While studying about kurtosis and extreme value theory, I came across the concept of tails of the distribution. So I wanted to ask that why is it such that distribution with higher number of moments definable have a lighter tail and vice-versa?
$\begingroup$
$\endgroup$
4
-
$\begingroup$ It's not true in general. It depends on your definition of "lighter tails." $\endgroup$– BigBendRegionCommented Jul 13, 2020 at 20:13
-
$\begingroup$ by lighter tails I mean to say that tails that decays faster. $\endgroup$– Devansh GandhiCommented Jul 14, 2020 at 18:04
-
$\begingroup$ Still not enough of a definition to make the statement mathematically true. You could define a notion of "decays faster" for a class of bounded distributions, all of whose moments are finite. Within that class, some distributions will be lighter tailed than others, yet all moments are defined for every distribution in the class. But even with infinite tails the statement is not mathematically true. Just because one distribution is lighter tailed than another does not mean that the former has a higher number of defined moments. $\endgroup$– BigBendRegionCommented Jul 15, 2020 at 19:39
-
1$\begingroup$ On the other hand, if one distribution has fewer defined moments than another, then one can easily argue that it is "heavier tailed." Eg, suppose $E(X^4) = \infty$ but $E(X^3)$ is finite. Suppose also that $E(Y^6) = \infty$ but $E(Y^5)$ is finite. Let $m_4 = E(Y^4)$. Draw the distribution of the pdf of $X^4$, and place a fulcrum at $m_4$. The distribution will not balance at $m_4$; instead it will fall to the right. In that sense, the distribution of $X$ is heavier tailed than that of $Y$. $\endgroup$– BigBendRegionCommented Jul 15, 2020 at 19:41
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
I decided to make my comment an answer.
If one distribution has fewer defined moments than another, then one can easily argue that it is "heavier tailed" as follows:
Suppose $E(X^4)=\infty $, but $E(X^3) < \infty$. Suppose also that $E(Y^6)=\infty$ but $E(Y^5)< \infty$. Then $X$ has fewer defined moments than $Y$.
Let $\mu_4=E(Y^4)$. Draw the distribution of the pdf of $X^4$, and place a fulcrum at $\mu_4$. The distribution will not balance at $\mu_4$; instead it will fall to the right. In that sense, the distribution of $X$ is heavier tailed than that of $Y$. (Large values of $X^4$, which make the graph fall to the right of the fulcrum at $\mu_4$, correspond to values of $X$ that are far into one or both tails of its distribution.)
This argument obviously generalizes to all other cases where $|X|$ has fewer (finite) moments than $|Y|$.
answered Jul 27, 2020 at 21:02