8

I'm reading about heavy-tailed distributions, the definition states that:

The distribution of a real-valued random variable $X$ is said to have a heavy right tail if the probabilities $\mathbb{P}(X > x)$ decay more slowly than those of any exponential distribution, i.e., if: $$\lim_{x \to \infty} e^{\lambda x} \cdot \mathbb{P}(X > x) = \infty,$$ for every $\lambda > 0$.

Can anyone please intuitively explain the mathematical expression? As for the first part of the definition, it was understood. However, I'm struggling to connect it with the mathematical definition.

kjetil b halvorsen
  • 63,378
  • 26
  • 142
  • 467
Blg Khalil
  • 608
  • 5
  • 12
  • 2
    Hint: Assume $X$ has an exponential distribution and compute the limit. Then try with $X$ a Pareto distribution, then with $X$ a Weibull distribution. Then make comparison plots of the (log) tail probabilities ... – kjetil b halvorsen Jun 15 '21 at 01:40
  • 2
    See also https://stats.stackexchange.com/questions/168851/example-of-heavy-tailed-distribution-that-is-not-long-tailed, – kjetil b halvorsen Jun 15 '21 at 01:56
  • For the exponential distribution I found: $$\lim_{x \to \infty} e^{\lambda x} \cdot \mathbb{P}(X > x) = \lim_{x \to \infty} e^{\lambda x} \cdot e^{-\lambda x} = 1.$$ – Blg Khalil Jun 15 '21 at 21:45
  • For the Pareto distribution I found: $$\lim_{x \to \infty} e^{\lambda x} \cdot \mathbb{P}(X > x) = \lim_{x \to \infty} e^{\lambda x} \cdot (x_m/x)^{\alpha} = \lim_{x \to \infty} e^{\lambda x}/x^{\alpha} \cdot x_m^{\alpha} = \infty \cdot x_m^{\alpha} = \infty.$$ – Blg Khalil Jun 15 '21 at 21:49
  • Are the demonstrations correct @kjetil b halvorsen? And thank you in advance. – Blg Khalil Jun 15 '21 at 22:00
  • Hi @kjetil b halvorsen, I'm really embarassed to ask you again. I'm incapable of reaching $\lambda - \mu$. I don't how to go from $\lim_{x \to \infty} e^{\lambda x} \cdot e^{-\mu x}$: $$\lim_{x \to \infty} e^{\lambda x} \cdot \mathbb{P}(X > x) = \lim_{x \to \infty} e^{\lambda x} \cdot e^{-\mu x} = ...$$ I really apologize for the additional question. – Blg Khalil Jun 15 '21 at 23:32
  • 1
    Sorry I was to fast. So the limit with details. $\lim_{x \to \infty} e^{\lambda x}\cdot e^{-\mu x}=\lim e^{(\lambda-\mu) x}=\begin{cases} 1&\mu=\lambda \\ \infty & \lambda>\mu \\ 0 & \lambda < \mu \end{cases}$. So, since it is **not** $\infty$ for all $\lambda$, the conclusion that the exponential is not heavy-tailed follows. – kjetil b halvorsen Jun 15 '21 at 23:48
  • Really well understood. Thank you so much for your time and patience! – Blg Khalil Jun 15 '21 at 23:54

1 Answers1

4

For a fixed $\lambda>0$, we can rewrite the product as a fraction, so we can see the complementary cdf (right tail) of $Y_\lambda\sim Exp(\lambda)$:

$$ P(Y_\lambda >x)=e^{-\lambda x}. $$

We have:

$$\lim_{x \to \infty} e^{\lambda x} \cdot P(X > x) = \lim_{x \rightarrow \infty} \frac{P(X > x)}{ P(Y_\lambda>x)} = \infty.$$

This infinite limit at infinity means that: for every $M>0$ there is an $N_M>0$ such that $$ \frac{P(X > x)}{ P(Y_\lambda >x)} >M \; \;\left[\iff P(X>x)> M P(Y_\lambda >x) \right]$$

for all $x>N_M.$ That is for any given positive factor, for sufficiently high $x$, $X$'s right tail (starting at $x$) is bigger than $Y_\lambda$'s right tail (starting at $x$) scaled by the given factor.

In particular, for, say, $M=1$ there is $N_1> 0$ such that

$$ P(X>x)> P(Y_\lambda>x) $$

for all $x>N_1$.

This has to hold for all $\lambda > 0$ values.

ir7
  • 326
  • 1
  • 5