Theorem: Excess Kurtosis of Binomial Distribution

Let $X$ be a discrete random variable with a binomial distribution with parameters $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$:
Then the excess kurtosis $\gamma_2$ of $X$ is given by: 
:$\gamma_2 = \dfrac {1 - 6 p q} {n p q}$
where $q = 1 - p$.

Proof:
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Binomial Distribution, we have: 
:$\mu = n p$
By Variance of Binomial Distribution, we have: 
:$\sigma = \sqrt {n p \paren {1 - p} }$ 
So: 
{{begin-eqn}}
{{eqn | l = \gamma_2
      | r = \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4} - 3
      | c = Kurtosis in terms of Non-Central Moments
}}
{{eqn | r = \frac {\expect {X^4} - 4 n p \paren {n p + 3 n^2 p^2 - 3 n p^2 + n^3 p^3 - 3 n^2 p^3 + 2 n p^3} + 6 n^2 p^2 \paren {n^2 p^2 + n p \paren {1 - p} } - 3 n^4 p^4} {n^2 p^2 \paren {1 - p}^2} - 3
      | c = Skewness of Binomial Distribution, Variance of Binomial Distribution
}}
{{end-eqn}}
To calculate $\gamma_2$, we must calculate $\expect {X^4}$.
We find this using the moment generating function of $X$, $M_X$. 
By Moment Generating Function of Binomial Distribution, this is given by: 
:$\map {M_X} t = \paren {1 - p + p e^t}^n$
From Moment in terms of Moment Generating Function:
:$\expect {X^4} = \map {M_X' ' ' '} 0$
So:
{{begin-eqn}}
{{eqn | l = \map {M_X'} t
      | r = n p e^t \paren {1 - p + p e^t}^{n - 1}
      | c = Chain Rule for Derivatives, Derivative of Power
}}
{{eqn | l = \map {M_X' '} t
      | r = n p e^t \paren {1 - p + p e^t}^{n - 1} + n \paren {n - 1} p^2 e^{2 t} \paren {1 - p + p e^t}^{n - 2}
      | c = Product Rule for Derivatives, Chain Rule for Derivatives, Derivative of Power
}}
{{eqn | l = \map {M_X' ' '} t
      | r = \map {M_X' '} t + 2 n \paren {n - 1} p^2 e^{2 t} \paren {1 - p + p e^t}^{n - 2} + n \paren {n - 1} \paren {n - 2} p^3 e^{3 t} \paren {1 - p + p e^t}^{n - 3}
      | c = Product Rule for Derivatives, Chain Rule for Derivatives, Derivative of Power
}}
{{eqn | l = \map {M_X' ' ' '} t
      | r = \map {M_X' ' '} t + 4 n \paren {n - 1} p^2 e^{2 t} \paren {1 - p + p e^t}^{n - 2} + 2 n \paren {n - 1} \paren {n - 2} p^3 e^{3 t} \paren {1 - p + p e^t}^{n - 3}
}}
{{eqn | o = +
      | r = 3 n \paren {n - 1} \paren {n - 2} p^3 e^{3 t} \paren {1 - p + p e^t}^{n - 3} + n \paren {n - 1} \paren {n - 2} \paren {n - 3} p^4 e^{4 t} \paren {1 - p + p e^t}^{n - 4}
      | c = Product Rule for Derivatives, Chain Rule for Derivatives, Derivative of Power
}}
{{end-eqn}}
Setting $t = 0$:
{{begin-eqn}}
{{eqn | l = \expect {X^4}
      | r = \map {M_X' ' '} 0 + 4 n \paren {n - 1} p^2 + 2 n \paren {n - 1} \paren {n - 2} p^3 + 3 n \paren {n - 1} \paren {n - 2} p^3 + n \paren {n - 1} \paren {n - 2} \paren {n - 3} p^4
}}
{{eqn | r = n p + 3 n^2 p^2 - 3 n p^2 + n^3 p^3 - 3 n^2 p^3 + 2 n p^3 + 4 n^2 p^2 - 4 n p^2 + 5 n^3 p^3 - 15 n^2 p^3 + 10 n p^3 + n^4 p^4 - 6 n^3 p^4 + 11 n^2 p^4 - 6 n p^4
      | c = Skewness of Binomial Distribution
}}
{{end-eqn}}
Plugging this result back into our equation above:
{{begin-eqn}}
{{eqn | l = \gamma_2
      | r = \frac 1 {n^2 p^2 \paren {1 - p}^2} \big (n p + 3 n^2 p^2 - 3 n p^2 + n^3 p^3 - 3 n^2 p^3 + 2 n p^3 + 4 n^2 p^2 - 4 n p^2 + 5 n^3 p^3 - 15 n^2 p^3 + 10 n p^3 + n^4 p^4 - 6 n^3 p^4 + 11 n^2 p^4 - 6 n p^4
}}
{{eqn | o = -
      | r = 4 n p \paren {n p + 3 n^2 p^2 - 3 n p^2 + n^3 p^3 - 3 n^2 p^3 + 2 n p^3} + 6 n^2 p^2 \paren {n^2 p^2 + n p \paren {1 - p} } - 3 n^4 p^4 \big ) - 3
}}
{{eqn | r = \frac 1 {n^2 p^2 \paren {1 - p}^2} \big (n p + 3 n^2 p^2 - 3 n p^2 + n^3 p^3 - 3 n^2 p^3 + 2 n p^3 + 4 n^2 p^2 - 4 n p^2 + 5 n^3 p^3 - 15 n^2 p^3 + 10 n p^3 + n^4 p^4 - 6 n^3 p^4 + 11 n^2 p^4 - 6 n p^4
}}
{{eqn | o = -
      | r = 4 n^2 p^2 - 12 n^3 p^3 + 12 n^2 p^3 - 4 n^4 p^4 + 12 n^3 p^4 - 8 n^2 p^4 + 6 n^4 p^4 + 6 n^3 p^3  - 6 n^3 p^4 - 3 n^4 p^4 - 3 n^2 p^2 + 6 n^2 p^3 - 3 n^2 p^4 \big )
}}
{{eqn | r = \frac {n p - 7 n p^2 + 12 n p^3 - 6 n p^4} {n^2 p^2 \paren {1 - p}^2}
}}
{{eqn | r = \frac {n p \paren {1 - p} \paren {6 p^2 - 6 p + 1} } {n^2 p^2 \paren {1 - p}^2}
}}
{{eqn | r = \frac {1 + 6 p \paren {p - 1} } {n p \paren {1 - p} }
}}
{{eqn | r = \frac {1 - 6 p q} {n p q}
      | c = $q = 1 - p$
}}
{{end-eqn}}
{{qed}}
Category:Kurtosis
Category:Binomial Distribution
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