Editing Bernoulli distribution (section)

==Higher moments and cumulants==
The raw moments are all equal because <math>1^k=1</math> and <math>0^k=0</math>.

:<math>\operatorname{E}[X^k] = \Pr(X=1)\cdot 1^k + \Pr(X=0)\cdot 0^k = p \cdot 1 + q\cdot 0 = p = \operatorname{E}[X].</math>

The central moment of order <math>k</math> is given by
:<math>
\mu_k =(1-p)(-p)^k +p(1-p)^k.
</math>
The first six central moments are
:<math>\begin{align}
\mu_1 &= 0, \\
\mu_2 &= p(1-p), \\
\mu_3 &= p(1-p)(1-2p), \\
\mu_4 &= p(1-p)(1-3p(1-p)), \\
\mu_5 &= p(1-p)(1-2p)(1-2p(1-p)), \\
\mu_6 &= p(1-p)(1-5p(1-p)(1-p(1-p))).
\end{align}</math>
The higher central moments can be expressed more compactly in terms of <math>\mu_2</math> and <math>\mu_3</math>
:<math>\begin{align}
\mu_4 &= \mu_2 (1-3\mu_2 ), \\
\mu_5 &= \mu_3 (1-2\mu_2 ), \\
\mu_6 &= \mu_2 (1-5\mu_2 (1-\mu_2 )).
\end{align}</math>
The first six cumulants are
:<math>\begin{align}
\kappa_1 &= p, \\
\kappa_2 &= \mu_2 , \\
\kappa_3 &= \mu_3 , \\
\kappa_4 &= \mu_2 (1-6\mu_2 ), \\
\kappa_5 &= \mu_3 (1-12\mu_2 ), \\
\kappa_6 &= \mu_2 (1-30\mu_2 (1-4\mu_2 )).
\end{align}</math>