Can some one show me the steps to derive kurtosis of Bernoulli distribution?

First $$E[X^i]=p \mathrm{\;\;\;for\;all\;}i$$ $$\begin{align}E[x-u/s]^4=&(EX^4-4uEX^3+6u^2EX^2-4u^3EX-u^4)/s^4 \\ =&(p-4p^2+6p^3-4p^4-p^4)/s^4\end{align}$$


2 Answers 2


The (excess) kurtosis $\kappa$ of a random variable $X$ is the fourth derivative of its cumulant generating function (cgf) evaluated at zero, divided by the square of the variance (which itself is the second derivative of the cgf, evaluated at zero). By definition, the cgf is

$$\psi_X(t) = \log\mathbb{E}\left(e^{itX}\right).$$

The values of $e^{itX}$ are $1$ when $Xt=0$ and $e^{it}$ when $X=1$; these happen with probabilities $1-p$ and $p,$ respectively. Thus, by the definition of expectation $\mathbb E,$

$$\mathbb{E}\left(e^{itX}\right) = (1-p) + pe^{it} = (1-p)\left(1 + e^u\right)$$

where $$u = it + \log\left(\frac{p}{1-p}\right).$$

Let's work out the derivatives of $g(u)=\log(1-p) + \log(1 + e^u) = \psi_X(t),$ because--up to powers of $i$, which will not matter in the end--these will be the same as its derivatives with respect to $t:$

$$\frac{d}{du}g(u) = \frac{e^u}{1+e^u} = f(u).\tag{1}$$

$$\frac{d^2}{du^2} g(u) = \frac{d}{du}f(u) = \frac{e^u}{(1+e^u)^2} = f(u)(1-f(u)).\tag{2}$$

This makes finding successive derivatives easy: apply the product rule and keep replacing $f^\prime(u)$ by $f(u)(1-f(u)).$ Let's abbreviate $f(u)$ simply as $f$:

$$\frac{d^2}{du^2} f = f(1-f)(1-2f);$$

$$\frac{d^4}{du^4} g(u) =\frac{d^3}{du^3} f = f(1-f)(1-6f(1-f)).\tag{3}$$

When $t=0$, $e^u = (1-p)/p$, implying (by $(1)$) $f(u) = 1-p.$ Consequently, $(2)$ asserts

$$\frac{d^2}{du^2} g\mid_{t=0} = f(1-f) = (1-p)p$$

(recovering the familiar expression for the variance of a Bernoulli$(p)$ variable) and $(3)$ asserts

$$\frac{d^4}{du^4} g\mid_{t=0} = \cdots = p(1-p)(1 - 6p(1-p)).$$

Dividing the latter by the square of the former gives

$$\kappa = \frac{p(1-p)(1 - 6p(1-p))}{(p(1-p))^2} = \frac{1}{p(1-p)} - 6.$$

This is equivalent to the expression given by Wikipedia at https://en.wikipedia.org/wiki/Bernoulli_distribution.


Alan Stuart & Keith Ord, Kendall's Advanced Theory of Statistics, Fifth Edition, Volume 1. 1986.


I will assume the self-study tag.

You may start with another equation for the kurtosis :

$$ \operatorname{Kurt}[X] = \operatorname{E}\left[\left(\frac{X - p}{\sigma}\right)^4\right] = \frac{p_4}{\sigma^4} = \frac{\operatorname{E}[(X-p)^4]}{(\operatorname{E}[(X-p)^2])^2}, $$

Using the binomial formula you can expand : $\operatorname{E}[(X-p)^4]$ and $\operatorname{E}[(X-p)^2]$, and finally use $\operatorname{E}[X^i]=p$

  • $\begingroup$ Woah, I would never have thought of doing it this way $\endgroup$ Apr 4, 2018 at 21:25

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