# Are the skew-normal distribution and the skew-Cauchy distribution heavy-tailed?

I think the title is self-explanatory. I understand that the skewness and the tail behavior of some distribution are completely unrelated as any symmetric distribution will have a skewness of zero irrespective of how heavy its tails are.

However, I was wondering (i) if the skew-normal distribution and the skew-Cauchy distribution are heavy-tailed.

(ii) Also, we can get skewed distribution using a selection procedure. That is, if $$X$$ and $$Y$$ are Gaussian (or Cauchy), $$\frac{d}{dz}P(X-kY\leq z|Y>0)$$ represents a density of a skewed version of normal (or Cauchy distribution). What about the tails of such constructed skewed distributions?

Note: I say the distribution of a random variable $$X$$ heavy-tailed if $$\lim \limits_{x\rightarrow \infty} e^{tx} P(X>x) = \infty$$ for all $$t>0$$.

• Which definition of heavy-tailed are you using? Jun 30, 2019 at 12:16
• Hi @Glen_b, I only know one definition of heavy-tailed distributions. That is, the distribution of a random variable $X$ is said to be heavy-tailed if $\lim \limits_{x\rightarrow \infty} e^{tx} P(X>x) = \infty$ for all $t>0$.
– Joy
Jun 30, 2019 at 12:19
• That's fine, though some people mean other things when they say "heavy tailed". Could you edit that into your question, please? Jun 30, 2019 at 12:20
• Are you using this skew Cauchy and this skew normal, or something else? Jun 30, 2019 at 12:24
• Yes, I am using the mentioned skew normal and skew Cauchy.
– Joy
Jun 30, 2019 at 12:27

Taking $$f(x)=2\phi (x)\Phi (\alpha x)$$, we can immediately see that $$f(x)<2\phi(x)$$, so $$P(X>x)<2(1-\Phi(x))$$. Indeed, we can see that $$\lim \limits_{x\rightarrow \infty} e^{tx} P(X>x)$$ is less than twice that of a standard normal (which we already know to be light tailed, so we know that integral is finite even before we try to evaluate it).
A somewhat similar approach to the one I used for the skew-normal can be used. Let's restrict consideration to the right half (i.e. consider only $$x>\mu$$). Then the density is proportional to the density of a Cauchy (because of the parameter $$\lambda$$), so $$P(X>x)$$ is proportional to that for a Cauchy. Consequently, the limit goes to infinity because the corresponding quantity for the Cauchy does.