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Original question (with confused terms):

Wikipedia and Wolfram Math World claim that the kurtosis of exponential distribution is equal to $6$. Whenever I calculate the kurtosis in math software (or manually) I get $9$, so I am slightly confused.

I calculate 4th central moment as:

$$ D^4X = \int_0^\infty (x-\lambda^{-1})^4 \lambda e^{-\lambda x} \, dx\,. $$

And kurtosis as:

$$ K = \frac{D^4X}{(D^2X)^2} $$

Is the approach and result correct (kurtosis equal to $9$)? I trust that the calculation of this very specific integral I shown is correct.

Comment:

I didn't know 'kurtosis' and 'excess kurtosis' are different terms. Thank you all for your help.

Szymon Brych
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    If a statement on MathWorld is confusing to you just ignore it. MathWorld isn't editable the way Wikipedia is so it's a lot easier for bizarre mistakes to sneak in; I've seen all sorts of weird nonstandard definitions and things like that on it. – Qiaochu Yuan Oct 24 '17 at 06:02
  • Thank you for the support. I probably got confused by wikipedia redirecting "excess kurtosis" to "kurtosis" and then explaining the difference within. – Szymon Brych Oct 24 '17 at 06:14

3 Answers3

15

Often "kurtosis" is taken to mean "excess kurtosis", i.e. the amount by which the kurtosis exceeds that of the normal distribution, thus the kurtosis minus $3.$

Subtraction of $3$ makes sense in some contexts even without thinking about the normal distribution. Let $\mu=\operatorname{E}(X)$ and note that the two functionals $$ \operatorname{A}(X) = \operatorname{E}\big((X-\mu)^4\big) \quad \text{and} \quad \operatorname{B}(X) = \Big(\operatorname{E}\big((X-\mu)^2\big) \Big)^2 $$ are $(1)$ homogenous of degree $4$ (i.e. multiplying $X$ by a scalar $c$ multiplies the value of the functional by $c^4,$ and $(2)$ translation-invariant. But they are not "cumulative", i.e. for independent random variables $X_1,\ldots,X_n$ we do not have $\operatorname{A}(X_1+\cdots+X_n) = \operatorname{A}(X_1)+\cdots+\operatorname{A}(X_n)$ nor $\operatorname{B}(X_1+\cdots+X_n) = \operatorname{B}(X_1)+\cdots+\operatorname{B}(X_n).$ But $\kappa = {\operatorname{A}}-{3\operatorname{B}}$ is homogenous of degree $4$ and translation invariant and cumulative. And for any coefficient besides $-3$ that doesn't work. This quantity $\kappa(X)$ is the fourth cumulant of the distribution of $X.$ The excess kurtosis is $$ \frac{\kappa(X)}{\sigma^4}. $$

Michael Hardy
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8

A more careful reading of the MathWorld article states that the kurtosis excess of an exponential distribution is $6$. This is not the same as kurtosis, since the former is defined as $$\gamma_2 = \frac{\mu_4}{\mu_2^2} - 3,$$ (see Equation 3), as opposed to the simple kurtosis $$\beta_2 = \frac{\mu_4}{\mu_2^2},$$ which is Equation (1). The reason for these two definitions is also explained in the article, since $\beta_2$ for a normal distribution is $3$, hence the definition of kurtosis excess representing in some sense the amount of kurtosis in excess of a normal distribution.

Therefore, there is no error. MathWorld is correct.

heropup
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I get for $\lambda=1$ $$\int_0^\infty(x-1)^4\exp(-x)\,dx=9.$$ In general I get $$\int_0^\infty(x-1)^n\exp(-x)\,dx=D_n,$$ the $n$-th derangement number.

Wikipedia is correctable.

Angina Seng
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