1 Answer 1


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.

  • $\begingroup$ 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, 2019 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, 2019 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.