# Talk:Exponentially modified Gaussian distribution

Using Mathematica 9.0, I find the Kurtosis excess to be $\frac{\frac{9}{\lambda ^4}+\frac{6 \sigma ^2}{\lambda ^2}+3 \sigma ^4}{\left(\frac{1}{\lambda ^2}+\sigma ^2\right)^2}-3$ (i.e., the factored constant at the front of the numerator is "2" in the article, but should be "3"). This alternative expression simplifies to $\frac{6}{\left(\lambda ^2 \sigma ^2+1\right)^2}$. Sadly, this is "original research". -- 139.78.143.6 (talk) 22:38, 22 July 2014 (UTC)