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Why is the kurtosis equation compared to a normal distribution? Are there cases where you would want to compare the tails versus some other types of distributions?

$$\gamma = \frac{M_4}{\sigma^4} -3$$

In the above equation would you simply replace the $3$ with the kurtosis of another distribution with which you would like to compare?

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3 Answers 3

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Subtracting $3$ has at least one other justification besides taking the normal distribution as the standard. The functional $\kappa_4$ given by $$ \kappa_4(X) = \operatorname{E}((X-\mu)^4) - 3(\operatorname{var}(X))^2, \quad \text{where } \mu = \operatorname{E}(X) $$ is the fourth cumulant of the distribution of $X$. It is

  • translation-invariant, i.e. $\kappa_4(X+c) = \kappa_4(X)$ if $c$ is constant (i.e. not random);
  • homogeneous of degree $4$, i.e. $\kappa_4(cX) = c^4\kappa_4(X)$;
  • "cumulative", i.e. $\kappa_4(X_1+\cdots+X_n) = \kappa_4(X_1)+\cdots+\kappa_4(X_n)$ if $X_1,\ldots,X_n$ are independent.

That last property holds only if the coefficient $-3$, rather than some other number, appears where it does.

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    $\begingroup$ Hi @Michael-Hardy can you expand on the bullets that you have listed. I'd like to understand what you have listed here, but not sure I fully understand the application of what you have stated. $\endgroup$
    – TsTeaTime
    Commented Jul 4, 2016 at 1:01
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    $\begingroup$ @TylerFurrer : Recall from an introductory statistics course: $$ \begin{align} \operatorname{var}(X) & = \operatorname{var}(X+c) & & \text{if $c$ is constant}; \\ \operatorname{var}(cX) & = c^2\operatorname{var}(X) & & \text{if $c$ is constant}; \\ \operatorname{var}(X_1+\cdots + X_n) & = \operatorname{var}(X_1)+\cdots + \operatorname{var}(X_n) & & \text{if } X_1,\ldots,X_n \text{ are independent}. \end{align} $$ The $4$th cumulant is like that, except it's $4$th degree rather than $2$nd degree. The $4$th cumulant is $\text{the }\ldots\qquad$ $\endgroup$ Commented Jul 4, 2016 at 3:17
  • $\begingroup$ $\ldots\,4$th central moment minus $3$ times the square of the variance. $\qquad$ $\endgroup$ Commented Jul 4, 2016 at 3:18
  • $\begingroup$ The difference is that $c^2$ appears when dealing with the variance and $c^4$ when dealing with the $4$th cumulant. The first two listed properties of the $4$th cumulant are true of the $4$th central moment and also of the square of the variance, but the third listed property is not true of those. $\qquad$ $\endgroup$ Commented Jul 4, 2016 at 3:24
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I think there are two conventions out there on what to call your $\gamma$: the first is "kurtosis" and the second is "excess kurtosis". The second term is more precise as it makes it clear that the Gaussian distribution is the benchmark. Also, the second convention implies that the "raw" kurtosis (without the -3 term) would not refer to any specific benchmark. So one answer is yes, you could make up your own "excess kurtosis with respect to distribution XYZ" by replacing the 3 with the kurtosis of that distribution. But in order to avoid confusion it's probably better to keep the Gaussian benchmark intact.

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    $\begingroup$ Given the generality of @Michael Hardy's answer, one might instead look at the "benchmark" as being unrelated to the Normal distribution. The Normal distribution, after all, is not the only one with a kurtosis of $3$. $\endgroup$
    – whuber
    Commented Jul 4, 2016 at 16:28
  • $\begingroup$ I agree that any distribution with kurtosis 3 could be regarded as an equally valid benchmark. But apart from the fact that you can of course construct such special cases, what are those other benchmark candidates, that is distributions that also have some practical appeal? (Honest question, don't know the answer right now.) Thanks. $\endgroup$
    – Sven S.
    Commented Jul 5, 2016 at 17:40
  • $\begingroup$ The point is more conceptual than practical, but it might be worth noting that many Beta distributions have an excess kurtosis of 0, as well as any Binomial$(n, (1\pm\sqrt{1/3})/2)$ distribution. It is difficult to find commonly used symmetric distributions with zero excess kurtosis. $\endgroup$
    – whuber
    Commented Jul 5, 2016 at 18:14
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Sure, you could use any distribution for comparison. Higher kurtosis for your distribution vs. the comparator implies your distribution has greater tail weight (or tail leverage; see below) than the comparison distribution. Lower kurtosis for your distribution vs. the comparator implies your distribution has less tail weight (leverage).

Logic is as follows. Consider your random variable $X$, having distribution $p_X(x)$. Assume finite fourth moment and let $V = \{(X-\mu)/\sigma\}^4$. Then the (non-excess) kurtosis of $X$ is $\kappa = E(V)$.

A standard way of understanding expectation is the "point of balance" of the distribution. Thus the distribution $p_V(v)$ balances at $\kappa$, which is the kurtosis of your distribution $p_X(x)$.

Now, suppose your comparator distribution has kurtosis $\kappa_0$. Locate $\kappa_0$ on the horizontal axis of the graph of $p_V(v)$ with a fulcrum. If $p_V(v)$ falls to the right of the fulcrum at $\kappa_0$, then $\kappa > \kappa_0$ and your distribution $p_X(x)$ is heavier-tailed than the comparator. If $p_V(v)$ falls to the left of the fulcrum at $\kappa_0$, then $\kappa < \kappa_0$ and your distribution $p_X(x)$ is lighter-tailed than the comparator.

Greater "heaviness" of tail refers to greater leverage of the tail rather than greater mass in the tail. You can have less mass in the tail with greater leverage (higher kurtosis), provided the mass is sufficiently distant from the mean.

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