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 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.  
