What's the kurtosis of exponential distribution? 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.
 A: 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.  
A: 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.
A: 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
$$
\frac{\kappa(X)}{\sigma^4}.
$$
