Show that $X$ has no mean

Question

$X$ is said to have a Cauchy distribution iff $X$ has a probability distribution of the form:

$$f(x)= \begin{cases} \frac{a}{\pi(ax^2+x^2)} & \text{,$x>0$} \\ \ \ \ \ 0 & \text {,otherwise} \end{cases}$$ where $a>0$

How can I show that $X$ has no mean?

http://stats.stackexchange.com/questions/36027/why-does-the-cauchy-distribution-have-no-mean — Niklas, Oct 14 '15 at 12:37
Incidentally, the pdf as written above is wrong. It should be $a^2$, not $ax^2$. — Clement C., Oct 14 '15 at 12:41
And Cauchy distributions have support the line (-oo,+oo), not (0,+oo). — Did, Oct 14 '15 at 12:58
Duplicate of Mean of a Cauchy Distribution and of Why does the Cauchy distribution have no mean if it's symmetric around 0?. Please search before asking: entering "mean of Cauchy distribution" is all it takes. — , Oct 14 '15 at 17:43

Clement C. · Answer 1 · 2015-10-14T13:36:11.080

1

Hint: Try to compute it: $\mathbb{E}[X]$ would be equal to $$ \int_{(-\infty,\infty)} xf(x) dx$$ Does this integral exist (in a proper sense)? (i.e., is the function $x\mapsto x f(x)$ integrable on $(-\infty,\infty)$?)

In particular, look at what asymptotically $xf(x)$ becomes when $x\to \infty$:$\frac{x}{a^2+x^2} \sim_{x\to\infty} ?$

edited Oct 14 '15 at 13:36

answered Oct 14 '15 at 12:37

Clement C.

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Show that $X$ has no mean

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