# Kurtosis and Skewness of Binomial Distribution

Let $X \thicksim B(n,p)$ then I would like to evaluate kurtosis and skewness of X.

First I want to use the fact that kurtosis $k_3(\dfrac{X-\mu}{ \sigma})=\dfrac{k_3(X)}{\sigma^3}$ and skewness kurtosis $k_4(\dfrac{X-\mu}{ \sigma})=\dfrac{k_4(X)}{\sigma^4}$.

To use above identity, one needs to derive 3 and 4-th cumulant of X.

mgf of bionomial $X$ is $M(t)=[(1-p)+pe^t]^n$ thus

$K(t) = \log M(t)=n\log [(1-p)+pe^t]$

My question is here:

How could one expand above log term into the form such as $\sum_{r=1}^{\infty}\dfrac{k_r(0)}{r!}t^r$?

textbook has denoted

$n\log [(1-p)+pe^t] = \sum_{r=1}^{\infty}\dfrac{t^r}{r!}\{n\sum_{k=1}^r\sum_{j_1+j_2...+j_k=r}\begin{pmatrix}r\\j_!,j_2...j_k\end{pmatrix}\dfrac{(-1)^{k-1}}{k}p^k\}$

How could above identity be derived?

• This is called a Taylor Series (sometimes, a Maclaurin series). Its terms can be found via differentiation, complex contour integration, and many other means. If you are unfamiliar with these techniques, then you will find it far easier to compute the first four moments directly: after all, each one is just the sum of two quantities!
– whuber
Jul 4, 2017 at 13:49
• slight typo in your question $k_3$ denotes skewness and $k_4$ denotes kurtosis Jul 7, 2017 at 0:46

To calculate the derivatives up to the 4th you can do them by hand and make sure you don't make any errors. To do this you'll need to use chain rule, quotient rule, product rule, and lots of organization and notebook paper. Or you can use an online differentiator-this is what I did using this one (http://www.derivative-calculator.net/). Where you can enter the function as $$nlog(1-p+pe^x)$$ because $x$ is the variable it is programmed to differentiate with respect to and you can get all 4 derivatives. Writing the function in the OP as $f(t)$ they are:

$$f^\prime(t)=\frac{npe^t}{1-p+pe^t}$$

$$f^{(2)}(t)=\frac{np(1-p)e^t}{(1-p+pe^t)^2}$$

$$f^{(3)}(t)=\frac{np(p-1)(1-p+pe^t)e^t}{(1-p+pe^t)^3}$$

$$f^{(4)}(t)=\frac{n(1-p)pe^t(p^2e^{2t}+(4p^2-4p)e^t +p^2-2p+1)}{(1-p+pe^t)^4}$$

now evaluating each of these at $t=0$ gives the first 4 cumulants, denoted $k_i$ for $i=1,2,3,4$. These cumulants are:

$$k_1=np$$

$$k_2=np(1-p)$$

$$k_3=-np(1-p)(2p-1)$$

$$k_4=np(1-p)(6p(p-1)+1)$$

and you can check that the skewness is:

$$\frac{k_3}{k_2^{3/2}}=\frac{-np(1-p)(2p-1)}{np(1-p)\sqrt{np(1-p)} }=\frac{-(2p-1)}{\sqrt{np(1-p)}}$$

and the kurtosis is:

$$\frac{k_4}{k_2^{2}}=\frac{np(1-p)(6p(p-1)+1)}{n^2p^2(1-p)^2}=\frac{6p(p-1)+1}{np(1-p)}=\frac{1-6p(1-p)}{np(1-p)}$$

The expression you have written for the infinite series is the taylor series expansion of the function $f(t)$.

• why does kurtosis is $\dfrac{k_4}{k_2^2}$ and skeness is $\dfrac{k_3}{k_2^{3/2}}$ Jul 7, 2017 at 17:11
• I'm guessing you are wondering about why the denominators are different powers? To match with the power in the numerator, $k_3=\mathbb{E}(x-\mu)^3$ and $k_2=\mathbb{E}(x-\mu)^2$ so the weird $3/2$ power on the skewness is to match the powers in the numerators. Was the answer to this post acceptable? If so, marking it as such would help close out the question. Jul 7, 2017 at 19:04